Question

1. let $g(x)=2x^{3}-5$. write an equation for $h(x)$, the inverse of $g$.
2. consider the function $h = f(g)=sqrt{2g - 1}$.

a. what is the independent variable of $f^{-1}$?
b. what is the dependent variable of $f^{-1}$.
c. find $f^{-1}$.

1. let $c = f(t)$ represent the number of calories burned after playing soccer for $t$ hours. interpret the meaning of the function $t = f^{-1}(c)$.
2. let $h(x)=\frac{2}{x + 1}$.

a. find $h^{-1}$.
b. use compositions to verify that $h$ and $h^{-1}$ are in fact inverses.

Explanation:

7.

Step1: Replace \(g(x)\) with \(y\)

Let \(y = 2x^{3}-5\).

Step2: Swap \(x\) and \(y\)

We get \(x = 2y^{3}-5\).

Step3: Solve for \(y\)

First, add 5 to both sides: \(x + 5=2y^{3}\). Then divide both sides by 2: \(\frac{x + 5}{2}=y^{3}\). Finally, take the cube - root of both sides: \(y=\sqrt[3]{\frac{x + 5}{2}}\). So \(h(x)=\sqrt[3]{\frac{x + 5}{2}}\).

8.

Given \(h = f(g)=\sqrt{2g - 1}\)
a. The independent variable of \(f^{-1}\) is the dependent variable of \(f\). For \(y = f(g)=\sqrt{2g - 1}\), the independent variable of \(f^{-1}\) is \(h\).
b. The dependent variable of \(f^{-1}\) is the independent variable of \(f\). So the dependent variable of \(f^{-1}\) is \(g\).
c.

Step1: Replace \(f(g)\) with \(y\)

Let \(y=\sqrt{2g - 1}\).

Step2: Swap \(y\) and \(g\)

We have \(g=\sqrt{2y - 1}\).

Step3: Solve for \(y\)

Square both sides: \(g^{2}=2y - 1\). Then add 1 to both sides: \(g^{2}+1 = 2y\). Divide both sides by 2: \(y=\frac{g^{2}+1}{2}\). So \(f^{-1}(g)=\frac{g^{2}+1}{2},g\geq0\).

9.

If \(c = f(t)\) represents the number of calories burned after playing soccer for \(t\) hours, then \(t = f^{-1}(c)\) represents the number of hours of playing soccer required to burn \(c\) calories.

10.

a.

Step1: Replace \(h(x)\) with \(y\)

Let \(y=\frac{2}{x + 1}\).

Step2: Swap \(x\) and \(y\)

We get \(x=\frac{2}{y + 1}\).

Step3: Solve for \(y\)

Cross - multiply: \(x(y + 1)=2\). Expand: \(xy+x = 2\). Then isolate \(y\): \(xy=2 - x\), and \(y=\frac{2 - x}{x},x
eq0\). So \(h^{-1}(x)=\frac{2 - x}{x},x
eq0\).
b.

Answer:

Step1: Find \(h(h^{-1}(x))\)

\[

$$\begin{align*} h(h^{-1}(x))&=h(\frac{2 - x}{x})\\ &=\frac{2}{\frac{2 - x}{x}+1}\\ &=\frac{2}{\frac{2 - x+x}{x}}\\ &=\frac{2}{\frac{2}{x}}\\ &=x \end{align*}$$

\]

Step2: Find \(h^{-1}(h(x))\)

\[

$$\begin{align*} h^{-1}(h(x))&=h^{-1}(\frac{2}{x + 1})\\ &=\frac{2-\frac{2}{x + 1}}{\frac{2}{x + 1}}\\ &=\frac{\frac{2(x + 1)-2}{x + 1}}{\frac{2}{x + 1}}\\ &=\frac{2x+2 - 2}{2}\\ &=x \end{align*}$$

\]
Since \(h(h^{-1}(x))=x\) and \(h^{-1}(h(x))=x\), \(h\) and \(h^{-1}\) are inverses.