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\problemgrouptitle{Inverse and Other Functions Review}

\item For the function $\displaystyle f(x)=3+\frac{1}{x^3}$\,:
\begin{enumerate}
\item Show the function is one-to-one.
\item Find the inverse of the function, $g(x)=f^{-1}(x)$\,.
\item Calculate the derivative $g'(11)$\,.
\end{enumerate}
\answer{
\begin{enumerate}
\item $\displaystyle f(x_1)=f(x_2)\implies 3+\frac{1}{x_1^3}=3+\frac{1}{x_2^3}\implies\frac{1}{x_1^3}=\frac{1}{x_2^3}\implies x_1^3=x_2^3\implies x_1=x_2$
\item $\displaystyle g(x)=f^{-1}(x)=\sqrt[3]{\frac{1}{x-3}}$\\
\item $\displaystyle g'(11)=\frac{1}{f'(g(11))}=\frac{1}{f'(1/2)}=\frac{1}{-3(1/2)^{-4}}=-\frac{1}{48}$
\end{enumerate}
}

\begin{problemblock}[2]{Find the indicated derivative.}

\item $\displaystyle f(x)=\frac {2\ln x+5}{e^{3x}+4}, f'(x)$
\answer{
$\displaystyle f'(x)=\frac{2 \left(e^{3x}+4\right)-3x e^{3x}(2\ln x+5)}{x \left( e^{3x}+4 \right)^2}$
}

\item $\displaystyle f(t)=\ln \left( t^4+e^{2t}+1 \right), f'(t)$
\answer{
$\displaystyle f'(t)=\frac{4t^3+2e^{2t}}{t^4+e^{2t}+1}$
}

\item $\displaystyle g(x)=\ln \sqrt [3]{\frac{x+1}{2x+4}}, g'(x)$
\answer{
$\displaystyle g'(x)=\frac 13 \left[\frac 1{x+1}-\frac 2{2x+4}\right]$
}

\item $\displaystyle F(y)=\sqrt{e^{4y}+e^{-4y}}+2e^y, F'(0)$
\answer{
$\displaystyle F'(0)=2$
}

\item $\displaystyle f(x)=x^2 e^{-x^2},  f''(x)$
\answer{
$\displaystyle f''(x)=2e^{-x^2}-10x^2 e^{-x^2}+4x^4 e^{-x^2}$
}

\item $\displaystyle y=\left( 2x+3 \right)^{4x}, y'$
\answer{
$\displaystyle y'=\left(2x+3\right)^{4x}\left[ 4\ln (2x+3)+\frac{8x}{2x+3}\right]$
}

\item $\displaystyle f(t)=10^{e^t}, f'(t)$
\answer{
$\displaystyle f'(t)=(\ln 10) (e^t) 10^{e^t}$
}

\item $\displaystyle y=(\ln x)^{\cos x}, y'$
\answer{
$\displaystyle y'=(\ln x)^{\cos x}\left[-\sin x \ln (\ln x)+\frac {\cos x}{x \ln x}\right]$
}

\item $\displaystyle \ln x+\ln y=e^{xy}, y'$
\answer{
$\displaystyle y'=\frac{xy^2e^{xy}-y}{x-x^2ye^{xy}}$
}

\item $\displaystyle f(x)=\tan^{-1} \left(e^x+\ln x \right), f'(x)$
\answer{
$\displaystyle f'(x)=\frac {x e^x+1}{x+x \left(e^x+\ln x \right)^2}$
}

\item $\displaystyle g(t)=\sin^{-1} (\tan t+3), g'(t)$
\answer{
$\displaystyle g'(t)=\frac {\sec^2\! t}{\sqrt{1-(\tan t+3)^2}}$
}

\item $\displaystyle h(x)=\log \left( \cos^{-1} x+1 \right), h'(x)$
\answer{
$\displaystyle h'(x)=-\frac 1{\ln 10} \frac 1{\left( \cos^{-1}x+1 \right) \sqrt{1-x^2}}$
}

\end{problemblock}

\begin{problemblock}[2]{Evaluate the given integral.}

\item $\displaystyle \int \frac {e^{4 \sqrt x}}{\sqrt x} \,dx$
\answer{
$\displaystyle \frac 12 e^{4\sqrt x}+C$
}

\item $\displaystyle \int \frac {4x^2}{2x+1} \,dx$
\answer{
$\displaystyle \frac 14  (2x+1)^2-(2x+1)+\frac 12 \ln |2x+1|+C$
}

\item $\displaystyle \int 4^x  e^{2x} \,dx$
\answer{
$\displaystyle \frac {4^x  e^{2x}}{\ln \left( 4 e^2\right)}+C$
}

\item $\displaystyle \int 4^x  \left( 4+\cos 4^x \right) \,dx$
\answer{
$\displaystyle \frac 1{\ln 4} \left[ 4\left(4^x\right)+\sin 4^x\right]+C$
}

\item $\displaystyle \int_0^1 \frac {e^{2x}}{1+e^{4x}} \,dx$
\answer{
$\displaystyle \frac 12 \left[\tan^{-1}\left(e^2\right)-\frac{\pi}4 \right]$
}

\item $\displaystyle \int \frac{\cos x}{\sqrt{3-\sin^2 x}}\,dx$
\answer{
$\displaystyle \sin^{-1} \left(\frac{\sin x}{\sqrt 3}\right)+C$
}

\item $\displaystyle \int \frac {\tan (\ln x)}{x} \,dx$
\answer{
$\displaystyle \ln |\sec (\ln x)|+C$
}

\item $\displaystyle \int \frac {x+2}{\sqrt{16-x^2}} \,dx$
\answer{
$\displaystyle -\sqrt{16-x^2}+2 \sin^{-1}\left(\frac x4\right)+C$
}

\item $\displaystyle \int \frac 2{3x\sqrt{x^4-1}} \,dx$   
\answer{
$\displaystyle \frac 13 \sec^{-1}\left(x^2\right)+C$
}

\item $\displaystyle \int_0^{\pi/2} \frac {\sin x}{1+\cos^2 x} \,dx$
\answer{
$\displaystyle \frac {\pi}4$
}

\end{problemblock}

\item Find the points on the graph of $y=\tan^{-1}(2x)$ at which the tangent line is parallel to the line $x-2y-8=0$\,.
\answer{
$\displaystyle \left( \frac {\sqrt 3}2, \frac{\pi}3\right), \left( -\frac {\sqrt 3}2, -\frac{\pi}3\right)$
}

\textbook{\clearpage}
\begin{problemblock}[2]{Evaluate the given limit.}
\item $\displaystyle\lim_{ x \to \infty}  \frac{x^3+4x+2}{\ln \left(x+2\right)}$
\answer{
$\displaystyle \infty$
}

\item $\displaystyle\lim_{ x \to 0}  \frac{\sin^{-1}(2x)}{\tan^{-1}(2x)}$
\answer{
$\displaystyle 1$
}

\item $\displaystyle\lim_{ x \to 1} (\ln x)^{x^2-1}$
\answer{
$\displaystyle 1$
}

\item $\displaystyle\lim_{ x \to 0}  \left(\frac 1{e^x-1}-\frac 1{e^x-e^{-x}} \right)$
\answer{
$\displaystyle \infty$
}

\item $\displaystyle\lim_{ x \to \infty}  \left( e^x+x\right)^{1/x}$
\answer{
$\displaystyle e$
}

\item $\displaystyle\lim_{ x \to 1}  \left( 1+2\ln x\right)^{\frac 1{x-1}}$
\answer{
$\displaystyle e^2$
}

\item $\displaystyle\lim_{ x \to \infty}  \left[ \ln (5x+2)-\ln(2x+5)\right]$
\answer{
$\displaystyle \ln \frac 52$
}

\end{problemblock}

\item A certain  bacteria culture  doubles in size every 5 minutes. If there are 1000 bacteria initially present, how many will there be in 30 minutes? 
\answer{
$64000$ bacteria
}

\item A drug used for sedation decays exponentially.  It is observed that the amount decays by 25\% after each hour. A patient receiving this drug should not drive until there is only 30 mg left in their system. If the initial dose is 400 mg, how long will it be until it is safe to drive?
\answer{
$\lambda=-k=-\ln(3/4)\ \frac{1}{\text{hr}}\implies t=\frac{1}{k}\ln(30/400)\approx9.0039$ hours
}

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