# 3.6: Sketching Graphs

- Page ID
- 89733

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)One of the most obvious applications of derivatives is to help us understand the shape of the graph of a function. In this section we will use our accumulated knowledge of derivatives to identify the most important qualitative features of graphs \(y=f(x)\text{.}\) The goal of this section is to highlight features of the graph \(y=f(x)\) that are easily

- determined from \(f(x)\) itself, and
- deduced from \(f'(x)\text{,}\) and
- read from \(f''(x)\text{.}\)

We will then use the ideas to sketch several examples.

## Domain, Intercepts and Asymptotes

Given a function \(f(x)\text{,}\) there are several important features that we can determine from that expression before examining its derivatives.

- The domain of the function — take note of values where \(f\) does not exist. If the function is rational, look for where the denominator is zero. Similarly be careful to look for roots of negative numbers or other possible sources of discontinuities.
- Intercepts — examine where the function crosses the \(x\)-axis and the \(y\)-axis by solving \(f(x)=0\) and computing \(f(0)\text{.}\)
- Vertical asymptotes — look for values of \(x\) at which \(f(x)\) blows up. If \(f(x)\) approaches either \(+\infty\) or \(-\infty\) as \(x\) approaches \(a\) (or possibly as \(x\) approaches \(a\) from one side) then \(x=a\) is a vertical asymptote to \(y=f(x)\text{.}\) When \(f(x)\) is a rational function (written so that common factors are cancelled), then \(y=f(x)\) has vertical asymptotes at the zeroes of the denominator.
- Horizontal asymptotes — examine the limits of \(f(x)\) as \(x\to+\infty\) and \(x\to-\infty\text{.}\) Often \(f(x)\) will tend to \(+\infty\) or to \(-\infty\) or to a finite limit \(L\text{.}\) If, for example, \(\lim\limits_{x\rightarrow+\infty}f(x)=L\text{,}\) then \(y=L\) is a horizontal asymptote to \(y=f(x)\) as \(x\rightarrow\infty\text{.}\)

Consider the function

\begin{align*} f(x) &= \frac{x+1}{(x+3)(x-2)} \end{align*}

- We see that it is defined on all real numbers except \(x=-3,+2\text{.}\)
- Since \(f(0)=-1/6\) and \(f(x)=0\) only when \(x=-1\text{,}\) the graph has \(y\)-intercept \((0,-1/6)\) and \(x\)-intercept \((-1,0)\text{.}\)
- Since the function is rational and its denominator is zero at \(x=-3,+2\) it will have vertical asymptotes at \(x=-3,+2\text{.}\) To determine the shape around those asymptotes we need to examine the limits
\begin{align*} \lim_{x\to -3} f(x) && \lim_{x\to2} f(x) \end{align*}

Notice that when \(x\) is close to \(-3\text{,}\) the factors \((x+1)\) and \((x-2)\) are both negative, so the sign of \(f(x) = \frac{x+1}{x-2} \cdot \frac{1}{x+3}\) is the same as the sign of \(x+3\text{.}\) Hence\begin{align*} \lim_{x\to -3^+} f(x) &= +\infty & \lim_{x\to -3^-} f(x) &= -\infty \end{align*}

A similar analysis when \(x\) is near \(2\) gives\begin{align*} \lim_{x\to 2^+} f(x) &= +\infty & \lim_{x\to 2^-} f(x) &= -\infty \end{align*}

- Finally since the numerator has degree 1 and the denominator has degree 2, we see that as \(x \to \pm \infty\text{,}\) \(f(x) \to 0\text{.}\) So \(y=0\) is a horizontal asymptote.
- Since we know the behaviour around the asymptotes and we know the locations of the intercepts (as shown in the left graph below), we can then join up the pieces and smooth them out to get the a good sketch of this function (below right).

## First Derivative — Increasing or Decreasing

Now we move on to the first derivative, \(f'(x)\text{.}\) This is a good time to revisit the mean-value theorem (Theorem 2.13.5) and some of its consequences (Corollary 2.13.12). In particular, let us assume that \(f(x)\) is continuous on an interval \([A,B]\) and differentiable on \((A,B)\text{.}\) Then

- if \(f'(x) \gt 0\) for all \(A \lt x \lt B\text{,}\) then \(f(x)\) is increasing on \((A,B)\)
— that is, for all \(A \lt a \lt b \lt B\text{,}\) \(f(a) \lt f(b)\text{.}\)

- if \(f'(x) \lt 0\) for all \(A \lt x \lt B\text{,}\) then \(f(x)\) is decreasing on \((A,B)\)
— that is, for all \(A \lt a \lt b \lt B\text{,}\) \(f(a) \gt f(b)\text{.}\)

Thus the sign of the derivative indicates to us whether the function is increasing or decreasing. Further, as we discussed in Section 3.5.1, we should also examine points at which the derivative is zero — critical points — and where the derivative does not exist — singular points. These points may indicate a local maximum or minimum.

After studying the function \(f(x)\) as described above, we should compute its derivative \(f'(x)\text{.}\)

- Critical points — determine where \(f'(x)=0\text{.}\) At a critical point, \(f\) has a horizontal tangent.
- Singular points — determine where \(f'(x)\) is not defined. If \(f'(x)\) approaches \(\pm\infty\) as \(x\) approaches a singular point \(a\text{,}\) then \(f\) has a vertical tangent there when \(f\) approaches a finite value as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side) and a vertical asymptote when \(f(x)\) approaches \(\pm\infty\) as \(x\) approaches \(a\) (or possibly approaches \(a\) from one side).
- Increasing and decreasing — where is the derivative positive and where is it negative. Notice that in order for the derivative to change sign, it must either pass through zero (a critical point) or have a singular point. Thus neighbouring regions of increase and decrease will be separated by critical and singular points.

Consider the function

\begin{align*} f(x) &= x^4-6x^3 \end{align*}

- Before we move on to derivatives, let us first examine the function itself as we did above.
- As \(f(x)\) is a polynomial its domain is all real numbers.
- Its \(y\)-intercept is at \((0,0)\text{.}\) We find its \(x\)-intercepts by factoring
\begin{align*} f(x) &=x^4-6x^3 = x^3(x-6) \end{align*}

So it crosses the \(x\)-axis at \(x=0,6\text{.}\) - Again, since the function is a polynomial it does not have any vertical asymptotes. And since
\begin{align*} \lim_{x \to \pm \infty} f(x) &= \lim_{x \to \pm \infty} x^4(1-6/x) = +\infty \end{align*}

it does not have horizontal asymptotes — it blows up to \(+\infty\) as \(x\) goes to \(\pm\infty\text{.}\) - We can also determine where the function is positive or negative since we know it is continuous everywhere and zero at \(x=0,6\text{.}\) Thus we must examine the intervals
\begin{align*} (-\infty,0)&& (0,6) && (6,\infty) \end{align*}

When \(x \lt 0\text{,}\) \(x^3 \lt 0\) and \(x-6 \lt 0\) so \(f(x) = x^3(x-6) = (\text{negative})(\text{negative}) \gt 0\text{.}\) Similarly when \(x \gt 6\text{,}\) \(x^3 \gt 0, x-6 \gt 0\) we must have \(f(x) \gt 0\text{.}\) Finally when \(0 \lt x \lt 6\text{,}\) \(x^3 \gt 0\) but \(x-6 \lt 0\) so \(f(x) \lt 0\text{.}\) Thus

interval \((-\infty,0)\) 0 \((0,6)\) 6 \((6,\infty)\) \(f(x)\) positive 0 negative 0 positive - Based on this information we can already construct a rough sketch.

- Now we compute its derivative
\begin{align*} f'(x) &= 4x^3-18x^2 = 2x^2(2x-9) \end{align*}

- Since the function is a polynomial, it does not have any singular points, but it does have two critical points at \(x=0, 9/2\text{.}\) These two critical points split the real line into 3 open intervals
\begin{align*} (-\infty, 0) && (0,9/2) && (9/2,\infty) \end{align*}

We need to determine the sign of the derivative in each intervals.

- When \(x \lt 0\text{,}\) \(x^2 \gt 0\) but \((2x-9) \lt 0\text{,}\) so \(f'(x) \lt 0\) and the function is decreasing.
- When \(0 \lt x \lt 9/2\text{,}\) \(x^2 \gt 0\) but \((2x-9) \lt 0\text{,}\) so \(f'(x) \lt 0\) and the function is still decreasing.
- When \(x \gt 9/2\text{,}\) \(x^2 \gt 0\) and \((2x-9) \gt 0\text{,}\) so \(f'(x) \gt 0\) and the function is increasing.

We can then summarise this in the following table

interval \((-\infty,0)\) 0 \((0,9/2)\) 9/2 \((9/2,\infty)\) \(f'(x)\) negative 0 negative 0 positive decreasing horizontal

tangentdecreasing minimum increasing Since the derivative changes sign from negative to positive at the critical point \(x=9/2\text{,}\) this point is a minimum. Its \(y\)-value is

\begin{align*} y&=f(9/2) = \frac{9^3}{2^3}\left(\frac{9}{2} - 6\right)\\ &= \frac{3^6}{2^3} \cdot \left(\frac{-3}{2} \right) = -\frac{3^7}{2^4} \end{align*}

On the other hand, at \(x=0\) the derivative does not change sign; while this point has a horizontal tangent line it is not a minimum or maximum.

- Putting this information together we arrive at a quite reasonable sketch.
To improve upon this further we will examine the second derivative.

## Second Derivative — Concavity

The second derivative \(f''(x)\) tells us the rate at which the derivative changes. Perhaps the easiest way to understand how to interpret the sign of the second derivative is to think about what it implies about the slope of the tangent line to the graph of the function. Consider the following sketches of \(y=1+x^2\) and \(y=-1-x^2\text{.}\)

- In the case of \(y = f(x) = 1+x^2\), \(f''(x) = 2 \gt 0\text{.}\) Notice that this means the slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) increases as \(x\) increases. Looking at the figure on the left above, we see that the graph always lies above the tangent lines.
- For \(y = f(x) = -1-x^2\), \(f''(x) = -2 \lt 0\text{.}\) The slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) decreases as \(x\) increases. Looking at the figure on the right above, we see that the graph always lies below the tangent lines.

Similarly consider the following sketches of \(y=x^{-1/2}\) and \(y=\sqrt{4-x}\text{:}\)

Both of their derivatives, \(-\frac{1}{2}x^{-3/2}\) and \(-\frac{1}{2}(4-x)^{-1/2}\text{,}\) are negative, so they are decreasing functions. Examining second derivatives shows some differences.

- For the first function, \(y''(x) = \frac{3}{4}x^{-5/2} \gt 0\text{,}\) so the slopes of tangent lines are increasing with \(x\) and the graph lies above its tangent lines.
- However, the second function has \(y''(x) = -\frac{1}{4}(4-x)^{-3/2} \lt 0\) so the slopes of the tangent lines are decreasing with \(x\) and the graph lies below its tangent lines.

More generally

Let \(f(x)\) be a continuous function on the interval \([a,b]\) and suppose its first and second derivatives exist on that interval.

- If \(f''(x) \gt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies above its tangent lines for \(a \lt x \lt b\) and it is said to be concave up.
- If \(f''(x) \lt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies below its tangent lines for \(a \lt x \lt b\) and it is said to be concave down.
- If \(f''(c)=0\) for some \(a \lt c \lt b\text{,}\) and the concavity of \(f\) changes across \(x=c\text{,}\) then we call \((c,f(c))\) an inflection point.

Note that one might also see the terms

- “convex” or “convex up” used in place of “concave up”, and
- “concave” or “convex down” used to mean “concave down”.

To avoid confusion we recommend the reader stick with the terms “concave up” and “concave down”.

Let's now continue Example 3.6.2 by discussing the concavity of the curve.

Consider again the function

\begin{align*} f(x) &= x^4-6x^3 \end{align*}

- Its first derivative is \(f'(x)=4x^3-18x^2\text{,}\) so
\begin{align*} f''(x) &= 12x^2 - 36x = 12x(x-3) \end{align*}

- Thus the second derivative is zero (and potentially changes sign) at \(x=0,3\text{.}\) Thus we should consider the sign of the second derivative on the following intervals
\begin{align*} (-\infty,0) && (0,3) && (3,\infty) \end{align*}

A little algebra gives us

interval \((-\infty,0)\) 0 \((0,3)\) 3 \((3,\infty)\) \(f''(x)\) positive 0 negative 0 positive concavity up inflection down inflection up Since the concavity changes at both \(x=0\) and \(x=3\text{,}\) the following are inflection points

\begin{align*} (0,0) && (3,3^4-6\times3^3)=(3,-3^4) \end{align*}

- Putting this together with the information we obtained earlier gives us the following sketch

## Symmetries

Before we proceed to some examples, we should examine some simple symmetries possessed by some functions. We'll look at three symmetries — evenness, oddness and periodicity. If a function possesses one of these symmetries then it can be exploited to reduce the amount of work required to sketch the graph of the function.

Let us start with even and odd functions.

A function \(f(x)\) is said to be even if \(f(-x)=f(x)\) for all \(x\text{.}\)

A function \(f(x)\) is said to be odd if \(f(-x)=-f(x)\) for all \(x\text{.}\)

Let \(f(x) = x^2\) and \(g(x)=x^3\text{.}\) Then

\begin{align*} f(-x) &= (-x)^2 = x^2 = f(x)\\ g(-x) &= (-x)^3 = -x^3 = -g(x) \end{align*}

Hence \(f(x)\) is even and \(g(x)\) is odd.

Notice any polynomial involving only even powers of \(x\) will be even

\begin{align*} f(x) &= 7x^6+2x^4-3x^2+5 & \text{remember that } 5=5x^0\\ f(-x) &= 7(-x)^6+2(-x)^4-3(-x)^2+5\\ &= 7x^6+2x^4-3x^2+5 = f(x) \end{align*}

Similarly any polynomial involving only odd powers of \(x\) will be odd

\begin{align*} g(x) &= 2x^5-8x^3-3x\\ g(-x) &= 2(-x)^5-8(-x)^3-3(-x)\\ &= -2x^5+8x^3+3x = -g(x) \end{align*}

Not all even and odd functions are polynomials. For example

\begin{align*} |x| && \cos x && \text{ and } (e^x + e^{-x}) \end{align*}

are all even, while

\begin{align*} \sin x && \tan x && \text{ and } (e^x-e^{-x}) \end{align*}

are all odd. Indeed, given any function \(f(x)\text{,}\) the function

\begin{align*} g(x) &= f(x)+f(-x) & \text{ will be even, and}\\ h(x) &= f(x)-f(-x) & \text{ will be odd.} \end{align*}

Now let us see how we can make use of these symmetries to make graph sketching easier. Let \(f(x)\) be an even function. Then

\begin{gather*} \text{the point } (x_0,y_0)\text{ lies on the graph of }y=f(x) \end{gather*}

if and only if \(y_0= f(x_0) = f(-x_0)\) which is the case if and only if

\begin{gather*} \text{the point }(-x_0,y_0)\text{ lies on the graph of }y=f(x). \end{gather*}

Notice that the points \((x_0,y_0)\) and \((-x_0,y_0)\) are just reflections of each other across the \(y\)-axis. Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it in the \(y\)–axis. Here is an example. The part with \(x\ge 0\) is on the left and the full graph is on the right.

Very similarly, when \(f(x)\) is an odd function then

\begin{gather*} (x_0,y_0)\text{ lies on the graph of }y=f(x) \end{gather*}

if and only if

\begin{gather*} (-x_0,-y_0)\text{ lies on the graph of }y=f(x) \end{gather*}

Now the symmetry is a little harder to interpret pictorially. To get from \((x_0,y_0)\) to \((-x_0,-y_0)\) one can first reflect \((x_0,y_0)\) in the \(y\)–axis to get to \((-x_0,y_0)\) and then reflect the result in the \(x\)–axis to get to \((-x_0,-y_0)\text{.}\) Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw the part of the graph with \(x\ge 0\) and then reflect it first in the \(y\)–axis and then in the \(x\)–axis. Here is an example. First, here is the part of the graph with \(x\ge 0\text{.}\)

Next, as an intermediate step (usually done in our heads rather than on paper), we add in the reflection in the \(y\)–axis.

Finally to get the full graph, we reflect the dashed line in the \(x\)–axis

and then remove the dashed line.

Let's do a more substantial example of an even function

Consider the function

\begin{align*} g(x) &= \frac{x^2-9}{x^2+3} \end{align*}

- The function is even since
\begin{align*} g(-x) &= \frac{(-x)^2-9}{(-x)^2+3} = \frac{x^2-9}{x^2+3} = g(x) \end{align*}

Thus it suffices to study the function for \(x\geq0\) because we can then use the even symmetry to understand what happens for \(x \lt 0\text{.}\) - The function is defined on all real numbers since its denominator \(x^2+3\) is never zero. Hence it has no vertical asymptotes.
- The \(y\)-intercept is \(g(0) = \frac{-9}{3} = -3\text{.}\) And \(x\)-intercepts are given by the solution of \(x^2-9=0\text{,}\) namely \(x=\pm 3\text{.}\) Note that we only need to establish \(x=3\) as an intercept. Then since \(g\) is even, we know that \(x=-3\) is also an intercept.
- To find the horizontal asymptotes we compute the limit as \(x\to+\infty\)
\begin{align*} \lim_{x\to \infty} g(x) &= \lim_{x\to \infty} \frac{x^2-9}{x^2+3}\\ &= \lim_{x\to \infty} \frac{x^2(1-9/x^2)}{x^2(1+3/x^2)}\\ &= \lim_{x\to \infty} \frac{1-9/x^2}{1+3/x^2} = 1 \end{align*}

Thus \(y=1\) is a horizontal asymptote. Indeed, this is also the asymptote as \(x\to-\infty\) since by the even symmetry\begin{align*} \lim_{x\to -\infty} g(x) &=\lim_{x\to \infty} g(-x) = \lim_{x\to \infty} g(x). \end{align*}

- We can already produce a quite reasonable sketch just by putting in the horizontal asymptote and the intercepts and drawing a smooth curve between them.
Note that we have drawn the function as never crossing the asymptote \(y=1\text{,}\) however we have not yet proved that. We could by trying to solve \(g(x)=1\text{.}\)

\begin{align*} \frac{x^2-9}{x^2+3} &= 1\\ x^2-9 &= x^2+3\\ -9=3 & \text{ so no solutions.} \end{align*}

Alternatively we could analyse the first derivative to see how the function approaches the asymptote.

- Now we turn to the first derivative:
\begin{align*} g'(x) &= \frac{(x^2+3)(2x) - (x^2-9)(2x)}{(x^2+3)^2}\\ &= \frac{24x}{(x^2+3)^2} \end{align*}

There are no singular points since the denominator is nowhere zero. The only critical point is at \(x=0\text{.}\) Thus we must find the sign of \(g'(x)\) on the intervals\begin{align*} (-\infty,0) && (0,\infty) \end{align*}

- When \(x \gt 0\text{,}\) \(24x \gt 0\) and \((x^2+3) \gt 0\text{,}\) so \(g'(x) \gt 0\) and the function is increasing. By even symmetry we know that when \(x \lt 0\) the function must be decreasing. Hence the critical point \(x=0\) is a local minimum of the function.
- Notice that since the function is increasing for \(x \gt 0\) and the function must approach the horizontal asymptote \(y=1\) from below. Thus the sketch above is quite accurate.
- Now consider the second derivative:
\begin{align*} g''(x) &= \dfrac{d}{dx} \frac{24x}{(x^2+3)^2}\\ &= \frac{(x^2+3)^2 \cdot 24 - 24x\cdot 2 (x^2+3)\cdot2x}{(x^2+3)^4}\\ \end{align*}

cancel a factor of \((x^2+3)\)

\begin{align*} &= \frac{(x^2+3) \cdot 24 - 96x^2}{(x^2+3)^3}\\ &= \frac{72(1-x^2)}{(x^2+3)^3} \end{align*} - It is clear that \(g''(x) = 0\) when \(x=\pm 1\text{.}\) Note that, again, we can infer the zero at \(x=-1\) from the zero at \(x=1\) by the even symmetry. Thus we need to examine the sign of \(g''(x)\) the intervals
\begin{align*} (-\infty,-1)&&(-1,1)&&(1,\infty) \end{align*}

- When \(|x| \lt 1\) we have \((1-x^2) \gt 0\) so that \(g''(x) \gt 0\) and the function is concave up. When \(|x| \gt 1\) we have \((1-x^2) \lt 0\) so that \(g''(x) \lt 0\) and the function is concave down. Thus the points \(x=\pm 1\) are inflection points. Their coordinates are \((\pm1, g(\pm1)) =(\pm 1,-2)\text{.}\)
- Putting this together gives the following sketch:

Another symmetry we should consider is periodicity.

A function \(f(x)\) is said to be periodic, with period \(P \gt 0\text{,}\) if \(f(x+P)=f(x)\) for all \(x\text{.}\)

Note that if \(f(x+P)=f(x)\) for all \(x\text{,}\) then replacing \(x\) by \(x+P\text{,}\) we have

\begin{gather*} f(x+2P)=f(x+P+P)=f(x+P)=f(x). \end{gather*}

More generally \(f(x+kP)=f(x)\) for all integers \(k\text{.}\) Thus if \(f\) has period \(P\text{,}\) then it also has period \(nP\) for all natural numbers \(n\text{.}\) The smallest period is called the fundamental period.

The classic example of a periodic function is \(f(x)=\sin x\text{,}\) which has period \(2\pi\) since \(f(x+2\pi)=\sin(x+2\pi)=\sin x=f(x)\text{.}\)

If \(f(x)\) has period \(P\) then

\begin{gather*} (x_0,y_0)\text{ lies on the graph of }y=f(x)\\ \end{gather*}

if and only if \(y_0=f(x_0)=f(x_0+P)\) which is the case if and only if

\begin{gather*} (x_0+P,y_0)\text{ lies on the graph of }y=f(x) \end{gather*}

and, more generally,

\begin{gather*} (x_0,y_0)\text{ lies on the graph of }y=f(x)\\ \end{gather*}

if and only if

\begin{gather*} (x_0+nP,y_0)\text{ lies on the graph of }y=f(x) \end{gather*}

for all integers \(n\text{.}\)

Note that the point \((x_0+P,y_0)\) can be obtained by translating \((x_0,y_0)\) horizontally by \(P\text{.}\) Similarly the point \((x_0+nP,y_0)\) can be found by repeatedly translating \((x_0,y_0)\) horizontally by \(P\text{.}\)

Consequently, to draw the graph \(y=f(x)\text{,}\) it suffices to draw one period of the graph, say the part with \(0\le x\le P\text{,}\) and then translate it repeatedly. Here is an example. Here is a sketch of one period

and here is the full sketch.

## A Checklist for Sketching

Above we have described how we can use our accumulated knowledge of derivatives to quickly identify the most important qualitative features of graphs \(y=f(x)\text{.}\) Here we give the reader a quick checklist of things to examine in order to produce an accurate sketch based on properties that are easily read off from \(f(x)\text{,}\) \(f'(x)\) and \(f''(x)\text{.}\)

### A Sketching Checklist

- Features of \(y = f(x)\) that are read off of \(f(x)\text{:}\)
- First check where \(f(x)\) is defined. Then
- y=f(x) is plotted only for \(x\)'s in the domain of \(f(x)\text{,}\) i.e. where \(f(x)\) is defined.
- \(y = f(x)\) has vertical asymptotes at the points where \(f(x)\) blows up to \(\pm\infty\text{.}\)
- Next determine whether the function is even, odd, or periodic.
- \(y=f(x)\) is first plotted for \(x\ge 0\) if the function is even or odd. The rest of the sketch is then created by reflections.
- \(y=f(x)\) is first plotted for a single period if the function is periodic. The rest of the sketch is then created by translations.
- Next compute \(f(0)\text{,}\) \(\lim_{x\rightarrow\infty} f(x)\) and \(\lim_{x\rightarrow-\infty} f(x)\) and look for solutions to \(f(x)=0\) that you can easily find. Then
- \(y = f(x)\) has \(y\)–intercept \(\big(0, f(0)\big)\text{.}\)
- \(y = f(x)\) has \(x\)–intercept \((a,0)\) whenever \(f(a)=0\)
- \(y = f(x)\) has horizontal asymptote \(y=Y\) if \(\lim_{x\rightarrow\infty} f(x)=L\) or \(\lim_{x\rightarrow-\infty} f(x)=L\text{.}\)

- Features of \(y=f(x)\) that are read off of \(f'(x)\text{:}\)
- Compute \(f'(x)\) and determine its critical points and singular points, then
- \(y=f(x)\) has a horizontal tangent at the points where \(f'(x)=0\text{.}\)
- \(y=f(x)\) is increasing at points where \(f'(x) \gt 0\text{.}\)
- \(y=f(x)\) is decreasing at points where \(f'(x) \lt 0\text{.}\)
- \(y=f(x)\) has vertical tangents or vertical asymptotes at the points where \(f'(x)=\pm\infty\text{.}\)

- Features of \(y=f(x)\) that are read off of \(f''(x)\text{:}\)
- Compute \(f''(x)\) and determine where \(f''(x)=0\) or does not exist, then
- \(y=f(x)\) is concave up at points where \(f''(x) \gt 0\text{.}\)
- \(y=f(x)\) is concave down at points where \(f''(x) \lt 0\text{.}\)
- \(y=f(x)\) may or may not have inflection points where \(f''(x)=0\text{.}\)

## Sketching Examples

- Reading from \(f(x)\text{:}\)
- The function is a polynomial so it is defined everywhere.
- Since \(f(-x) = -x^3+3x+1 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
- The \(y\)-intercept is \(y=1\text{.}\) The \(x\)-intercepts are not easily computed since it is a cubic polynomial that does not factor nicely
^{1}. So for this example we don't worry about finding them. - Since it is a polynomial it has no vertical asymptotes.
- For very large \(x\text{,}\) both positive and negative, the \(x^3\) term in \(f(x)\) dominates the other two terms so that
\begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ -\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}

and there are no horizontal asymptotes.

- We now compute the derivative:
\begin{align*} f'(x) &= 3x^2-3 = 3(x^2-1)=3(x+1)(x-1) \end{align*}

- The critical points (where \(f'(x)=0\)) are at \(x=\pm 1\text{.}\) Further since the derivative is a polynomial it is defined everywhere and there are no singular points. The critical points split the real line into the intervals \((-\infty,-1),(-1,1)\) and \((1,\infty)\text{.}\)
- When \(x \lt -1\text{,}\) both factors \((x+1),(x-1) \lt 0\) so \(f'(x) \gt 0\text{.}\)
- Similarly when \(x \gt 1\text{,}\) both factors \((x+1),(x-1) \gt 0\) so \(f'(x) \gt 0\text{.}\)
- When \(-1 \lt x \lt 1\text{,}\) \((x-1) \lt 0\) but \((x+1) \gt 0\) so \(f'(x) \lt 0\text{.}\)
- Summarising all this
\((-\infty,-1)\) -1 (-1,1) 1 \((1,\infty)\) \(f'(x)\) positive 0 negative 0 positive increasing maximum decreasing minimum increasing So \((-1,f(-1))=(-1,3)\) is a local maximum and \((1,f(1))=(1,-1)\) is a local minimum.

- Compute the second derivative:
\begin{gather*} f''(x) = 6x \end{gather*}

- The second derivative is zero when \(x=0\text{,}\) and the problem is quite easy to analyse. Clearly, \(f''(x) \lt 0\) when \(x \lt 0\) and \(f''(x) \gt 0\) when \(x \gt 0\text{.}\)
- Thus \(f\) is concave down for \(x \lt 0\text{,}\) concave up for \(x \gt 0\) and has an inflection point at \(x=0\text{.}\)

Putting this all together gives:

- Reading from \(f(x)\text{:}\)
- The function is a polynomial so it is defined everywhere.
- Since \(f(-x) = x^4+4x^3 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
- The \(y\)-intercept is \(y=f(0)=0\text{,}\) while the \(x\)-intercepts are given by the solution of
\begin{align*} f(x)=x^4-4x^3 &= 0\\ x^3(x-4)&=0 \end{align*}

Hence the \(x\)-intercepts are \(0,4\text{.}\) - Since \(f\) is a polynomial it does not have any vertical asymptotes.
- For very large \(x\text{,}\) both positive and negative, the \(x^4\) term in \(f(x)\) dominates the other term so that
\begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ +\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}

and the function has no horizontal asymptotes.

- Now compute the derivative \(f'(x)\text{:}\)
\begin{gather*} f'(x) = 4x^3-12x^2 = 4(x-3)x^2 \end{gather*}

- The critical points are at \(x=0,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,0)\text{,}\) \((0,3)\) and \((3,\infty)\text{.}\)
- When \(x \lt 0\text{,}\) \(x^2 \gt 0\) and \(x-3 \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
- When \(0 \lt x \lt 3\text{,}\) \(x^2 \gt 0\) and \(x-3 \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
- When \(3 \lt x\text{,}\) \(x^2 \gt 0\) and \(x-3 \gt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)
- Summarising all this
\((-\infty,0)\) 0 (0,3) 3 \((3,\infty)\) \(f'(x)\) negative 0 negative 0 positive decreasing horizontal

tangentdecreasing minimum increasing So the point \((3,f(3))=(3,-27)\) is a local minimum. The point \((0,f(0))=(0,0)\) is neither a minimum nor a maximum, even though \(f'(0)=0\text{.}\)

- Now examine \(f''(x)\text{:}\)
\begin{gather*} f''(x) = 12x^2-24x=12x(x-2) \end{gather*}

- So \(f''(x)=0\) when \(x=0,2\text{.}\) This splits the real line into the intervals \((-\infty,0),(0,2)\) and \((2,\infty)\text{.}\)
- When \(x \lt 0\text{,}\) \(x-2 \lt 0\) and so \(f''(x) \gt 0\text{.}\)
- When \(0 \lt x \lt 2\text{,}\) \(x \gt 0\) and \(x-2 \lt 0\) and so \(f''(x) \lt 0\text{.}\)
- When \(2 \lt x\text{,}\) \(x \gt 0\) and \(x-2 \gt 0\) and so \(f''(x) \gt 0\text{.}\)
- Thus the function is convex up for \(x \lt 0\text{,}\) then convex down for \(0 \lt x \lt 2\text{,}\) and finally convex up again for \(x \gt 2\text{.}\) Hence \((0,f(0))=(0,0)\) and \((2,f(2))=(2,-16)\) are inflection points.

Putting all this information together gives us the following sketch.

- Reading from \(f(x)\text{:}\)
- The function is a polynomial so it is defined everywhere.
- Since \(f(-x) = -x^3-6x^2-9x-54 \neq \pm f(x)\text{,}\) it is not even or odd. Nor is it periodic.
- The \(y\)-intercept is \(y=f(0)=-54\text{,}\) while the \(x\)-intercepts are given by the solution of
\begin{align*} f(x)=x^3-6x^2+9x-54 &= 0\\ x^2(x-6) + 9(x-6) &=0\\ (x^2+9)(x-6) &= 0 \end{align*}

Hence the only \(x\)-intercept is \(6\text{.}\) - Since \(f\) is a polynomial it does not have any vertical asymptotes.
- For very large \(x\text{,}\) both positive and negative, the \(x^3\) term in \(f(x)\) dominates the other term so that
\begin{align*} f(x)\rightarrow\begin{cases}+\infty &\text{as }x\rightarrow+\infty\\ -\infty &\text{as }x\rightarrow-\infty \end{cases} \end{align*}

and the function has no horizontal asymptotes.

- Now compute the derivative \(f'(x)\text{:}\)
\begin{align*} f'(x) &= 3x^2-12x+9\\ &= 3(x^2-4x+3) = 3(x-3)(x-1) \end{align*}

- The critical points are at \(x=1,3\text{.}\) Since the function is a polynomial there are no singular points. The critical points split the real line into the intervals \((-\infty,1)\text{,}\) \((1,3)\) and \((3,\infty)\text{.}\)
- When \(x \lt 1\text{,}\) \((x-1) \lt 0\) and \((x-3) \lt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)
- When \(1 \lt x \lt 3\text{,}\) \((x-1) \gt 0\) and \((x-3) \lt 0\text{,}\) so \(f'(x) \lt 0\text{.}\)
- When \(3 \lt x\text{,}\) \((x-1) \gt 0\) and \((x-3) \gt 0\text{,}\) so \(f'(x) \gt 0\text{.}\)
- Summarising all this
\((-\infty,1)\) 1 (1,3) 3 \((3,\infty)\) \(f'(x)\) positive 0 negative 0 positive increasing maximum decreasing minimum increasing So the point \((1,f(1))=(1,-50)\) is a local maximum. The point \((3,f(3))=(3,-54)\) is a local minimum.

- Now examine \(f''(x)\text{:}\)
\begin{gather*} f''(x) = 6x-12 \end{gather*}

- So \(f''(x)=0\) when \(x=2\text{.}\) This splits the real line into the intervals \((-\infty,2)\) and \((2,\infty)\text{.}\)
- When \(x \lt 2\text{,}\) \(f''(x) \lt 0\text{.}\)
- When \(x \gt 2\text{,}\) \(f''(x) \gt 0\text{.}\)
- Thus the function is convex down for \(x \lt 2\text{,}\) then convex up for \(x \gt 2\text{.}\) Hence \((2,f(2))=(2,-52)\) is an inflection point.

Putting all this information together gives us the following sketch.

and if we zoom in around the interesting points (minimum, maximum and inflection point), we have

An example of sketching a simple rational function.

- Reading from \(f(x)\text{:}\)
- The function is rational so it is defined except where its denominator is zero — namely at \(x=\pm2\text{.}\)
- Since \(f(-x) = \dfrac{-x}{x^2-4} = - f(x)\text{,}\) it is odd. Indeed this means that we only need to examine what happens to the function for \(x \geq 0\) and we can then infer what happens for \(x\leq 0\) using \(f(-x) = -f(x)\text{.}\) In practice we will sketch the graph for \(x\geq0\) and then infer the rest from this symmetry.
- The \(y\)-intercept is \(y=f(0)=0\text{,}\) while the \(x\)-intercepts are given by the solution of \(f(x)=0\text{.}\) So the only \(x\)-intercept is \(0\text{.}\)
- Since \(f\) is rational, it may have vertical asymptotes where its denominator is zero — at \(x=\pm 2\text{.}\) Since the function is odd, we only have to analyse the asymptote at \(x=2\) and we can then infer what happens at \(x=-2\) by symmetry.
\begin{align*} \lim_{x\to 2^+} f(x) &= \lim_{x\to 2^+} \frac{x}{(x-2)(x+2)} = + \infty\\ \lim_{x\to 2^-} f(x) &= \lim_{x\to 2^-} \frac{x}{(x-2)(x+2)} = - \infty \end{align*}

- We now check for horizontal asymptotes:
\begin{align*} \lim_{x\to +\infty} f(x) &= \lim_{x\to +\infty} \frac{x}{x^2-4}\\ &= \lim_{x\to +\infty} \frac{1}{x-4/x} = 0 \end{align*}

- Now compute the derivative \(f'(x)\text{:}\)
\begin{align*} f'(x) &= \frac{(x^2-4)\cdot 1 - x\cdot 2x}{(x^2-4)^2}\\ &= \frac{-(x^2+4)}{(x^2-4)^2} \end{align*}

- Hence there are no critical points. There are singular points where the denominator is zero, namely \(x=\pm2\text{.}\) Before we proceed, notice that the numerator is always negative and the denominator is always positive. Hence \(f'(x) \lt 0\) except at \(x=\pm 2\) where it is undefined.
- The function is decreasing except at \(x=\pm 2\text{.}\)
- We already know that at \(x = 2\) we have a vertical asymptote and that \(f'(x) \lt 0\) for all \(x\text{.}\) So
\begin{gather*} \lim_{x\rightarrow 2} f'(x) = -\infty \end{gather*}

- Summarising all this
[0,2) 2 \((2,\infty)\) \(f'(x)\) negative DNE negative decreasing vertical

asymptotedecreasing Remember — we will draw the graph for \(x\geq 0\) and then use the odd symmetry to infer the graph for \(x \lt 0\text{.}\)

- Now examine \(f''(x)\text{:}\)
\begin{align*} f''(x) &=- \frac{(x^2-4)^2\cdot(2x) - (x^2+4)\cdot2\cdot 2x\cdot(x^2-4)}{(x^2-4)^4}\\ &=- \frac{(x^2-4)\cdot(2x) - (x^2+4)\cdot4x}{(x^2-4)^3}\\ &=- \frac{2x^3-8x - 4x^3-16x}{(x^2-4)^3}\\ &= \frac{2x(x^2+12)}{(x^2-4)^3} \end{align*}

- So \(f''(x)=0\) when \(x=0\) and does not exist when \(x=\pm 2\text{.}\) This splits the real line into the intervals \((-\infty,-2), (-2,0), (0,2)\) and \((2,\infty)\text{.}\) However we only need to consider \(x \geq 0\) (because of the odd symmetry).
- When \(0 \lt x \lt 2\text{,}\) \(x \gt 0, (x^2+12) \gt 0\) and \((x^2-4) \lt 0\) so \(f''(x) \lt 0\text{.}\)
- When \(x \gt 2\text{,}\) \(x \gt 0, (x^2+12) \gt 0\) and \((x^2-4) \gt 0\) so \(f''(x) \gt 0\text{.}\)

Putting all this information together gives the following sketch for \(x \geq 0\text{:}\)

We can then draw in the graph for \(x \lt 0\) using \(f(-x) = -f(x)\text{:}\)

Notice that this means that the concavity changes at \(x=0\text{,}\) so the point \((0,f(0))=(0,0)\) is a point of inflection (as indicated).

This final example is more substantial since the function has singular points (points where the derivative is undefined). The analysis is more involved.

- Reading from \(f(x)\text{:}\)
- First notice that we can rewrite
\begin{align*} f(x) &= \root{3}\of{\frac{x^2}{(x-6)^2}} = \root{3}\of{\frac{x^2}{x^2\cdot(1-6/x)^2}} = \root{3}\of{\frac{1}{(1-6/x)^2}} \end{align*}

- The function is the cube root of a rational function. The rational function is defined except at \(x=6\text{,}\) so the domain of \(f\) is all reals except \(x=6\text{.}\)
- Clearly the function is not periodic, and examining
\begin{align*} f(-x) &= \root{3}\of{\frac{ 1}{(1-6/(-x))^2}}\\ &= \root{3}\of{\frac{1}{(1+6/x)^2}} \neq \pm f(x) \end{align*}

shows the function is neither even nor odd. - To compute horizontal asymptotes we examine the limit of the portion of the function inside the cube-root
\begin{gather*} \lim_{x\rightarrow\pm\infty} \frac{1}{(1-\frac{6}{x})^2} =1 \end{gather*}

This means we have\begin{gather*} \lim_{x\rightarrow\pm\infty} f(x)=1 \end{gather*}

That is, the line \(y=1\) will be a horizontal asymptote to the graph \(y=f(x)\) both for \(x\rightarrow+\infty\) and for \(x\rightarrow-\infty\text{.}\) - Our function \(f(x)\rightarrow+\infty\) as \(x\rightarrow 6\text{,}\) because of the \((1-6/x)^2\) in its denominator. So \(y=f(x)\) has \(x=6\) as a vertical asymptote.

- First notice that we can rewrite
- Now compute \(f'(x)\text{.}\) Since we rewrote
\begin{align*} f(x) &= \root{3}\of{\frac{1}{(1-6/x)^2}} =\left(1-\frac{6}{x}\right)^{-\frac{2}{3}} \end{align*}

we can use the chain rule

\begin{gather*} f'(x) = -\frac{2}{3}{\left(1-\frac{6}{x}\right)}^{-\frac{5}{3}}\frac{6}{x^2}\\ =-4 {\left(\frac{x-6}{x}\right)}^{-\frac{5}{3}}\frac{1}{x^2}\\ =-4 {\left(\frac{1}{x-6}\right)}^{\frac{5}{3}}\frac{1}{x^{\frac{1}{3}}} \end{gather*}

- Notice that the derivative is nowhere equal to zero, so the function has no critical points. However there are two places the derivative is undefined. The terms
\begin{align*} \left(\frac{1}{x-6}\right)^{\frac{5}{3}} && \frac{1}{x^{\frac{1}{3}}} \end{align*}

are undefined at \(x=6,0\) respectively. Hence \(x=0,6\) are singular points. These split the real line into the intervals \((-\infty,0), (0,6)\) and \((6,\infty)\text{.}\) - When \(x \lt 0\text{,}\) \((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \lt 0\) and so \(f'(x)=-4 \cdot (\text{negative})\cdot(\text{negative}) \lt 0 \text{.}\)
- When \(0 \lt x \lt 6\text{,}\) \((x-6) \lt 0\text{,}\) we have that \((x-6)^{-\frac53} \lt 0\) and \(x^{-\frac13} \gt 0\) and so \(f'(x) \gt 0\text{.}\)
- When \(x \gt 6\text{,}\) \((x-6) \gt 0\text{,}\) we have that \((x-6)^{-\frac53} \gt 0\) and \(x^{-\frac13} \gt 0\) and so \(f'(x) \lt 0\text{.}\)
- We should also examine the behaviour of the derivative as \(x \to 0\) and \(x\to 6\text{.}\)
\begin{align*} \lim_{x \to 0^-} f'(x) &= -4 \left( \lim_{x \to 0^-} (x-6)^{-\frac53} \right) \left( \lim_{x \to 0^-} x^{-\frac13} \right) = -\infty\\ \lim_{x \to 0^+} f'(x) &= -4 \left( \lim_{x \to 0^+} (x-6)^{-\frac53} \right) \left( \lim_{x \to 0^+} x^{-\frac13} \right) = +\infty\\ \lim_{x \to 6^-} f'(x) &= -4 \left( \lim_{x \to 6^-} (x-6)^{-\frac53} \right) \left( \lim_{x \to 6^-} x^{-\frac13} \right) = +\infty\\ \lim_{x \to 6^+} f'(x) &= -4 \left( \lim_{x \to 6^+} (x-6)^{-\frac53} \right) \left( \lim_{x \to 6^+} x^{-\frac13} \right) = -\infty \end{align*}

We already know that \(x=6\) is a vertical asymptote of the function, so it is not surprising that the lines tangent to the graph become vertical as we approach 6. The behavior around \(x=0\) is less standard, since the lines tangent to the graph become vertical, but \(x=0\) is not a vertical asymptote of the function. Indeed the function takes a finite value \(y=f(0)=0\text{.}\) - Summarising all this
\((-\infty,0)\) 0 (0,6) 6 \((6,\infty)\) \(f'(x)\) negative DNE positive DNE negative decreasing vertical

tangentsincreasing vertical

asymptotedecreasing

- Notice that the derivative is nowhere equal to zero, so the function has no critical points. However there are two places the derivative is undefined. The terms
- Now look at \(f''(x)\text{:}\)
\begin{align*} f''(x)&=-4\dfrac{d}{dx} \left[{\left(\frac{1}{x-6}\right)}^{\frac{5}{3}} \frac{1}{x^{\frac{1}{3}}} \right]\\ &=-4 \left[-\frac{5}{3}{\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{1}{3}}} -\frac{1}{3} {\left(\frac{1}{x-6}\right)}^{\frac{5}{3}} \frac{1}{x^{\frac{4}{3}}}\right]\\ &=\frac{4}{3} {\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{4}{3}}}\ \left[5x +(x-6)\right]\\ &=8 {\left(\frac{1}{x-6}\right)}^{\frac{8}{3}} \frac{1}{x^{\frac{4}{3}}}\ \left[x-1\right] \end{align*}

Oof!

- Both of the factors \({\Big(\frac{1}{x-6}\Big)}^{\frac{8}{3}} ={\Big(\frac{1}{\root{3}\of{x-6}}\Big)}^8\) and \(\frac{1}{x^{\frac{4}{3}}} =\Big(\frac{1}{\root{3}\of{x}}\Big)^4\) are even powers and so are positive (though possibly infinite). So the sign of \(f''(x)\) is the same as the sign of the factor \(x-1\text{.}\) Thus
\((-\infty,1)\) 1 \((1,\infty)\) \(f''(x)\) negative 0 positive concave down inflection

pointconcave up

- Both of the factors \({\Big(\frac{1}{x-6}\Big)}^{\frac{8}{3}} ={\Big(\frac{1}{\root{3}\of{x-6}}\Big)}^8\) and \(\frac{1}{x^{\frac{4}{3}}} =\Big(\frac{1}{\root{3}\of{x}}\Big)^4\) are even powers and so are positive (though possibly infinite). So the sign of \(f''(x)\) is the same as the sign of the factor \(x-1\text{.}\) Thus

Here is a sketch of the graph \(y=f(x)\text{.}\)

It is hard to see the inflection point at \(x=1\text{,}\) \(y=f(1)=\frac{1}{ \root{3}\of{25} }\) in the above sketch. So here is a blow up of the part of the sketch around \(x=1\text{.}\)

And if we zoom in even more we have

## Exercises

### Exercises for § 3.6.1

Stage 1

Suppose \(f(x)\) is a function given by

\[ f(x)= \frac{g(x)}{x^2-9} \nonumber \]

where \(g(x)\) is also a function. True or false: \(f(x)\) has a vertical asymptote at \(x=-3\text{.}\)

Stage 2

Match the functions \(f(x)\text{,}\) \(g(x)\text{,}\) \(h(x)\text{,}\) and \(k(x)\) to the curves \(y=A(x)\) through \(y=D(x)\text{.}\)

\begin{align*} f(x) & =\sqrt{x^2+1} & g(x) &=\sqrt{x^2-1} \\ h(x) &=\sqrt{x^2+4} & k(x) &=\sqrt{x^2-4} \end{align*}

Below is the graph of

\[ y=f(x)=\sqrt{\log^2(x+p)} \nonumber \]

- What is \(p\text{?}\)
- What is \(b\) (marked on the graph)?
- What is the \(x\)-intercept of \(f(x)\text{?}\)

Remember \(\log(x+p)\) is the natural logarithm of \(x+p\text{,}\) \(\log_e(x+p)\text{.}\)

Find all asymptotes of \(f(x)=\dfrac{x(2x+1)(x-7)}{3x^3-81}\text{.}\)

Find all asymptotes of \(f(x)=10^{3x-7}\text{.}\)

### Exercises for § 3.6.2

Stage 1

Match each function graphed below to its *derivative* from the list. (For example, which function on the list corresponds to \(A'(x)\text{?}\))

The \(y\)-axes have been scaled to make the curve's behaviour clear, so the vertical scales differ from graph to graph.

\(l(x)=(x-2)^4\)

\(m(x)=(x-2)^4(x+2)\)

\(n(x)=(x-2)^2(x+2)^2\)

\(o(x)=(x-2)(x+2)^3\)

\(p(x)=(x+2)^4\)

Stage 2

Find the interval(s) where \(f(x)=\dfrac{e^x}{x+3}\) is increasing.

Find the interval(s) where \(f(x)=\dfrac{\sqrt{x-1}}{2x+4}\) is increasing.

Find the interval(s) where \(f(x)=2\arctan (x) - \log(1+x^2)\) is increasing.

### Exercises for § 3.6.3

Stage 1

On the graph below, mark the intervals where \(f''(x) \gt 0\) (i.e. \(f(x)\) is concave up) and where \(f''(x) \lt 0\) (i.e. \(f(x)\) is concave down).

Sketch a curve that is:

- concave up when \(|x| \gt 5\text{,}\)
- concave down when \(|x| \lt 5\text{,}\)
- increasing when \(x \lt 0\text{,}\) and
- decreasing when \(x \gt 0\text{.}\)

Suppose \(f(x)\) is a function whose second derivative exists and is continuous for all real numbers.

True or false: if \(f''(3)=0\text{,}\) then \(x=3\) is an inflection point of \(f(x)\text{.}\)

Remark: compare to Question 3.6.7.7

Stage 2

Find all inflection points for the graph of \(f(x)=3x^5-5x^4+13x\text{.}\)

Stage 3

Questions 3.6.7.5 through 3.6.7.7 ask you to show that certain things are true. Give a clear explanation using concepts and theorems from this semester.

Let

\[ f(x)=\frac{x^5}{20}+\frac{5x^3}{6}-10x^2+500x+1000 \nonumber \]

Show that \(f(x)\) has exactly one inflection point.

Let \(f(x)\) be a function whose first two derivatives exist everywhere, and \(f''(x) \gt 0\) for all \(x\text{.}\)

- Show that \(f(x)\) has at most one critical point and that any critical point is an absolute minimum for \(f(x)\text{.}\)
- Show that the maximum value of \(f(x)\) on any finite interval occurs at one of the endpoints of the interval.

Suppose \(f(x)\) is a function whose second derivative exists and is continuous for all real numbers, and \(x=3\) is an inflection point of \(f(x)\text{.}\) Use the Intermediate Value Theorem to show that \(f''(3)=0\text{.}\)

Remark: compare to Question 3.6.7.3.

### Exercises for § 3.6.4

Stage 1

What symmetries (even, odd, periodic) does the function graphed below have?

What symmetries (even, odd, periodic) does the function graphed below have?

Suppose \(f(x)\) is an even function defined for all real numbers. Below is the curve \(y=f(x)\) when \(x \lt 0\text{.}\) Complete the sketch of the curve.

Suppose \(f(x)\) is an odd function defined for all real numbers. Below is the curve \(y=f(x)\) when \(x \lt 0\text{.}\) Complete the sketch of the curve.

Stage 2

In Questions 3.6.7.7 through 3.6.7.10, find the symmetries of a function from its equation.

\[ f(x)=\frac{x^4-x^6}{e^{x^2}} \nonumber \]

Show that \(f(x)\) is even.

\[ f(x)=\sin(x)+\cos\left(\frac{x}{2}\right) \nonumber \]

Show that \(f(x)\) is periodic.

\[ f(x)=x^4+5x^2+\cos\left(x^3\right) \nonumber \]

What symmetries (even, odd, periodic) does \(f(x)\) have?

\[ f(x)=x^5+5x^4 \nonumber \]

What symmetries (even, odd, periodic) does \(f(x)\) have?

\[ f(x)=\tan\left(\pi x\right) \nonumber \]

What is the period of \(f(x)\text{?}\)

Stage 3

\[ f(x)=\tan\left(3 x\right)+\sin\left(4 x\right) \nonumber \]

What is the period of \(f(x)\text{?}\)

### Exercises for § 3.6.6

Stage 1

In Questions 3.6.7.2 through 3.6.7.4, you will sketch the graphs of rational functions.

In Questions 3.6.7.6 and 3.6.7.7, you will sketch the graphs of functions with an exponential component. In the next section, you will learn how to find their horizontal asymptotes, but for now these are given to you.

In Questions 3.6.7.8 and 3.6.7.9, you will sketch the graphs of functions that have a trigonometric component.

Let \(f(x) = x\sqrt{3 - x}\text{.}\)

- Find the domain of \(f(x)\text{.}\)
- Determine the \(x\)-coordinates of the local maxima and minima (if any) and intervals where \(f(x)\) is increasing or decreasing.
- Determine intervals where \(f(x)\) is concave upwards or downwards, and the \(x\) coordinates of inflection points (if any). You may use, without verifying it, the formula \(f''(x) = (3x -12)(3 - x)^{-3/2}/4\text{.}\)
- There is a point at which the tangent line to the curve \(y = f(x)\) is vertical. Find this point.
- Sketch the graph \(y = f(x)\text{,}\) showing the features given in items (a) to (d) above and giving the \((x, y)\) coordinates for all points occurring above.

Sketch the graph of

\[ f(x)= \dfrac{x^3-2}{x^4}. \nonumber \]

Indicate the critical points, local and absolute maxima and minima, vertical and horizontal asymptotes, inflection points and regions where the curve is concave upward or downward.

The first and second derivatives of the function \(f(x)=\dfrac{x^4}{1+x^3}\) are:

\[ f'(x)=\frac{4x^3+x^6}{(1+x^3)^2}\qquad\hbox{and}\qquad f''(x)=\frac{12x^2-6x^5}{(1+x^3)^3} \nonumber \]

Graph \(f(x)\text{.}\) Include local and absolute maxima and minima, regions where \(f(x)\) is increasing or decreasing, regions where the curve is concave upward or downward, and any asymptotes.

The first and second derivatives of the function \(f(x)=\dfrac{x^3}{1-x^2}\) are:

\[ f'(x)=\frac{3x^2-x^4}{(1-x^2)^2}\qquad\hbox{and}\qquad f''(x)=\frac{6x+2x^3}{(1-x^2)^3} \nonumber \]

Graph \(f(x)\text{.}\) Include local and absolute maxima and minima, regions where the curve is concave upward or downward, and any asymptotes.

The function \(f(x)\) is defined by

\[ f(x) = \left\{\begin{array}{lc} e^x &x \lt 0\\ \frac{x^2+3}{3(x+1)} & x \ge 0 \end{array}\right. \nonumber \]

- Explain why \(f(x)\) is continuous everywhere.
- Determine all of the following if they are present:
- \(x\)--coordinates of local maxima and minima, intervals where \(f(x)\) is increasing or decreasing;
- intervals where \(f(x)\) is concave upwards or downwards;
- equations of any horizontal or vertical asymptotes.

- Sketch the graph of \(y = f(x)\text{,}\) giving the \((x, y)\) coordinates for all points of interest above.

The function \(f(x)\) and its derivative are given below:

\[ f(x)=(1+2x)e^{-x^2}\qquad\hbox{and}\qquad f'(x)=2(1-x-2x^2)e^{-x^2} \nonumber \]

Sketch the graph of \(f(x)\text{.}\) Indicate the critical points, local and/or absolute maxima and minima, and asymptotes. Without actually calculating the inflection points, indicate on the graph their approximate location.

Note: \(\ds\lim_{x \to \pm\infty}f(x)=0\text{.}\)

Consider the function \(f(x) = xe^{-x^2/2}\text{.}\)

Note: \(\ds\lim_{x \to \pm\infty}f(x)=0\text{.}\)

- Find all inflection points and intervals of increase, decrease, convexity up, and convexity down. You may use without proof the formula \(f''(x) = (x^3-3x)e^{-x^2/2}\text{.}\)
- Find local and global minima and maxima.
- Use all the above to draw a graph for \(f\text{.}\) Indicate all special points on the graph.

Use the techniques from this section to sketch the graph of \(f(x)=x+2\sin x\text{.}\)

\[ f(x) = 4\sin x - 2\cos 2x \nonumber \]

Graph the equation \(y = f(x)\text{,}\) including all important features. (In particular, find all local maxima and minima and all inflection points.) Additionally, find the maximum and minimum values of \(f(x)\) on the interval \([0,\pi]\text{.}\)

Sketch the curve \(y=\sqrt[3]{\dfrac{x+1}{x^2}}\text{.}\)

You may use the facts \(y'(x)=\dfrac{-(x+2)}{3x^{5/3}(x+1)^{2/3}}\) and \(y''(x)=\dfrac{4x^2+16x+10}{9x^{8/3}(x+1)^{5/3}}\text{.}\)

Stage 3

A function \(f(x)\) defined on the whole real number line satisfies the following conditions

\[ f(0)=0\qquad f(2)=2\qquad \lim_{x\rightarrow+\infty}f(x)=0\qquad f'(x)=K(2x-x^2)e^{-x} \nonumber \]

for some positive constant \(K\text{.}\) (Read carefully: you are given the *derivative* of \(f(x)\text{,}\) not \(f(x)\) itself.)

- Determine the intervals on which \(f\) is increasing and decreasing and the location of any local maximum and minimum values of \(f\text{.}\)
- Determine the intervals on which \(f\) is concave up or down and the \(x\)--coordinates of any inflection points of \(f\text{.}\)
- Determine \(\lim\limits_{x\rightarrow-\infty}f(x)\text{.}\)
- Sketch the graph of \(y=f(x)\text{,}\) showing any asymptotes and the information determined in parts 3.6.7.11.a and 3.6.7.11.b.

Let \(f(x) = e^{-x}\), \(x \ge 0\text{.}\)

- Sketch the graph of the equation \(y = f(x)\text{.}\) Indicate any local extrema and inflection points.
- Sketch the graph of the inverse function \(y = g (x)=f^{-1}(x)\text{.}\)
- Find the domain and range of the inverse function \(g(x)= f^{-1}(x)\text{.}\)
- Evaluate \(g'(\half)\text{.}\)

- Sketch the graph of \(y=f(x)=x^5-x\text{,}\) indicating asymptotes, local maxima and minima, inflection points, and where the graph is concave up/concave down.
- Consider the function \(f(x)=x^5-x+k\text{,}\) where \(k\) is a constant, \(-\infty \lt k \lt \infty\text{.}\) How many roots does the function have? (Your answer might depend on the value of \(k\text{.}\))

The hyperbolic trigonometric functions \(\sinh(x)\) and \(\cosh(x)\) are defined by

\[ \sinh(x)=\dfrac{e^x-e^{-x}}{2}\qquad \cosh(x)=\dfrac{e^x+e^{-x}}{2} \nonumber \]

They have many properties that are similar to corresponding properties of \(\sin(x)\) and \(\cos(x)\text{.}\) In particular, it is easy to see that

\[ \dfrac{d}{dx} \sinh(x)=\cosh(x)\qquad \dfrac{d}{dx} \cosh(x)=\sinh(x)\qquad \cosh^2(x)-\sinh^2(x)=1 \nonumber \]

You may use these properties in your solution to this question.

- Sketch the graphs of \(\sinh(x)\) and \(\cosh(x)\text{.}\)
- Define inverse hyperbolic trigonometric functions \(\sinh^{-1}(x)\) and \(\cosh^{-1}(x)\text{,}\) carefully specifing their domains of definition. Sketch the graphs of \(\sinh^{-1}(x)\) and \(\cosh^{-1}(x)\text{.}\)
- Find \(\dfrac{d}{dx}\left\{ \cosh^{-1}(x)\right\}\text{.}\)