1.3: Continuity
- Page ID
- 204092
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1.3 Continuity
- Apply the three conditions for continuity to determine whether a function is continuous or discontinuous at a given point.
- Identify and describe the three types of continuity in functions.
- Use the Intermediate Value Theorem to demonstrate the existence of a solution within a given interval for continuous functions.
A function \( f \) is continuous at \( a \) if the following conditions are satisfied:
1. \( f(a) \) is defined
2. \( \displaystyle{\lim_{x \to a} f(x)} \) exists
3. \( \displaystyle{\lim_{x \to a} f(x) = f(a)} \)
When can we say that \( f\) is not continuous (or that it is discontinuous) at \(a\)?
Determine whether the following functions are continuous at the given value:
- \( f(x)=x^2+5\), \(a=1\)
- \( f(x) = \dfrac{x^2 - 9}{x - 3} \), \( a = 3 \)
-
\(
f(x) =
\begin{cases}
\frac{x^2 - 9}{x - 3} & \text{if } x < 4 \\
5 & \text{if } x = 4 \\
x + 2 & \text{if } x > 4
\end{cases}
\qquad a = 4
\) -
\(
f(x) =
\begin{cases}
\dfrac{x^2 + 2x}{x}, & \text{if } x \neq 0 \\
2, & \text{if } x = 0
\end{cases}
\qquad a = 0
\) -
\(
f(x) =
\begin{cases}
\dfrac{1 - \cos x}{x}, & \text{if } x \neq 0 \\
0, & \text{if } x = 0
\end{cases}
\qquad a = 0
\) - \( f(x) = \dfrac{1}{(x - 2)^2}, \qquad a = 2\)
Types of Discontinuities
Figure \(\PageIndex{8}\): Types of discontinuities (a) removable, (b) jump, or (c) infinite
Note that if \( f \) has a discontinuity at \( a \), then:
1. The discontinuity at \( a \) is removable if \( \displaystyle{\lim_{x \to a} f(x)} \) exists.
2. The discontinuity at \( a \) is a jump discontinuity if \( \displaystyle{\lim_{x \to a^-} f(x)} \) and \( \displaystyle{\lim_{x \to a^+} f(x)} \) both exist, but they are not equal.
3. The discontinuity at \( a \) is an infinite discontinuity if \( \displaystyle{\lim_{x \to a^-} f(x) = \pm \infty} \) and/or \( \displaystyle{\lim_{x \to a^+} f(x) = \pm \infty} \).
The graph of the function \( f(x)=f(x)=\frac{x^2-1}{x-1} \) is given below. Determine whether the function is continuous at \( x = 1 \). If it is not continuous, what type of discontinuity occurs at that point?
Figure \(\PageIndex{9}\): Graph of \(f(x) \) (from Desmos)
The graph of the function \( f(x)=\begin{cases}
-x^2+2x+1, & x< 2 \\
x+1, & x\geq 2
\end{cases}\) is given below. Determine whether the function is continuous at \( x = 2 \). If it is not continuous, what type of discontinuity occurs at that point?
Figure \(\PageIndex{10}\): Graph of \( f(x) \)(from Desmos)
The graph of the function \( f(x)\) is given below. Determine whether the function is continuous at \( x = 1 \). If it is not continuous, what type of discontinuity occurs at that point?
Figure \(\PageIndex{11}\): Graph of \( f(x) \)(from Desmos)
The graph of the function \( f(x)\) is given below. Determine whether the function is continuous at \( x = -2 \). If it is not continuous, what type of discontinuity occurs at that point?
Figure \(\PageIndex{12}\): Graph of \( f(x) \)(from Desmos)
Classify the discontinuities (if any) of the following functions:
- \( f(x)=x^2+5\), \(a=1\)
- \( f(x) = \dfrac{x^2 - 9}{x - 3} \), \( a = 3 \)
-
\(
f(x) =
\begin{cases}
\frac{x^2 - 9}{x - 3} & \text{if } x < 4 \\
5 & \text{if } x = 4 \\
x + 2 & \text{if } x > 4
\end{cases}
\qquad a = 4
\) -
\(
f(x) =
\begin{cases}
\dfrac{x^2 + 2x}{x}, & \text{if } x \neq 0 \\
2, & \text{if } x = 0
\end{cases}
\qquad a = 0
\) -
\(
f(x) =
\begin{cases}
\dfrac{1 - \cos x}{x}, & \text{if } x \neq 0 \\
0, & \text{if } x = 0
\end{cases}
\qquad a = 0
\) - \( f(x) = \dfrac{1}{(x - 2)^2}, \qquad a = 2\)
A function \( f \) is continuous on an interval if it is continuous at every number in the interval. Note that at an endpoint, you must check one-sided continuity — left-hand continuity at a right endpoint and right-hand continuity at a left endpoint.
Polynomial, rational, root, and trigonometric functions are continuous at every number in their domains.
State the interval(s) over which the following functions are continuous:
- \(y=3x^3+6x^2+2\)
- \( y= \dfrac{2x+7}{x^2-3x+2}\)
- \(y=\sqrt{x+3}\)
- \( y= \tan x\)
Suppose that \( f \) is continuous on the interval \( [a, b] \), and let \( z \) be any number between \( f(a) \) and \( f(b) \), where \( f(a) \neq f(b) \). Then there exists a number \( c \) in \( (a, b) \) such that \( f(c) = z \).
Use the Intermediate Value Theorem to show that:
- \( f(x) = \ln(x) - x + 2 \) has at least one zero.
- \( f(x) = x^3 - 4x + 1 \) has a zero over the interval \( [1, 2] \).