Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 373: 39

$ a.\quad f(g(x))=x$ and $g(f(x))=x$ $ b.\quad$ see image $ c.\quad$ At $(1,1),$ the tangent to the graph of $f$ has slope $1$ the tangent to the graph of $g$ has slope $1/3$ At $(-1,-1),$ the tangent to the graph of $f$ has slope $-1$ the tangent to the graph of $g$ has slope $1/3$ $ d.\quad$ At $(0,0),$ the tangent to the graph of $f$ is the x-axis the tangent to the graph of $g$ is the y-axis

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