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$f\left(x\right)=\sqrt[3]{x-4}$
$f\left(x\right)={x}^{3}+1$
a. ${f}^{\mathrm{-1}}\left(x\right)=\sqrt[3]{x-1}$ b. Domain: all real numbers, range: all real numbers
$f\left(x\right)={\left(x-1\right)}^{2},x\le 1$
$f\left(x\right)=\sqrt{x-1}$
a. ${f}^{\mathrm{-1}}\left(x\right)={x}^{2}+1,$ b. Domain: $x\ge 0,$ range: $y\ge 1$
$f\left(x\right)=\frac{1}{x+2}$
For the following exercises, use the graph of $f$ to sketch the graph of its inverse function.
For the following exercises, use composition to determine which pairs of functions are inverses.
$f\left(x\right)=8x,g\left(x\right)=\frac{x}{8}$
These are inverses.
$f\left(x\right)=8x+3,g\left(x\right)=\frac{x-3}{8}$
$f\left(x\right)=5x-7,g\left(x\right)=\frac{x+5}{7}$
These are not inverses.
$f\left(x\right)=\frac{2}{3}x+2,g\left(x\right)=\frac{3}{2}x+3$
$f\left(x\right)=\frac{1}{x-1},x\ne 1,g\left(x\right)=\frac{1}{x}+1,x\ne 0$
These are inverses.
$f\left(x\right)={x}^{3}+1,g\left(x\right)={\left(x-1\right)}^{1\text{/}3}$
$f(x)={x}^{2}+2x+1,x\ge \mathrm{-1},\phantom{\rule{1em}{0ex}}g(x)=\mathrm{-1}+\sqrt{x},x\ge 0$
These are inverses.
$f\left(x\right)=\sqrt{4-{x}^{2}},0\le x\le 2,g\left(x\right)=\sqrt{4-{x}^{2}},0\le x\le 2$
For the following exercises, evaluate the functions. Give the exact value.
${\text{tan}}^{\mathrm{-1}}\left(\frac{\sqrt{3}}{3}\right)$
$\frac{\pi}{6}$
${\text{cos}}^{\mathrm{-1}}\left(-\frac{\sqrt{2}}{2}\right)$
${\text{sin}}^{\mathrm{-1}}\left(\mathrm{-1}\right)$
${\text{cos}}^{\mathrm{-1}}\left(\frac{\sqrt{3}}{2}\right)$
$\frac{\pi}{6}$
$\text{cos}\left({\text{tan}}^{\mathrm{-1}}\left(\sqrt{3}\right)\right)$
$\text{sin}\left({\text{cos}}^{\mathrm{-1}}\left(\frac{\sqrt{2}}{2}\right)\right)$
$\frac{\sqrt{2}}{2}$
${\text{sin}}^{\mathrm{-1}}\left(\text{sin}\left(\frac{\pi}{3}\right)\right)$
${\text{tan}}^{\mathrm{-1}}\left(\text{tan}\left(-\frac{\pi}{6}\right)\right)$
$-\frac{\pi}{6}$
The function $C=T\left(F\right)=(5\text{/}9)\left(F-32\right)$ converts degrees Fahrenheit to degrees Celsius.
[T] The velocity V (in centimeters per second) of blood in an artery at a distance x cm from the center of the artery can be modeled by the function $V=f(x)=500(0.04-{x}^{2})$ for $0\le x\le 0.2.$
a. $x={f}^{\mathrm{-1}}\left(V\right)=\sqrt{0.04-\frac{V}{500}}$ b. The inverse function determines the distance from the center of the artery at which blood is flowing with velocity V . c. 0.1 cm; 0.14 cm; 0.17 cm
A function that converts dress sizes in the United States to those in Europe is given by $D\left(x\right)=2x+24.$
[T] The cost to remove a toxin from a lake is modeled by the function
$C(p)=75p\text{/}(85-p),$ where $C$ is the cost (in thousands of dollars) and $p$ is the amount of toxin in a small lake (measured in parts per billion [ppb]). This model is valid only when the amount of toxin is less than 85 ppb.
a. $31,250, $66,667, $107,143 b. $\left(p=\frac{85C}{C+75}\right)$ c. 34 ppb
[T] A race car is accelerating at a velocity given by
$v(t)=\frac{25}{4}t+54,$
where v is the velocity (in feet per second) at time t .
[T] An airplane’s Mach number M is the ratio of its speed to the speed of sound. When a plane is flying at a constant altitude, then its Mach angle is given by $\mu =2{\text{sin}}^{\mathrm{-1}}\left(\frac{1}{M}\right).$
Find the Mach angle (to the nearest degree) for the following Mach numbers.
a. $~92\text{\xb0}$ b. $~42\text{\xb0}$ c. $~27\text{\xb0}$
[T] Using $\mu =2{\text{sin}}^{\mathrm{-1}}\left(\frac{1}{M}\right),$ find the Mach number M for the following angles.
[T] The temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function
$T\left(x\right)=5+18\phantom{\rule{0.1em}{0ex}}\text{sin}\left[\frac{\pi}{6}\left(x-4.6\right)\right],$
where $x$ is time in months and $x=1.00$ corresponds to January 1. Determine the month and day when the temperature is $21\text{\xb0}\text{C}.$
$x\approx 6.69,8.51;$ so, the temperature occurs on June 21 and August 15
[T] The depth (in feet) of water at a dock changes with the rise and fall of tides. It is modeled by the function
$D\left(t\right)=5\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\frac{\pi}{6}t-\frac{7\pi}{6}\right)+8,$
where $t$ is the number of hours after midnight. Determine the first time after midnight when the depth is 11.75 ft.
[T] An object moving in simple harmonic motion is modeled by the function
$s\left(t\right)=\mathrm{-6}\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\frac{\pi t}{2}\right),$
where $s$ is measured in inches and $t$ is measured in seconds. Determine the first time when the distance moved is 4.5 ft.
$~1.5\phantom{\rule{0.2em}{0ex}}\text{sec}$
[T] A local art gallery has a portrait 3 ft in height that is hung 2.5 ft above the eye level of an average person. The viewing angle $\theta $ can be modeled by the function
$\theta ={\text{tan}}^{\mathrm{-1}}\frac{5.5}{x}-{\text{tan}}^{\mathrm{-1}}\frac{2.5}{x},$
where $x$ is the distance (in feet) from the portrait. Find the viewing angle when a person is 4 ft from the portrait.
[T] Use a calculator to evaluate ${\text{tan}}^{\mathrm{-1}}\left(\text{tan}(2.1)\right)$ and ${\text{cos}}^{\mathrm{-1}}\left(\text{cos}(2.1)\right).$ Explain the results of each.
${\text{tan}}^{\mathrm{-1}}\left(\text{tan}(2.1)\right)\approx -1.0416;$ the expression does not equal 2.1 since $2.1>1.57=\frac{\pi}{2}$ —in other words, it is not in the restricted domain of $\text{tan}\phantom{\rule{0.1em}{0ex}}x.\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{\mathrm{-1}}\left(\text{cos}(2.1)\right)=2.1,$ since 2.1 is in the restricted domain of $\text{cos}\phantom{\rule{0.1em}{0ex}}x.$
[T] Use a calculator to evaluate $\text{sin}({\text{sin}}^{\mathrm{-1}}(\mathrm{-2}))$ and $\text{tan}({\text{tan}}^{\mathrm{-1}}(\mathrm{-2})).$ Explain the results of each.
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