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

(c) \(\lim _{x\to 4^+} f(x)=\begin{prompt}\answer {\infty }\end{prompt}\)

(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

(c) Given \(f(x)=\frac {1}{x}, f^{-1}(x)=\frac {1}{x}\).

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 \(a\).

There are no points There is at least one point
\(a=\answer {3}\)
(b) Both limits \(\lim _{x\to a^+} f(x)\) and \(\lim _{x\to a^-}\) exist, but the limit \(\lim \limits _{x\to a} f(x)\) does not exist.
There are no points There is at least one point
\(a=\answer {0}\)
(c) \(\lim _{x\to a} f(x)=-2\).
There are no points There is at least one point
\(a=\answer {-4}\)
(d) \(\lim _{x\to a^+} f(x)=0\).
There are no points There is at least one point
\(a=\answer {-8}\)
(e) \(\lim _{x\to a^-} f(x)=0\).
There are no points There is at least one point
\(a=\answer {8}\)

(iv) Find the following value or say ‘DNE’

(a)
\(f(0)=\begin{prompt}\answer [format=string]{DNE}\end{prompt}\)
(b)
\(f(3)=\begin{prompt}\answer {0}\end{prompt}\)
(c)
\(f^{-1}(0)=\begin{prompt}\answer {3}\end{prompt}\)
(d)
\(f^{-1}(2)=\begin{prompt}\answer {-8}\end{prompt}\)
(e)
\(f^{-1}(-4)=\begin{prompt}\answer [format=string]{DNE}\end{prompt}\)
(f)
\(f^{-1}(-2)=\begin{prompt}\answer {-4}\end{prompt}\)

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

(a) Find the domain and range of \(f\)

Domain: \(\begin{prompt} \Big [\answer {-4},\answer {0}\Big ) \cup \Big (\answer {0},\answer {7}\Big ] \end{prompt}\)

Range: \(\begin{prompt} \Big [\answer {-4},\answer {-2}\Big ] \cup \Big (\answer {0},\answer {2}\Big ) \cup \Big (\answer {2},\answer {6}\Big ] \end{prompt}\)

(b) Which of the following represents the graph of \(f=\frac {f(x-2)}{2}+4\)

(c) Find the following limits or say that a limit does not exist (DNE).

(i) \(\lim _{x\to 0} f(x)=\begin{prompt}\answer {2}\end{prompt}\)

(ii) \(\lim _{x\to 3^-} f(x)=\begin{prompt}\answer {0}\end{prompt}\)

(iii) \(\lim _{x\to 3^+} f(x)=\begin{prompt}\answer {-2}\end{prompt}\)

(iv) \(\lim _{x\to 3} f(x)=\begin{prompt}\answer {DNE}\end{prompt}\)

(d) List all the intervals of continuity: \(\begin{prompt} \Big [\answer {-4},\answer {0}\Big )\text { and } \Big (\answer {0},\answer {3}\Big )\text { and } \Big [\answer {3},\answer {7}\Big ] \end{prompt}\)

(e) Find the following values or expressions, or say ‘DNE’.

(i) \(f(0)=\begin{prompt}\answer {DNE}\end{prompt}\)

(ii) For \(-4<x<0\), \(f(x)=\begin{prompt}\answer {2-x}\end{prompt}\)

(f) Find the domain of \(f^{-1}(x)\): \(\begin{prompt} \Big [\answer {-4},\answer {-2}\Big ] \cup \Big (\answer {0},\answer {2}\Big )\text { and } \Big (\answer {2},\answer {6}\Big ] \end{prompt}\)

(g) Find the expression for \(f^{-1}(x)\), for \(2<x<6\). \(f^{-1}(x)=\begin{prompt}\answer {2-x}\end{prompt}\)

A function \(f\) is an even function, defined on \([-6,6]\). Part of the graph of \(f\) is shown below

Choose the correct (complete) graph of \(f\).

(a) A function \(f\) is defined on the interval \((0,7)\). \(f(1)=3\), and the following inequality holds:

\[\ln (x)\leq f(x) \leq x-1, x\neq 1, 0<x<7\]

Select the correct limit, and justification:

\(\lim _{x\to 1} f(x)=\)

1, using the Intermediate Value Theorem 0, using the Squeeze Theorem 1, using the Squeeze Theorem Not enough information to determine the limit

(b) A function \(f\) is defined on the interval \((0,2)\), and the following inequality holds:

\[\cos (\frac {\pi }{2}x)\leq f(x) \leq 1+(x-1)^2, x\neq 1, 0<x<2\]

Select the correct limit and justification:

\(\lim _{x\to 1} f(x)=\)

1, using the Intermediate Value Theorem 1, using the Squeeze Theorem 0, using the Mean Value Theorem Not enough information to determine the limit

The function \(f\) is defined by \(f(x)=\frac {x}{\sqrt {x^2-9}}\).

(a) Is the function \(f\) defined on \([-3,3]\):

Yes No

(b) Find the domain of \(f\): \(\begin{prompt}\Big (\answer {-\infty },\answer {-3}\Big ) \cup \Big (\answer {3},\answer {\infty }\Big )\end{prompt}\)

(c) Is the function \(f\) odd, even, or neither:

Odd Even Neither

(d) Find all horizontal asymptotes. \(y=\answer {-1},\answer {1}\)

(e) Find all vertical asymptotes. \(x=\answer {-3},\answer {3}\)

Let

\[ f(x)= \begin{cases} \frac {e^x}{x+1} & \text { if } x\leq 0, x\neq -1\\ \ln (x) & \text { if } x>0 \end{cases} \]

(i) Determine if the following limits exist. If they do, compute them analytically using the limit laws and techniques discussed in class. If they don’t, say ‘DNE’. [You may not use a table of values, a graph, or L’Hospitals rule to justify your answer.]

(a) \(\lim _{x\to -1^+} f(x)=\begin{prompt}\answer {\infty }\end{prompt}\)

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

(c) \(\lim _{x\to \infty } f(x)=\begin{prompt}\answer {\infty }\end{prompt}\)

(d) \(\lim _{x\to -\infty } f(x)=\begin{prompt}\answer {0}\end{prompt}\)

(ii) Find all vertical asymptotes of \(f\) or say ‘none’: \(x=\begin{prompt}\answer {-1},\answer {0}\end{prompt}\)

(iii) Find all horizontal asymptotes of \(f\) or say ‘none’: \(y=\begin{prompt} \answer {0}\end{prompt}\)

(iv) Find the (largest) intervals of continuity of \(f\): \(\begin{prompt}\Big (\answer {-\infty },\answer {-1}\Big ) \text { and} \Big (\answer {-1},\answer {0}\Big )\text { and } \Big (\answer {0},\answer {\infty }\Big )\end{prompt}\)

Let

\[ f(x)= \begin{cases} \frac {e^x}{x-1} & \text { if } x\leq 0\\ \frac {6x+j}{2x+5} & \text { if } x>0 \end{cases} \]

(a) Find the value \(j\) so that the function \(f\) is continuous at \(x=0\): \(j=\begin{prompt}\answer {-5}\end{prompt}\)

(ii) Find all vertical asymptotes of \(f\) or say ‘none’: \(x=\begin{prompt}\answer [format=string]{none}\end{prompt}\)

(iii) Find all horizontal asymptotes of \(f\) or say ‘none’: \(y=\begin{prompt}\answer {0},\answer {3}\end{prompt}\)

(iv) Find the (largest) interval(s) of continuity of \(f\) (assuming \(j\) is equal to the value you found in part (a)): \(\Big (\answer {-\infty },\answer {\infty }\Big )\)