Practice questions

No. For example, ${f(0)=0}$ and ${f(4)=16-16=0}$. Two different inputs give the same output.

Compose in both orders:

$\begin{align*}&g\big(f(x)\big)\\&=\dfrac{(4x-7)+7}{4}=x\end{align*}$

$\begin{align*}&f\big(g(x)\big)\\&=4\cdot\dfrac{x+7}{4}-7=x\end{align*}$

Both give ${x}$, so they are inverses.

Write ${y=\sqrt[3]{x+1}}$. Cube both sides, and solve for ${x}$:

$\begin{align*}y^3&=x+1\\x&=y^3-1\end{align*}$

Swap ${x}$ and ${y}$: ${f^{-1}(x)=x^3-1}$.

Write ${y=\dfrac{x-2}{x+5}}$, and multiply by ${x+5}$:

$\begin{align*}y(x+5)&=x-2\\xy+5y&=x-2\\xy-x&=-5y-2\\x(y-1)&=-5y-2\\x&=\dfrac{-5y-2}{y-1}=\dfrac{5y+2}{1-y}\end{align*}$

Swap ${x}$ and ${y}$:

${f^{-1}(x)=\dfrac{5x+2}{1-x}}$

Check: ${f(0)=-\dfrac{2}{5}}$. Then

$\begin{align*}&f^{-1}\left(-\dfrac{2}{5}\right)\\&=\dfrac{-2+2}{1+\frac{2}{5}}=0\end{align*}$

The graph of ${y=3^{-x}}$ is the graph of ${y=3^x}$ reflected across the ${y}$-axis, so it falls. The ${+2}$ moves it up ${2}$. The asymptote is ${y=2}$. The ${y}$-intercept is ${3^0+2=3}$.

${f(0)=C=5}$. Then ${5a^3=40}$, so ${a^3=8}$, and ${a=2}$. The function is ${f(x)=5\cdot 2^x}$.

Quarterly means ${n=4}$. Each quarter adds ${\dfrac{0.04}{4}=0.01}$, and there are ${4\cdot 6=24}$ quarters:

${A=2500(1.01)^{24}\approx 3174.34}$

You have ${\$3174.34}$.

Here ${rt=0.04\cdot 6=0.24}$:

${A=2500e^{0.24}\approx 3178.12}$

You would have ${\$3178.12}$.

${3^4=81}$, so ${\log_3 81=4}$.

${5^{-3}=\dfrac{1}{125}}$, so ${\log_5\dfrac{1}{125}=-3}$.

${16^{1/2}=4}$, so ${\log_{16}4=\dfrac{1}{2}}$.

${\ln e^{-2}=-2}$, by the inverse property.

${\log_6 x=3}$ means ${6^3=x}$. And ${10^{-2}=0.01}$ means ${\log 0.01=-2}$.

The input must be positive: ${4-x^2>0}$, so ${x^2<4}$. This holds for ${-2<x<2}$. The domain is ${(-2,2)}$.

Write ${\sqrt{y}=y^{1/2}}$. Use the quotient, product, and power laws. Then ${\log_3 9=2}$:

$\begin{align*}&\log_3 9+\log_3 x^2-\log_3 y^{1/2}\\&=2+2\log_3 x-\dfrac{1}{2}\log_3 y\end{align*}$

Move the coefficients inside, then combine:

$\begin{align*}&\ln x^2-\ln(x+1)+\ln y^3\\&=\ln\dfrac{x^2y^3}{x+1}\end{align*}$

${\log_6 4+\log_6 9=\log_6 36=2}$, because ${6^2=36}$.

Use the change of base formula:

$\log_3 20=\dfrac{\ln 20}{\ln 3}\approx 2.727$

${125=5^3}$, so ${x-2=3}$, and ${x=5}$.

$\begin{align*}e^{2x}&=5\\2x&=\ln 5\\x&=\dfrac{\ln 5}{2}\approx 0.805\end{align*}$

Take ${\ln}$ of both sides, and collect the ${x}$-terms:

$\begin{align*}x\ln 3&=(x+2)\ln 2\\x\ln 3-x\ln 2&=2\ln 2\\x(\ln 3-\ln 2)&=2\ln 2\end{align*}$

$x=\dfrac{2\ln 2}{\ln 3-\ln 2}\approx 3.419$

Combine with the product law, and write in exponential form:

$\begin{align*}\log_2\big(x(x-2)\big)&=3\\x^2-2x&=8\\x^2-2x-8&=0\\(x-4)(x+2)&=0\end{align*}$

So ${x=4}$ or ${x=-2}$. But ${\log_2(-2)}$ is undefined, so ${x=-2}$ is extraneous.

Check ${x=4}$: ${\log_2 4+\log_2 2=2+1=3}$. The solution is ${x=4}$.

Write in exponential form: ${x+1=e^2}$. So ${x=e^2-1\approx 6.389}$.

In ${t}$ years, the population doubles ${\dfrac{t}{8}}$ times, so ${n(t)=5000\cdot 2^{t/8}}$. After ${20}$ years:

${n(20)=5000\cdot 2^{2.5}\approx 28{,}284}$

The model is ${m(t)=200\left(\dfrac{1}{2}\right)^{t/12}}$. After ${30}$ days:

$m(30)=200\left(\dfrac{1}{2}\right)^{2.5}\approx 35.4$ g

From ${200}$ g to ${25}$ g is ${\dfrac{1}{8}=\left(\dfrac{1}{2}\right)^3}$ of the sample. That takes ${3}$ half-lives, or ${36}$ days.

The model is ${T(t)=22+58e^{-kt}}$, since ${80-22=58}$. From ${T(5)=70}$:

${e^{-5k}=\dfrac{48}{58}}$

We do not need ${k}$ itself. Since ${e^{-15k}=\left(e^{-5k}\right)^3}$:

$\begin{align*}&T(15)\\&=22+58\left(\dfrac{48}{58}\right)^3\\&\approx 54.9\end{align*}$

After ${15}$ minutes, the soup is at about ${54.9}$°C.