Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 373: 40

$ a.\quad h(k(x))=x$ and $k(h(x))=x$ $ b.\quad$ see image $ c.\quad$ At $(2,2),$ the tangent to the graph of $h$ has slope $ 3$ the tangent to the graph of $k$ has slope $\displaystyle \frac{1}{3}$ At $(-2,-2),$ the tangent to the graph of $h$ has slope $ 3$ the tangent to the graph of $k$ has slope $\displaystyle \frac{1}{3}$ $ d.\quad$ At $(0,0),$ the tangent to the graph of $h$ is the x-axis the tangent to the graph of $k$ is the y-axis

$ a.\quad$ $h$ and $k$ are inverses $\Leftrightarrow h(k(x))=x$ and $k(h(x))=x$ $h(k(x))=\displaystyle \frac{[g(x)]^{3}}{4}=\frac{[(4x)^{1/3}]^{3}}{4}=\frac{4x}{4}=4$ $k(h(x))=[4h(x)]^{1/3}=[4\displaystyle \cdot\frac{x^{3}}{4}]^{1/3}=[x^{3}]^{1/3}=x$ $h$ and $k$ are inverses of each other. $ b.\quad$ See attached image. $ c.\quad$ $h(x)=\displaystyle \frac{1}{4}x^{3},\qquad g(x)=(4x)^{1/3}$ $h'(x)=\displaystyle \frac{3}{4}x^{2},\qquad g'(x)=\frac{1}{3}(4x)^{-2/3}\cdot 4=\frac{4}{3\sqrt[3]{16x^{2}}}$ At $(2,2)$ ... $x=2,$ the tangent to the graph of $h$ has slope $h'(2)=\displaystyle \frac{3\cdot 2^{2}}{4}=3$ the tangent to the graph of $k$ has slope $k'(2)=\displaystyle \frac{4}{3\sqrt[3]{16\cdot 4}}=\frac{1}{3}$ At $(-2,-2)$ ... $x=-1,$ the tangent to the graph of $h$ has slope $h'(-2)=\displaystyle \frac{3\cdot(-2)^{2}}{4}=3$ the tangent to the graph of $k$ has slope $k'(-2)=\displaystyle \frac{4}{3\sqrt[3]{16\cdot 4}}=\frac{1}{3}$ $ d.\quad$ At $(0,0)$ ... $x=0,$ the tangent to the graph of $h$ has slope $f'(0)=0$ the tangent to the graph of $k$ has slope $g'(0)=$ undefined. So the tangent to f is a horizontal line passing through the origin and the x-axis. The tangent to g is a vertical line passing through the origin and the y-axis.