## University Calculus: Early Transcendentals (3rd Edition)

a) $h(x)=x^3/4$ and $k(x)=(4x)^{1/3}$ To show that $h$ and $k$ are inverses of one another, we plug the formula for $h(x)$ for every $x$ in $k(x)$ and vice versa: - Plug $h(x)$ for every $x$ in $k(x)$: $$(koh)(x)=k(h(x))=\Big(4\Big(\frac{x^3}{4}\Big)\Big)^{1/3}=(x^3)^{1/3}=x^1=x$$ - Plug $k(x)$ for every $x$ in $h(x)$: $$(hok)(x)=h(k(x))=\frac{((4x)^{1/3})^3}{4}=\frac{(4x)^1}{4}=\frac{4x}{4}=x$$ Since they both result in $x$, we conclude that $h$ and $k$ are inverses of one another. b) The graphs of $h$ and $k$ are shown below. c) First, we need to find $h'$ and $k'$: $$h'(x)=\Big(\frac{x^3}{4}\Big)'=\frac{1}{4}\times3x^2=\frac{3x^2}{4}$$ $$k'(x)=((4x)^{1/3})'=\frac{1}{3}\times(4x)^{-2/3}(4x)'=\frac{1}{3\times (4x)^{2/3}}\times4=\frac{4}{3\sqrt[3]{(4x)^2}}$$ - At the point $(2,2)$, we have: The slope of $h(x)$ is $$h'(2)=\frac{3\times2^2}{4}=\frac{3\times4}{4}=3$$ The slope of $k(x)$ is $$k'(2)=\frac{4}{3\sqrt[3]{(4\times2)^2}}=\frac{4}{3\sqrt[3]{64}}=\frac{4}{3\times4}=\frac{1}{3}$$ - At the point $(-2,-2)$, we have: The slope of $h(x)$ is $$h'(-2)=\frac{3\times(-2)^2}{4}=\frac{3\times4}{4}=3$$ The slope of $k(x)$ is $$k'(-2)=\frac{4}{3\sqrt[3]{(4\times(-2))^2}}=\frac{4}{3\sqrt[3]{64}}=\frac{4}{3\times4}=\frac{1}{3}$$ d) At the origin, we have: $$h'(0)=\frac{3\times0^2}{4}=0$$ The tangent line to $h(x)$ at the origin is $$y-0=0(x-0)=0$$ $$y=0$$ $$k'(0)=\frac{4}{3\sqrt[3]{(4\times0)^2}}=\frac{4}{3\sqrt[3]{0}}=\frac{4}{0}=\infty$$ This means $k(x)$ has a vertical tangent here. Since this point is the origin, that vertical tangent is $x=0$.