3.3: Shapes of Curves

Learning Objectives Explain how the sign of the first derivative affects the shape of a function's graph. State the First Derivative Test for critical points. Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function's graph. Explain the Concavity Test for a function over an open interval. Explain the relationship between a function and its first and second derivatives. State the Second Derivative Test for local extrema.

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Earlier in this chapter we stated that if a function \(f\) has a local extremum at a point \( (c, f(x)) \), then \(c\) must be a critical number of \(f\). However, a function is not guaranteed to have a local extremum at a critical number. For example, \(f(x)=x^3\) has a critical point at \( (0,0) \) since \(f^{\prime}(x)=3x^2\) is zero at \(x=0\), but \(f\) does not have a local extremum at \(x=0\). Using the results from the previous section, we can now determine whether a critical point of a function corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or downward.

Increasing/Decreasing Functions

We begin this section by allowing for one final corollary from the Mean Value Theorem. This corollary discusses when a function is increasing and when it is decreasing. Recall that a function \(f\) is increasing over \(I\) if \(f(x_1) \lt f(x_2)\) whenever \(x_1 \lt x_2\), whereas \(f\) is decreasing over \(I\) if \(f(x_1) \gt f(x_2)\) whenever \(x_1 \lt x_2\). Using the Mean Value Theorem, we can show that if the derivative of a function is positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure \(\PageIndex{1}\)). We use this fact throughout the remainder of this chapter, where we show how to use the derivative of a function to locate the local maximum and minimum values of the function and how to determine the shape of the graph.

Corollary: Increasing and Decreasing Functions

Let \(f\) be continuous over the closed interval \([a,b]\) and differentiable over the open interval \((a,b)\).

1. If \(f^{\prime}(x) \gt 0\) for all \(x \in (a,b)\), then \(f\) is an increasing function over \([a,b]\).
2. If \(f^{\prime}(x) \lt 0\) for all \(x \in (a,b)\), then \(f\) is a decreasing function over \([a,b]\).

Proof (by contradiction)

We will prove i. by contradiction; the proof of ii. is similar.

Assume \(f\) is not an increasing function on \(I\). Then there exist \(a\) and \(b\) in \(I\) such that \(a \lt b\), but \(f(a) \geq f(b)\). Since \(f\) is a differentiable function over \(I\), by the Mean Value Theorem there exists \(c \in (a,b)\) such that\[f^{\prime}(c) = \dfrac{f(b)−f(a)}{b−a}. \nonumber \]Since \(f(a) \geq f(b)\), we know that \(f(b) − f(a) \leq 0\). Also, \(a \lt b\) tells us that \(b − a \gt 0\). We conclude that\[f^{\prime}(c) = \dfrac{f(b)−f(a)}{b−a} \leq 0. \nonumber \]However, \(f^{\prime}(x) \gt 0\) for all \(x \in I\). This is a contradiction. The only assumption we made was that \( f \) was not an increasing function on \( I \). Therefore, \(f\) must be an increasing function over \(I\).

Q.E.D.

Figure \( \PageIndex{1} \) demonstrates this corollary perfectly.

Figure \(\PageIndex{1}\): If a function has a positive derivative over some interval \(I\), then the function increases over that interval \(I\); if the derivative is negative over some interval \(I\), then the function decreases over that interval \(I\).

The First Derivative Test

The previous corollary shows that if the derivative of a function is positive over an interval \(I\), then the function is increasing over \(I\). On the other hand, if the function's derivative is negative over an interval \(I\), then the function is decreasing over \(I\), as shown in the following figure.

Figure \(\PageIndex{2}\): Both functions are increasing over the interval \((a,b)\). At each point \(x\), the derivative \(f^{\prime}(x)>0\). Both functions decrease over the interval \((a,b)\). At each point \(x\), the derivative \(f^{\prime}(x)<0\).

A continuous function \(f\) has a local maximum at point \(c\) if and only if \(f\) switches from increasing to decreasing at point \(c\). Similarly, \(f\) has a local minimum at \(c\) if and only if \(f\) switches from decreasing to increasing at \(c\). If \(f\) is a continuous function over an interval \(I\) containing \(c\) and differentiable over \(I\), except possibly at \(c\), the only way \(f\) can switch from increasing to decreasing (or vice versa) at point \(c\) is if \(f^{\prime}\) changes sign as \(x\) increases through \(c\). If \(f\) is differentiable at \(c\), the only way that \(f^{\prime}\) can change sign as \(x\) increases through \(c\) is if \(f^{\prime}(c)=0\). Therefore, for a function \(f\) that is continuous over an interval \(I\) containing \(c\) and differentiable over \(I\), except possibly at \(c\), the only way \(f\) can switch from increasing to decreasing (or vice versa) is if \(f^{\prime}(c)=0\) or \(f^{\prime}(c)\) is undefined. Consequently, to locate local extrema for a function \(f\), we look for numbers \(c\) in the domain of \(f\) such that \(f^{\prime}(c)=0\) or \(f^{\prime}(c)\) is undefined. Recall that such numbers are called critical numbers of \(f\).

Note that \(f\) need not have a local extrema at a critical number. The critical points are candidates for local extrema only. In Figure \(\PageIndex{3}\), we show that if a continuous function \(f\) has a local extremum, it must occur at a critical point. Still, it can be the case that a function does not have a local extremum at a critical point. We show that if \(f\) has a local extremum at a critical point, then the sign of \(f^{\prime}\) switches as \(x\) increases through that point.

Figure \(\PageIndex{3}\): The function \(f\) has four critical points: \(a,b,c\),and \(d\). The function \(f\) has local maxima at \(a\) and \(d\), and a local minimum at \(b\). The function \(f\) does not have a local extremum at \(c\). The sign of \(f^{\prime}\) changes at all local extrema.

Using Figure \(\PageIndex{3}\), we summarize the main results regarding local extrema.

• If a continuous function \(f\) has a local extremum, it must occur at a critical number \(c\).
• The function has a local extremum at the critical number \(c\) if and only if the derivative, \(f^{\prime}\), switches sign as \(x\) increases through \(c\).
• Therefore, to test whether a function has a local extremum at a critical number \(c\), we must determine the sign of \(f^{\prime}(x)\) to the left and right of \(c\).

This result is known as the First Derivative Test.

Theorem: First Derivative Test

Suppose that \(f\) is a continuous function over an interval \(I\) containing a critical number \(c\). If \(f\) is differentiable over \(I\), except possibly at \(c\), then \(f(c)\) satisfies one of the following descriptions:

1. If \(f^{\prime}\) changes sign from positive when \(x \lt c\) to negative when \(x \gt c\), then \(f(c)\) is a local maximum of \(f\).
2. If \(f^{\prime}\) changes sign from negative when \(x \lt c\) to positive when \(x \gt c\), then \(f(c)\) is a local minimum of \(f\).
3. If \(f^{\prime}\) has the same sign for \(x \lt c\) and \(x \gt c\), then \(f(c)\) is neither a local maximum nor a local minimum of \(f\)

Now let's look at how to use this strategy to locate all local extrema for particular functions.

Example \(\PageIndex{1}\)

Use the First Derivative Test to find the location of all local extrema for \(f(x)=x^3−3x^2−9x−1\). Use graphing technology to confirm your results.

Solution

The derivative is \(f^{\prime}(x)=3x^2−6x−9\). To find the critical numbers, we need to find where \(f^{\prime}(x)=0\). Factoring the polynomial, we conclude that the critical numbers must satisfy\[3(x^2−2x−3)=3(x−3)(x+1)=0. \nonumber \]Therefore, the critical numbers are \(x=3\) and \(x = -1\). Now divide the interval \((−\infty,\infty)\) into the smaller intervals \((−\infty,−1)\), \((−1,3)\), and \((3,\infty)\).

Since \(f^{\prime}\) is a continuous function, to determine the sign of \(f^{\prime}(x)\) over each subinterval, it suffices to choose a point over each of the intervals \((−\infty,−1)\), \((−1,3)\), and \((3,\infty)\), and determine the sign of \(f^{\prime}\) at each of these points. For example, let’s choose \(x=−2\), \(x=0\), and \(x=4\) as test points.

Table: \(\PageIndex{1}\): First Derivative Test for \(f(x)=x^3−3x^2−9x−1\).

Interval	Test Point	Sign of \(f^{\prime}(x)=3(x−3)(x+1)\) at Test Point	Conclusion
\((−\infty,−1)\)	\(x=−2\)	(+)(−)(−)=+	\(f\) is increasing.
\((−1,3)\)	\(x=0\)	(+)(−)(+)=-	\(f\) is decreasing.
\((3,\infty)\)	\(x=4\)	(+)(+)(+)=+	\(f\) is increasing.

Since \(f^{\prime}\) switches sign from positive to negative as \(x\) increases through \(-1\), \(f\) has a local maximum at \(x=−1\). Since \(f^{\prime}\) switches sign from negative to positive as \(x\) increases through \(3\), \(f\) has a local minimum at \(x=3\). These analytical results agree with the following graph.

Figure \(\PageIndex{4}\): The function \(f\) has a maximum at \(x=−1\) and a minimum at \(x=3\)

Checkpoint \(\PageIndex{1}\)

Use the First Derivative Test to locate all local extrema for \(f(x)=−x^3+\frac{3}{2}x^2+18x\).

Answer

\(f\) has a local minimum at \(−2\) and a local maximum at \(3\).

Example \(\PageIndex{2}\)

Use the First Derivative Test to find the location of all local extrema for \(f(x)=5x^{1/3}−x^{5/3}\). Use graphing technology to confirm your results.

Solution

The derivative is\[f^{\prime}(x)=\dfrac{5}{3}x^{−2/3}−\dfrac{5}{3}x^{2/3}=\dfrac{5}{3x^{2/3}}−\dfrac{5x^{2/3}}{3}=\dfrac{5−5x^{4/3}}{3x^{2/3}}=\dfrac{5(1−x^{4/3})}{3x^{2/3}}.\nonumber \]The derivative \(f^{\prime}(x)=0\) when \(1−x^{4/3}=0\). Therefore, \(f^{\prime}(x)=0\) at \(x= \pm 1\). The derivative \(f^{\prime}(x)\) is undefined at \(x=0\). Therefore, we have three critical numbers: \(x=0\), \(x=1\), and \(x=−1\). Consequently, divide the interval \((−\infty,\infty)\) into the smaller intervals \((−\infty,−1)\), \((−1,0)\), \((0,1)\), and \((1,\infty)\).

Since \(f^{\prime}\) is continuous over each subinterval, it suffices to choose a test point \(x\) in each of the intervals and determine the sign of \(f^{\prime}\) at each of these points. The points \(x=−2\), \(x=−\frac{1}{2}\), \(x=\frac{1}{2}\), and \(x=2\) are test points for these intervals.

Table: \(\PageIndex{2}\): First Derivative Test for \(f(x)=5x^{1/3}−x^{5/3}\).

Interval	Test Point	Sign of \(f^{\prime}(x)=\frac{5(1−x^{4/3})}{3x^{2/3}}\) at Test Point	Conclusion
\((−\infty,−1)\)	\(x=−2\)	\(\frac{(+)(−)}{+}=−\)	\(f\) is decreasing.
\((−1,0)\)	\(x=−\frac{1}{2}\)	\(\frac{(+)(+)}{+}=+\)	\(f\) is increasing.
\((0,1)\)	\(x=\frac{1}{2}\)	\(\frac{(+)(+)}{+}=+\)	\(f\) is increasing.
\((1,\infty)\)	\(x=2\)	\(\frac{(+)(−)}{+}=−\)	\(f\) is decreasing.

Since \(f\) is decreasing over the interval \((−\infty,−1)\) and increasing over the interval \((−1,0)\), \(f\) has a local minimum at \(x=−1\). Since \(f\) is increasing over the interval \((−1,0)\) and the interval \((0,1)\), \(f\) does not have a local extremum at \(x=0\). Since \(f\) is increasing over the interval \((0,1)\) and decreasing over the interval \((1,\infty)\), \(f\) has a local maximum at \(x=1\). The analytical results agree with the following graph.

Figure \(\PageIndex{5}\): The function \(f\) has a local minimum at \(x=−1\) and a local maximum at \(x=1\)

Checkpoint \(\PageIndex{2}\)

Use the First Derivative Test to find all local extrema for \(f(x)=\frac{3}{x−1}\).

Answer

\(f\) has no local extrema because \(f^{\prime}\) does not change sign at \(x=1\).

Concavity and Points of Inflection

We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is called the concavity of the function.

Figure \(\PageIndex{6a}\) shows a function \(f\) with a graph that curves upward. As \(x\) increases, the slope of the tangent line increases. Thus, since the derivative increases as \(x\) increases, \(f^{\prime}\) is an increasing function. We say this function \(f\) is concave up. Figure \(\PageIndex{6b}\) shows a function \(f\) that curves downward. As \(x\) increases, the slope of the tangent line decreases. Since the derivative decreases as \(x\) increases, \(f^{\prime}\) is a decreasing function. We say this function \(f\) is concave down.

Definition: Concavity

Let \(f\) be a function that is differentiable over an open interval \(I\). If \(f^{\prime}\) is increasing over \(I\), we say \(f\) is concave up over \(I\). If \(f^{\prime}\) is decreasing over \(I\), we say \(f\) is concave down over \(I\).

Figure \(\PageIndex{6}\): (a), (c) Since \(f^{\prime}\) is increasing over the interval \((a,b)\), we say \(f\) is concave up over \((a,b). (b), (d)\) Since \(f^{\prime}\) is decreasing over the interval \((a,b)\), we say \(f\) is concave down over \((a,b)\).

Note: Another Viewpoint

An equivalent way to define concavity is that a differentiable (and, therefore, continuous) function is concave up over an open interval \( I \) if it lies above all of its tangent lines on the interval. Likewise, the function is said to be concave down on \( I \) if it lies below all of its tangent lines on \( I \).

In general, without having the graph of a function \(f\), how can we determine its concavity? By definition, a function \(f\) is concave up if \(f^{\prime}\) is increasing. From our previous corollary, we know that if \(f^{\prime}\) is a differentiable function, then \(f^{\prime}\) is increasing if its derivative \(f^{\prime\prime}(x) \gt 0\). Therefore, a function \(f\) that is twice differentiable is concave up when \(f^{\prime\prime}(x) \gt 0\). Similarly, a function \(f\) is concave down if \(f^{\prime}\) is decreasing. We know that a differentiable function \(f^{\prime}\) is decreasing if its derivative \(f^{\prime\prime}(x) \lt 0\). Therefore, a twice-differentiable function \(f\) is concave down when \(f^{\prime\prime}(x) \lt 0\). Applying this logic is known as the Concavity Test.

Theorem: Concavity Test

Let \(f\) be a function that is twice differentiable over an interval \(I\).

1. If \(f^{\prime\prime}(x) \gt 0\) for all \(x \in I\), then \(f\) is concave up over \(I\)
2. If \(f^{\prime\prime}(x) \lt 0\) for all \(x \in I\), then \(f\) is concave down over \(I\).

We conclude that we can determine the concavity of a function \(f\) by looking at the second derivative of \(f\). In addition, we observe that a function \(f\) can switch concavity (Figure \(\PageIndex{7}\)). However, a continuous function can switch concavity only at a point \(x\) if \(f^{\prime\prime}(x) = 0\) or \(f^{\prime\prime}(x)\) is undefined. Consequently, to determine the intervals where a function \(f\) is concave up and concave down, we look for those values of \(x\) where \(f^{\prime\prime}(x) = 0\) or \(f^{\prime\prime}(x)\) is undefined. When we have determined these points, we divide the domain of \(f\) into smaller intervals and determine the sign of \(f^{\prime\prime}\) over each of these smaller intervals. If \(f^{\prime\prime}\) changes sign as we pass through a point \(x\), then \(f\) changes concavity. It is important to remember that a function \(f\) may not change concavity at a point \(x\) even if \(f^{\prime\prime}(x) = 0\) or \(f^{\prime\prime}(x)\) is undefined. If, however, \(f\) does change concavity at a point \(a\) and \(f\) is continuous at \(a\), we say the point \((a,f(a))\) is an inflection point of \(f\).

Definition: Inflection Point

If \(f\) is continuous at \(a\) and \(f\) changes concavity at \(a\), the point \((a, \,f(a))\) is an inflection point of \(f\).

Figure \(\PageIndex{7}\): Since \(f^{\prime\prime}(x) \gt 0\) for \(x \lt a\), the function \(f\) is concave up over the interval \((−\infty,a)\). Since \(f^{\prime\prime}(x) \lt 0\) for \(x \gt a\), the function \(f\) is concave down over the interval \((a,\infty)\). The point \((a,f(a))\) is an inflection point of \(f\).

Example \(\PageIndex{3}\)

For the function \(f(x)=x^3−6x^2+9x+30\), determine all intervals where \(f\) is concave up and all intervals where \(f\) is concave down. List all inflection points for \(f\). Use a graphing utility to confirm your results.

Solution

To determine concavity, we need to find the second derivative \(f^{\prime\prime}(x)\). The first derivative is \(f^{\prime}(x)=3x^2−12x+9\), so the second derivative is \(f^{\prime\prime}(x)=6x−12\). If the function changes concavity, it occurs either when \(f^{\prime\prime}(x)=0\) or \(f^{\prime\prime}(x)\) is undefined. Since \(f^{\prime\prime}\) is defined for all real numbers \(x\), we need only find where \(f^{\prime\prime}(x)=0\). Solving the equation \(6x−12=0\), we see that \(x=2\) is the only place where \(f\) could change concavity. We now test points over the intervals \((−\infty,2)\) and \((2,\infty)\) to determine the concavity of \(f\). The points \(x=0\) and \(x=3\) are test points for these intervals.

Table: \(\PageIndex{3}\): Test for Concavity for \(f(x)=x^3−6x^2+9x+30\).

Interval	Test Point	Sign of \(f^{\prime\prime}(x)=6x−12\) at Test Point	Conclusion
\((−\infty,2)\)	\(x=0\)	−	\(f\) is concave down
\((2,\infty)\)	\(x=3\)	+	\(f\) is concave up

We conclude that \(f\) is concave down over the interval \((−\infty,2)\) and concave up over the interval \((2,\infty)\). Since \(f\) changes concavity at \(x=2\), the point \((2,f(2))=(2,32)\) is an inflection point. Figure \(\PageIndex{8}\) confirms the analytical results.

Figure \(\PageIndex{8}\): The given function has a point of inflection at \((2,32)\) where the graph changes concavity.

Checkpoint \(\PageIndex{3}\)

For \(f(x)=−x^3+\frac{3}{2}x^2+18x\), find all intervals where \(f\) is concave up and all intervals where \(f\) is concave down.

Answer

\(f\) is concave up over the interval \((−\infty,\frac{1}{2})\) and concave down over the interval \((\frac{1}{2},\infty)\)

We now summarize, in Table \(\PageIndex{4}\), the information that the first and second derivatives of a function \(f\) provide about the graph of \(f\) and illustrate this information in Figure \(\PageIndex{9}\).

Table: \(\PageIndex{4}\): What Derivatives Tell Us about Graphs

Sign of \(f^{\prime}\)	Sign of \(f^{\prime\prime}\)	Is \(f\) increasing or decreasing?	Concavity
Positive	Positive	Increasing	Concave up
Positive	Negative	Increasing	Concave down
Negative	Positive	Decreasing	Concave up
Negative	Negative	Decreasing	Concave down

Figure \(\PageIndex{9}\):Consider a twice-differentiable function \(f\) over an open interval \(I\). If \(f^{\prime}(x)>0\) for all \(x \in I\), the function is increasing over \(I\). If \(f^{\prime}(x)<0\) for all \(x \in I\), the function is decreasing over \(I\). If \(f^{\prime\prime}(x)>0\) for all \(x \in I\), the function is concave up. If \(f^{\prime\prime}(x)<0\) for all \(x \in I\), the function is concave down on \(I\).

The Second Derivative Test

The First Derivative Test provides an analytical tool for finding local extrema. Still, the second derivative can also be used to locate extreme values. Using the second derivative can sometimes be a more straightforward method than using the first derivative.

We know that if a continuous function has a local extremum, it must occur at a critical point. However, a function does not need a local extremum at a critical point. Here, we examine how the Second Derivative Test can determine whether a function has a local extremum at a critical point. Let \(f\) be a twice-differentiable function such that \(f^{\prime}(a)=0\) and \(f^{\prime\prime}\) is continuous over an open interval \(I\) containing \(a\). Suppose \(f^{\prime\prime}(a)<0\). Since \(f^{\prime\prime}\) is continuous over \(I\), \(f^{\prime\prime}(x)<0\) for all \(x \in I\) (Figure \(\PageIndex{10}\)). Then \(f^{\prime}\) is a decreasing function over \(I\). Since \(f^{\prime}(a)=0\), we conclude that for all \(x \in I\), \(f^{\prime}(x)>0\) if \(x<a\) and \(f^{\prime}(x)<0\) if \(x>a\). Therefore, by the First Derivative Test, \(f\) has a local maximum at \(x=a\).

On the other hand, suppose there exists a point \(b\) such that \(f^{\prime}(b)=0\) but \(f^{\prime\prime}(b)>0\). Since \(f^{\prime\prime}\) is continuous over an open interval \(I\) containing \(b\), \(f^{\prime\prime}(x)>0\) for all \(x \in I\) (Figure \(\PageIndex{10}\)). Then \(f^{\prime}\) is an increasing function over \(I\). Since \(f^{\prime}(b)=0\), we conclude that for all \(x \in I\), \(f^{\prime}(x)<0\) if \(x<b\) and \(f^{\prime}(x)>0\) if \(x>b\). Therefore, by the First Derivative Test, \(f\) has a local minimum at \(x=b\).

Figure \(\PageIndex{10}\): Consider a twice-differentiable function \(f\) such that \(f^{\prime\prime}\) is continuous. Since \(f^{\prime}(a)=0\) and \(f^{\prime\prime}(a)<0\), there is an interval \(I\) containing \(a\) such that for all \(x\) in \(I\), \(f\) is increasing if \(x<a\) and \(f\) is decreasing if \(x>a\). As a result, \(f\) has a local maximum at \(x=a\). Since \(f^{\prime}(b)=0\) and \(f^{\prime\prime}(b)>0\), there is an interval \(I\) containing \(b\) such that for all \(x\) in \(I\), \(f\) is decreasing if \(x<b\) and \(f\) is increasing if \(x>b\). As a result, \(f\) has a local minimum at \(x=b\).

Theorem: Second Derivative Test

Suppose \(f^{\prime}(c)=0\) and \(f^{\prime\prime}\) is continuous over an interval containing \(c\).

1. If \(f^{\prime\prime}(c)>0\), then \(f\) has a local minimum at \(c\).
2. If \(f^{\prime\prime}(c)<0\), then \(f\) has a local maximum at \(c\).
3. If \(f^{\prime\prime}(c)=0\), then the test is inconclusive.

Note, when \(f^{\prime\prime}(c)=0\) (case iii), \(f\) may have a local maximum, local minimum, or neither at \(c\). For example, the functions \(f(x)=x^3\), \(f(x)=x^4\), and \(f(x)=−x^4\) all have critical points at \((0,0)\). In each case, the second derivative is zero at \(x=0\). However, the function \(f(x)=x^4\) has a local minimum at \(x=0\) whereas the function \(f(x)=−x^4\) has a local maximum at \(x=0\), and the function \(f(x)=x^3\) does not have a local extremum at \(x=0\).

Let's now look at how to use the Second Derivative Test to determine whether \(f\) has a local maximum or local minimum at a critical number \(c\) where \(f^{\prime}(c)=0\).

Example \(\PageIndex{4}\)

Use the second derivative to find the location of all local extrema for \(f(x)=x^5−5x^3\).

Solution

To apply the Second Derivative Test, we first need to find critical numbers \(c\) where \(f^{\prime}(c)=0\). The derivative is \(f^{\prime}(x)=5x^4−15x^2\). Therefore, \(f^{\prime}(x)=5x^4−15x^2=5x^2(x^2−3)=0\) when \(x=0,\, \pm \sqrt{3}\).

To determine whether \(f\) has a local extremum at any of these points, we need to evaluate the sign of \(f^{\prime\prime}\) at these points. The second derivative is\[f^{\prime\prime}(x)=20x^3−30x=10x(2x^2−3).\nonumber \]In the following table, we evaluate the second derivative at each critical point and use the Second Derivative Test to determine whether \(f\) has a local maximum or minimum at any of these points.

Table: \(\PageIndex{5}\): Second Derivative Test for \(f(x)=x^5−5x^3\).

\(x\)	\(f^{\prime\prime}(x)\)	Conclusion
\(−\sqrt{3}\)	\(−30\sqrt{3}\)	Local maximum
\(0\)	\(0\)	Second Derivative Test is inconclusive
\(\sqrt{3}\)	\(30\sqrt{3}\)	Local minimum

By the Second Derivative Test, we conclude that \(f\) has a local maximum at \(x=−\sqrt{3}\) and \(f\) has a local minimum at \(x=\sqrt{3}\). The Second Derivative Test is inconclusive at \(x=0\). We apply the First Derivative Test to determine whether \(f\) has a local extremum at \(x=0\). To evaluate the sign of \(f^{\prime}(x)=5x^2(x^2−3)\) for \(x \in (−\sqrt{3},0)\) and \(x \in (0,\sqrt{3})\), let \(x=−1\) and \(x=1\) be the two test points. Since \(f^{\prime}(−1)<0\) and \(f^{\prime}(1)<0\), we conclude that \(f\) is decreasing on both intervals and, therefore, \(f\) does not have a local extrema at \(x=0\) as shown in the following graph.

Figure \(\PageIndex{11}\):The function \(f\) has a local maximum at \(x=−\sqrt{3}\) and a local minimum at \(x=\sqrt{3}\)

Checkpoint \(\PageIndex{4}\)

Consider the function \(f(x)=x^3−(\frac{3}{2})x^2−18x\). The points \(c=3,\,−2\) satisfy \(f^{\prime}(c)=0\). Use the Second Derivative Test to determine whether \(f\) has a local maximum or local minimum at those points.

Answer

\(f\) has a local maximum at \(−2\) and a local minimum at \(3\).

We have now developed the tools to determine where a function is increasing and decreasing and understand the graph's basic shape. The next section discusses what happens to a function as \(x \to \pm \infty\). At that point, we have enough tools to provide accurate graphs of various functions.