Exam One Review

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Review questions for exam 1.

Find the following values or, if the value is not defined, say ‘DNE’: $\begin {array}{l l l} (a) \sin \left (\frac {17\pi }{6}\right )=\answer {\frac {1}{2}} & (b) \sin ^{-1}\left (\frac {17\pi }{6}\right )=\answer [format=string]{DNE} & (c) \sin ^{-1}\left (\frac {-\sqrt {3}}{2}\right )=\answer {-\frac {\pi }{3}} \\ (d) e^{-2\ln (3)}=\answer {3^{-2}} & (e) \ln \left (-e^2\right )=\answer [format=string]{DNE} & (f) \ln \left (e^{21}\right )=\answer {21} \\ (g) 10^{3\log _{10}(4)}=\answer {4^3} & (h) \ln \left (1\right )=\answer {0} & (i) \tan ^{-1}\left (-\sqrt {3}\right ) =\answer {-\frac {\pi }{3}} \\ (j) \cot \left (\frac {\pi }{3}\right )=\answer {\frac {1}{\sqrt {3}}} & (k) \cot ^{-1}\left (-\sqrt {3}\right )=\answer {-\frac {\pi }{6}} & \end {array}$

Consider the functions: $g(x)=-5|x-1|$ , and $h(x)=(x-1)^2$ .

(i) Find the limit. (Possible answers include $+\infty , -\infty$ or ‘DNE’)

(a) $\lim _{x\to 3} \frac {g(x)-g(3)}{x-3}=\begin {prompt}{\answer {-5}}\end {prompt}$

(b) $\lim _{x\to 1^-} \frac {g(x)}{h(x)}=\begin {prompt} \answer {-\infty }\end {prompt}$

(ii) Let $f$ be a function such that $g(x)\leq f(x)\leq h(x), 0<x<2$ .

(a) Find the limit or say ‘DNE’: $\lim _{x\to 1} f(x)= \begin {prompt}\answer {0}\end {prompt}$

Evaluate the following limits. Note: You may not use a table of values, a graph, or L’Hospital’s Rule to justify your answer. $\begin {array}{l l} (a) \lim _{x\to 3} \frac {\sqrt {2x-2}-2}{x-3}=\begin {prompt}\answer {.5}\end {prompt} & (b) \lim _{x\to -\infty } \frac {x^6-2x^5-2}{x^5-3}=\begin {prompt}\answer {-\infty }\end {prompt} \\ (c) \lim _{x\to 0^+} x\cos (\frac {\pi }{x})=\begin {prompt}\answer {0}\end {prompt} & (d) \lim _{x\to 0} \frac {(x^2+3)^2-9}{x^2}=\begin {prompt}\answer {6}\end {prompt}\\ (e) \lim _{x\to -\infty } \frac {\sqrt {9x^2+1}}{6x-5}=\begin {prompt}\answer {-.5}\end {prompt} & (f) \lim _{x\to 7^-} \frac {|x-7|}{x-7}=\begin {prompt}\answer {-1}\end {prompt}\\ (g) \lim _{x\to \infty } \frac {6e^x+e^{-x}}{3e^x+4e^{-x}}=\begin {prompt}\answer {2}\end {prompt}& (h) \lim _{x\to 4^-} \frac {\sin (\frac {\pi }{3}x)}{\ln (5-x)}=\begin {prompt}\answer {-\infty }\end {prompt}\\ (i) \lim _{x\to 0} \frac {\ln (\cos (\sqrt {3x^4+x^2}))}{\sin (x)+\cos (x)}=\begin {prompt}\answer {0}\end {prompt} & (j) \lim _{x\to 0^-} \frac {e^{-x}}{x^2+4x}=\begin {prompt}\answer {-\infty }\end {prompt} \end {array}$

Let $f(x) = \begin {cases} \frac {x^2-x-12}{x+3} &\text {if $x<4$ and $x\ne -3$}\\ 5 &\text {if $x=-3$}\\ \frac {x}{x-4} &\text {if $x>4$} \end {cases}$ (i) Determine if the following limits exist. If they do not, say ‘DNE’. Note: You may not use a table of values, a graph, or L’Hospital’s Rule to justify your answer.

(a) $\lim _{x\to -3} f(x)=\begin {prompt}\answer {-7}\end {prompt}$

(b) $\lim _{x\to 4^-} f(x)=\begin {prompt}\answer {0}\end {prompt}$

(d) $\lim _{x\to 4} f(x)=\begin {prompt}\answer [format=string]{DNE}\end {prompt}$

(ii) Find all vertical asymptotes of f: $x=\begin {prompt}\answer {4}\end {prompt}$

(iii) Find all horizontal asymptotes of f: $y=\begin {prompt}\answer {1}\end {prompt}$

(iv) List the (largest) intervals of continuity of f: $\begin {prompt}\left (\answer {-\infty },\answer {-3}\right ),\left (\answer {-3},\answer {4}\right ),\left (\answer {4},\answer {\infty }\right )\end {prompt}$

Fill in the blanks to explain how the Intermediate Value Theorem can be used to show that the equation $x^3+4x+2=0$ has a solution on the interval $[-1,0]$ .

Let $f(x)=x^3+4x+2$ . Since $f$ is a

monotonecontinuous function for all $x$ in the interval $[-1,0]$ $(-1,0)$ , and 01 is between -1f(-1)=-3 and 0f(0)=2 , then the IVT guarantees the existence of at least one number uf(u) in $[-1, 0]$ such that f(u) = 0f(0) = u .

(i) Evaluate the following expressions:

(a) $\sin ^{-1}(\sin (\frac {4\pi }{9}))=\begin {prompt}\answer {\frac {4\pi }{9}}\end {prompt}$

(b) $\sin ^{-1}(\sin (\frac {4\pi }{5}))=\begin {prompt}\answer {\frac {\pi }{5}}\end {prompt}$

(ii) Determine if the following statements are true or false:

(a) Given a one-to-one function $f$ and its inverse $f^{-1}$ , $f^{-1}(f(x))=x$ , where x is in the domain of $f$ .

True False

(b) $\sin ^{-1}(\sin (\frac {2\pi }{3}))=\frac {2\pi }{3}$ .

True False

(iii) Find the inverse of $f(x)=\sqrt [3]{x-2}+4$ . $f^{-1}(x)=\begin {prompt}\answer {x^3-12x^2+48x-62}\end {prompt}$

(iv) Use a right triangle to simplify $\tan (\cos ^{-1}(x))$ . $\begin {prompt}\answer {\frac {\sqrt {1-x^2}}{x}}\end {prompt}$

A table of values for $f(x)$ , along with a graph of a function $g(x)$ is shown below. $\begin {array}{c|c} x & f(x)\\ \hline 1 & 2 \\ \hline 2 & 3 \\ \hline 3 & 4 \\ \hline \end {array}$

(a)

Find expressions for $g(x)$ on the following intervals:

(i): For $0<x<1$ , $g(x)=\begin {prompt}\answer {3x}\end {prompt}$
(ii): For $1<x<4$ , $g(x)=\begin {prompt}\answer {4-x}\end {prompt}$

(b)

Find the values or write DNE. $\begin {array}{ l l l} \text { at } x=1, f(x)g(x)= \begin {prompt}\answer {6}\end {prompt} & \text { at } x=3, f(g(x))= \begin {prompt}\answer {2}\end {prompt} \\ \text { at } x=2, g(f(x))= \begin {prompt}\answer {1}\end {prompt} & \text { at } x=1.5, g(x)= \begin {prompt}\answer {2.5}\end {prompt} \\ \lim _{x\to 1} g(x)=\begin {prompt}\answer {3}\end {prompt} & \lim _{x\to 1^-} \frac {g(x)-g(1)}{x-1}=\begin {prompt}\answer {3}\end {prompt}\\ \lim _{x\to 1^+} \frac {g(x)-g(1)}{x-1}=\begin {prompt}\answer {-1}\end {prompt} & \lim _{x\to 1} \frac {g(x)-g(1)}{x-1}=\begin {prompt}\answer [format=string]{DNE}\end {prompt} \end {array}$

The (entire) graph of a function $f$ is given in the figure below.

(i) Find the domain and range of $f$ . Write your answer in interval notation.

Domain of $f$ : $\Big [\answer {-8},\answer {0}\Big )\cup \Big (\answer {0},\answer {8}\Big )$

Range of $f$ : $\Big (\answer {-4},\answer {6}\Big )$

(ii) List the largest intervals of continuity for $f$ : $\Big (\answer {-8},\answer {0}\Big )$ and $\Big (\answer {0},\answer {3}\Big )$ and $\Big (\answer {3},\answer {8}\Big )$

(iii) Determine if there are any points $a$ in the interval $[-8,8]$ for which the following statements are true. If there are any such points, find all of them.

(a) $\lim _{x\to a} f(x)$ exists, but the function $f$ is NOT continuous at $a$ .