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

$$f(x)=2x^2\hspace{1cm}(x\ge0)$$ a) To find $f^{-1}(x)$, we substitute $x$ with $f^{-1}x$ and $f(x)$ with $x$: $$x=2(f^{-1}(x))^2$$ Now solve for $f^{-1}(x)$: $$(f^{-1}(x))^2=\frac{x}{2}$$ $$f^{-1}(x)=\pm\sqrt{\frac{x}{2}}$$ b) The graphs of $f(x)$ and $f^{-1}(x)$ are enclosed below. c) - Find $df/dx$: $$\frac{df}{dx}=\frac{d}{dx}\Big(2x^2\Big)=4x$$ At $x=a=5$, $$df/dx=4\times5=20$$ - Find $df^{-1}/dx$: We would use implicit differentiation for $x=2(f^{-1}(x))^2$: $$1=4f^{-1}(x)\frac{df^{-1}}{dx}$$ $$\frac{df^{-1}}{dx}=\frac{1}{4f^{-1}(x)}$$ At $x=f(5)=2\times5^2=50$, we have $f^{-1}(50)=\pm\sqrt{(50/2)}=\pm\sqrt{25}=\pm5$ For $f^{-1}(50)=5$: $$df^{-1}/dx=\frac{1}{4\times5}=\frac{1}{20}$$ For $f^{-1}(50)=-5$: $$df^{-1}/dx=\frac{1}{4\times(-5)}=-\frac{1}{20}$$ Looking at the results of $df/dx$ and $df^{-1}/dx$, we find that $df^{-1}/dx=1/(df/dx)$, but only if we limit $f^{-1}(x)$ to the positive.