Practice

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Try these problems.

Consider $a(x) = \frac {x^2+x-6}{x^2-x-12}.$ Find all vertical asymptotes.

There are vertical asymptotes There are no vertical asymptotes

$x=\answer {4}$

Consider $s(x) = \frac {x^2-2 x-15}{x^2+3 x+2}.$ Find all vertical asymptotes.

There are vertical asymptotes There are no vertical asymptotes

Write your answers from least to greatest: $x=\answer {-2}\qquad \text {and}\qquad x=\answer {-1}$

Give intervals on which each of the following functions are continuous. Write combinations of intervals going from left to right on the number line.

(a): $\frac {1}{e^x+1}$ is continuous on $(\answer {-\infty },\answer {\infty })$ .
(b): $\frac {1}{x^2-1}$ is continuous on $(\answer {-\infty },\answer {-1})$ and $(\answer {-1},\answer {1})$ and $(\answer {1},\answer {\infty })$ .
(c): $\sqrt {5-x}$ is continuous on $(\answer {-\infty },\answer {5})$ .
(d): $\sqrt {5-x^2}$ is continuous on $(\answer {-\sqrt {5}},\answer {\sqrt {5}})$ .

Let $f(x) = \begin {cases} x^2-1, &x < 3 \\ x+5, &x\geq 3. \end {cases}$ Is $f$ continuous everywhere?

Yes No

Let $g(x)=\begin {cases} -2e^{x} & \text {if $x<0$}\\ \frac {x+6}{x-3} & \text {if $x\ge 0$}. \end {cases}$

Find

(a): $\lim _{x\to 0^{-}} g(x)\begin {prompt} = \answer {-2}\end {prompt}$
(b): $\lim _{x\to 0^+} g(x)\begin {prompt} = \answer {-2}\end {prompt}$
(c): $\lim _{x\to 3^{-}} g(x)\begin {prompt} = \answer {-\infty }\end {prompt}$
(d): $\lim _{x\to 3^{+}} g(x)\begin {prompt} = \answer {\infty }\end {prompt}$
(e): $\lim _{x\to -\infty } g(x)\begin {prompt} = \answer {0}\end {prompt}$
(f): $\lim _{x\to +\infty } g(x)\begin {prompt} = \answer {1}\end {prompt}$

The Intermediate Value Theorem states: If $f$ is a continuous function for all $x$ in the closed interval $[a,b]$ and $r$ is between af(a) and bf(b) , then there is a number uf(u) in $[a, b]$ such that f(u) = rf(r) = u .

Consider the following graph:

Is the function $f(x)=\left \{\begin {array}{ccc} \frac {x^2-64}{x^2-11 x+24}, & & x\ne 8\\ 5, & & x=8 \end {array}\right .$ continuous at $x=0$ or $x=8$ ?

$f$ is continuous at both $x=0$ and $x=8$ . $f$ is continuous at $x=0$ but not at $x=8$ . $f$ is continuous at at $x=8$ but not at $x=0$ . $f$ is not continuous at $x=0$ and $x=8$ .

Let $f$ be continuous on $\left [1,5\right ]$ where $f(1)=-2$ and $f(5)=-10$ . Does a value $1<c<5$ exist such that $f(c)=-9$ ?

There does not exist a value. Yes, by the Intermediate Value Theorem Yes, by the Mean Value Theorem There does not necessarily exist such a value