OBJECTIVES

When you finish this section, you should be able to:

1. Determine whether a function is continuous at a number (p. 93)
2. Determine intervals on which a function is continuous (p. 96)
3. Use properties of continuity (p. 98)
4. Use the Intermediate Value Theorem (p. 100)

spanDEFINITIONspan Continuity at a Number

A function \(f\) is continuous at a number \(c\) if the following three conditions are met:

• \(f(c)\) is defined (that is, \(c\) is in the domain of \(f\))
• \(\lim\limits_{x\rightarrow c}f(x)\) exists
• \(\lim\limits_{x\rightarrow c}f(x)=f(c)\)

Determining Whether a Function Is Continuous at a Number

1. Determine whether \(f(x)=3x^{2}-5x+4\) is continuous at \(1\).
2. Determine whether \(g(x)=\dfrac{x^{2}+9}{x^{2}-4}\) is continuous at \(2\).

Solution (a) We begin by checking the conditions for continuity. First, \(1\) is in the domain of \(f\) and \(f( 1) =2.\) Second, \(\lim\limits_{x\rightarrow 1}f(x)=\lim\limits_{x\rightarrow 1}(3x^{2}-5x+4)=2,\) so \(\lim\limits_{x\rightarrow 1}f( x) \) exists. Third, \(\lim\limits_{x\rightarrow 1}f( x) =f( 1)\). Since the three conditions are met, \(f\) is continuous at \(1\).

(b) Since \(2\) is not in the domain of \(g\), the function \(g\) is discontinuous at \(2\).

NOW WORK

Problem 19.

Determining Whether a Function Is Continuous at a Number

Determine if \(f(x)=\dfrac{\sqrt{x^{2}+2}}{x^{2}-4}\) is continuous at the numbers \(-2\), \(0\), and \(2\).

Figure 23 \(f\) is continuous at 0; \(f\) is discontinuous at \(-2\) and \(2\).

Solution The domain of \(f\) is \(\{x|x\neq -2\), \(x\neq 2\}\). Since \(f\) is not defined at \(-2\) and \(2\), the function \(f\) is not continuous at \(-2\) and at \(2\). The number \(0\) is in the domain of \(f\). That is, \(f\) is defined at \(0\), and \( f(0)=-\dfrac{\sqrt{2}}{4}\). Also, \begin{eqnarray*} \lim_{x\rightarrow 0}f(x)&=&\lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+2}}{ x^{2}-4}=\frac{\lim\limits_{x\rightarrow 0}\sqrt{x^{2}+2}}{ \lim\limits_{x\rightarrow 0}(x^{2}-4)}=\frac{\sqrt{\lim\limits_{x\rightarrow 0}(x^{2}+2)}}{\lim\limits_{x\rightarrow 0}x^{2}-\lim\limits_{x\rightarrow 0}4 }\\[5pt] &=&\frac{\sqrt{0+2}}{0-4}=-\frac{\sqrt{2}}{4}=f(0) \end{eqnarray*}

The three conditions of continuity at a number are met. So, the function \(f\) is continuous at \(0\).

Figure 23 shows the graph of \(f\).

NOW WORK

Problem 21.

Determining Whether a Piecewise-Defined Function Is Continuous

Determine if the function \begin{equation*} f(x)=\left\{ \begin{array}{c@{\qquad}l} \dfrac{x^{2}-9}{x-3} & \hbox{if }x\lt 3 \\ 9 & \hbox{if }x=3 \\[5pt] x^{2}-3 & \hbox{if }x>3 \end{array} \right. \end{equation*}

is continuous at \(3\).

Figure 24 \(f\) is discontinuous at 3.

Solution Since \(f(3)=9\), the function \(f\) is defined at \(3\). To check the second condition, we investigate the one-sided limits. \[ \begin{eqnarray*} \lim_{x\rightarrow 3^{-}}f(x) &=&\lim_{x\rightarrow 3^{-}}\frac{x^{2}-9}{x-3 }=\lim_{x\rightarrow 3^{-}}\frac{( x-3) ( x+3) }{x-3} \underset{\underset{\color{#0066A7}{\hbox{Divide out \(x-3\)}}}{\color{#0066A7}{\uparrow}}} {=}\lim_{x\rightarrow 3^{-}}(x+3)=6 \\ \lim_{x\rightarrow 3^{+}}f(x) &=&\lim_{x\rightarrow 3^{+}} (x^{2}-3) =9-3=6 \end{eqnarray*} \]

Since \(\lim\limits_{x\rightarrow 3^{-}}f(x)=\lim\limits_{x\rightarrow 3^{+}}f(x)\), then \(\lim\limits_{x\rightarrow 3}f(x)\) exists. But, \(\lim\limits_{x\rightarrow 3}f(x)=6\) and \(f(3)=9\), so the third condition of continuity is not satisfied. The function \(f\) is discontinuous at \(3\).

Figure 24 shows the graph of \(f\).

NOW WORK

Problem 25.

spanDEFINITIONspan Removable Discontinuity

Let \(f\) be a function that is defined everywhere in an open interval containing \(c\), except possibly at \(c\). The number \(c\) is called a removable discontinuity of \(f\) if the function is discontinuous at \(c\) but \(\lim\limits_{x\rightarrow c}f( x) \) exists. The discontinuity is removed by defining (or redefining) the value of \(f\) at \(c\) to be \(\lim\limits_{x\rightarrow c}f( x)\).

NOW WORK

Problems 13 and 35.

NEED TO REVIEW?

The floor function is discussed in Section P.2, p. 17.

Determining Whether a Function Is Continuous at a Number

Determine if the floor function \(f(x)=\lfloor x\rfloor\) is continuous at \(1\).

Solution The floor function \(f( x) =\lfloor x\rfloor =\) the greatest integer \(\leq x\). The floor function \(f\) is defined at \(1\) and \(f(1)=1\). But \begin{equation*} \lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{-}}\lfloor x\rfloor =0\hbox{ and }\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1^{+}}\lfloor x\rfloor =1 \end{equation*}

Figure 25 \(f(x)=\lfloor x\rfloor\)

So, \(\lim\limits_{x\rightarrow 1}\lfloor x\rfloor\) does not exist. Since \(\lim\limits_{x\rightarrow 1}\lfloor x\rfloor\) does not exist, \(f\) is discontinuous at \(1\).

NOW WORK

Problem 53.

spanDEFINITIONspan One-Sided Continuity at a Number

Let \(f\) be a function defined on the interval \((a,c]\). Then \(f\) is continuous from the left at the number \(c\) if \begin{equation*} \lim_{^{{}}x\rightarrow c^{-}}f(x)=f(c) \end{equation*}

Let \(f\) be a function defined on the interval \([c,b)\). Then \(f\) is continuous from the right at the number \(c\) if \begin{equation*} \lim_{^{{}}x\rightarrow c^{+}}f(x)=f(c) \end{equation*}

spanDEFINITIONspan Continuity on an Interval

• A function \(f\) is continuous on an open interval \((a,b)\) if \(f\) is continuous at every number in \((a,b)\).
• A function \(f\) is continuous on an interval \([a,b)\) if \(f\) is continuous on the open interval \((a,b)\) and continuous from the right at the number \(a\).
• A function \(f\) is continuous on an interval \(( a,b]\) if \(f\) is continuous on the open interval \((a,b)\) and continuous from the left at the number \(b\).
• A function \(f\) is continuous on a closed interval \([a,b]\) if \(f\) is continuous on the open interval \((a,b)\), continuous from the right at \(a\), and continuous from the left at \(b\).

Determining Whether a Function Is Continuous on a Closed Interval

Is the function \(f(x)=\sqrt{4-x^{2}}\) continuous on the closed interval \([-2,2]\)?

Solution The domain of \(f\) is \(\{x|-2\leq x\leq 2\}\). So, \(f\) is defined for every number in the closed interval \([-2,2]\).

For any number \(c\) in the open interval \((-2,2)\), \begin{equation*} \lim_{x\rightarrow c}f(x)=\lim_{x\rightarrow c}\sqrt{4-x^{2}}=\sqrt{ \lim\limits_{x\rightarrow c}(4-x^{2})}=\sqrt{4-c^{2}}=f(c) \end{equation*}

So, \(f\) is continuous on the open interval \((-2,2)\).

To determine whether \(f\) is continuous on \([-2,2]\), we investigate the limit from the right at \(-2\) and the limit from the left at \(2\). Then, \begin{equation*} \lim_{x\rightarrow -2^{+}}f(x)=\lim_{x\rightarrow -2^{+}}\sqrt{4-x^{2}} =0=f(-2) \end{equation*}

So, \(f\) is continuous from the right at \(-2\). Similarly, \begin{equation*} \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{-}}\sqrt{4-x^{2}} =0=f(2) \end{equation*}

So, \(f\) is continuous from the left at \(2\). We conclude that \(f\) is continuous on the closed interval \([-2,2]\).

NOW WORK

Problem 37.

spanDEFINITIONspan Continuity on a Domain

A function \(f\) is continuous on its domain if it is continuous at every number \(c\) in its domain.

Determining Whether \({f(x)=\sqrt{x^2(x-1)}}\) Is Continuous on Its Domain

Determine if the function \(f(x)=\sqrt{x^{2}(x-1)}\) is continuous on its domain.

Solution The domain of \(f(x)=\sqrt{x^{2}(x-1)}\) is \(\{x|x=0\}\cup \{ x|x\geq 1\}\). We need to determine whether \(f\) is continuous at the number \(0\) and whether \(f\) is continuous on the interval \([ 1,\infty)\).

At the number \(0\), there is an open interval containing \(0\) that contains no other number in the domain of \(f\). [For example, use the interval \(\left( -\dfrac{1}{2},\dfrac{1}{2}\right)\!.]\) This means \(\lim\limits_{x\rightarrow 0}f(x)\) does not exist. So, \(f\) is discontinuous at \(0\).

For all numbers \(c\) in the open interval \((1,\infty )\) we have \begin{equation*} f(c)=\sqrt{c^{2}(c-1)} \end{equation*}

and \begin{equation*} \lim_{x\rightarrow c}\sqrt{x^{2}(x-1)}=\sqrt{\lim\limits_{x\rightarrow c} [x^{2}(x-1)] }=\sqrt{c^{2}(c-1)}=f(c) \end{equation*}

So, \(f\) is continuous on the open interval \((1,\infty )\).

Now, at the number \(1\), \begin{equation*} f(1)=0\hbox{ and }\lim_{x\rightarrow 1^{+}}\sqrt{x^{2}(x-1)}=0 \end{equation*}

So, \(f\) is continuous from the right at \(1\).

The function \(f(x)=\sqrt{x^{2}(x-1)}\) is continuous on the interval \([1,\infty )\), but it is discontinuous at \(0\). So, \(f\) is not continuous on its domain.

Figure 27 \(f\) is discontinuous at \(0\); \(f\) is continuous on \([1, \infty)\).

THEOREM

• A polynomial function is continuous for all real numbers.
• A rational function is continuous on its domain.

Proof

If \(P\) is a polynomial function, its domain is the set of real numbers. For a polynomial function, \begin{equation*} \lim_{x\rightarrow c}P(x)=P(c) \end{equation*}

for any number \(c\). That is, a polynomial function is continuous at every real number.

If \(R(x)=\dfrac{p(x)}{q(x)}\) is a rational function, then \(p(x)\) and \(q(x)\) are polynomials and the domain of \(R\) is \(\{x|q(x)\neq 0\}\). The Limit of a Rational Function (p. 87) states that for all \(c\) in the domain of a rational function, \[ \begin{equation*} \lim_{x\rightarrow c}R(x)=R( c) \end{equation*} \]

So a rational function is continuous at every number in its domain.

• If a function is continuous on an interval, its graph has no holes or gaps on that interval.
• If a function is continuous on its domain, it will be continuous at every number in its domain; its graph may have holes or gaps at numbers that are not in the domain.

THEOREM Continuity of a Sum, Difference, Product, and Quotient

If the functions \(f\) and \(g\) are continuous at a number \(c\), and if \(k\) is a real number, then the functions \(f+g,\) \(f-g,\) \(f\cdot g\), and \(kf\) are also continuous at \(c\). If \(g( c) \neq 0,\) the function \(\dfrac{f}{g}\) is continuous at \(c\).

Identifying Where Functions Are Continuous

Determine where each function is continuous:

1. \(F(x) = x^{2}+5\) \(-\dfrac{x}{x^{2}+4}\)
2. \(G( x) =x^{3}+2x+\dfrac{x^{2}}{x^{2}-1}\)

Solution First we determine the domain of each function.

1. \(F\) is the difference of the two functions \(f( x) =x^{2}+5\) and \(g( x) =\dfrac{x}{x^{2}+4}\), each of whose domain is the set of all real numbers. So, the domain of \(F\) is the set of all real numbers. Since \(f\) and \(g\) are continuous on their domains, the difference function \(F\) is continuous on its domain.
2. \(G\) is the sum of the two functions \(f( x) =x^{3}+2x\), whose domain is the set of all real numbers, and \(g( x) =\dfrac{x^{2}}{x^{2}-1}\), whose domain is \(\left\{ x|x\neq -1,\hbox{ }x\neq 1\right\}\). Since \(f\) and \(g\) are continuous on their domains, \(G\) is continuous on its domain, \(\left\{ x|x\neq -1,\hbox{ }x\neq 1\right\}\).

NOW WORK

Problem 45.

NEED TO REVIEW?

Composite functions are discussed in Section P.3, pp. 25-27.

THEOREM Continuity of a Composite Function

If a function \(g\) is continuous at \(c\) and a function \(f\) is continuous at \(g( c)\), then the composite function \((f\circ g)(x)=f(g(x))\) is continuous at \(c\).

Identifying Where Functions Are Continuous

Determine where each function is continuous:

1. \(F(x) =\sqrt{x^{2}+4}\)
2. \(G(x) =\sqrt{x^{2}-1}\)
3. \(H(x) =\dfrac{x^{2}-1}{x^{2}-4}+\sqrt{x-1}\)

Solution (a) \(F=f\circ g\) is the composite of \(f( x) =\sqrt{x}\) and \(g( x) =x^{2}+4\). \(f\) is continuous for \(x\geq 0\) and \(g\) is continuous for all real numbers. The domain of \(F\) is all real numbers and \(F=( f\circ g) ( x) =\sqrt{x^{2}+4}\) is continuous for all real numbers. That is, \(F\) is continuous on its domain.

(b) \(G\) is the composite of \(f( x) =\sqrt{x}\) and \( g( x) =x^{2}-1\). \(f\) is continuous for \(x\geq 0\) and \(g\) is continuous for all real numbers. The domain of \(G\) is \(\{ x|x\geq 1\} \cup \{ x|x\leq -1\} \) and \(G=( f\circ g) ( x) =\sqrt{x^{2}-1}\) is continuous on its domain.

(c) \(H\) is the sum of \(f( x) =\) \(\dfrac{x^{2}-1}{ x^{2}-4}\) and the function \(g( x) =\sqrt{x-1}\). The domain of \(f\) is \(\{ x|x\neq -2, x\neq 2\} \); \(f\) is continuous on its domain. The domain of \(g\) is \(x\geq 1\). The domain of \(H\) is \(\{ x|1\leq x\lt 2\} \cup \{ x|x>2\}\); \(H\) is continuous on its domain.

NOW WORK

Problem 47.

NEED TO REVIEW?

Inverse functions are discussed in Section P.4, pp. 32-37.

THEOREM Continuity of an Inverse Function

If \(f\) is a one-to-one function that is continuous on its domain, then its inverse function \(f^{-1}\) is also continuous on its domain.

THEOREM The Intermediate Value Theorem

Let \(f\) be a function that is continuous on a closed interval \([a,b]\) and \(f(a)\neq f(b)\). If \(N\) is any number between \(f(a)\) and \(f(b)\), then there is at least one number \(c\) in the open interval \((a,b)\) for which \(f(c)=N\).

Using the Intermediate Value Theorem

Figure 33 \(f(x) = x^3 + x^2 - x - 2\)

Use the Intermediate Value Theorem to show that \begin{equation*} f(x)=x^{3}+x^{2}-x-2 \end{equation*}

has a zero between \(1\) and \(2\).

Solution Since \(f\) is a polynomial, it is continuous on the closed interval \([1,2]\). Because \(f(1)=-1\) and \(f(2)=8\) have opposite signs, the Intermediate Value Theorem states that \(f(c)=0\) for at least one number \(c\) in the interval \((1,2)\). That is, \(f\) has at least one zero between \(1\) and \(2.\) Figure 33 shows the graph of \(f\) on a graphing utility.

NOW WORK

Problem 59.

Using the Intermediate Value Theorem to Approximate a Real Zero of a Function

Figure 34

The function \(f(x)=x^{3}+x^{2}-x-2\) has a zero in the interval \((1,2)\). Use the Intermediate Value Theorem to approximate the zero correct to three decimal places.

Solution Using the TABLE feature on a graphing utility, we subdivide the interval \([1,2]\) into \(10\) subintervals, each of length \(0.1\). Then we find the subinterval whose endpoints have opposite signs, or the endpoint whose value equals \(0\) (in which case, the exact zero is found). From Figure 34, since \(f ( 1.2) =-0.032\) and \(f (1.3) =0.587\), by the Intermediate Value Theorem, a zero lies in the interval \(( 1.2,1.3)\). Correct to one decimal place, the zero is \(1.2\).

Repeat the process by subdividing the interval \([ 1.2,1.3]\) into \(10\) subintervals, each of length \(0.01.\) See Figure 35. We conclude that the zero is in the interval \((1.20, 1.21)\), so correct to two decimal places, the zero is \(1.20\).

Figure 35

Figure 36

Now subdivide the interval \(\left[ 1.20,1.21\right]\) into \(10\) subintervals, each of length \(0.001.\) See Figure 36.

We conclude that the zero of the function \(f\) is \(1.205,\) correct to three decimal places.

NOW WORK

Problem 65.