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

$$f(x)=2x+3$$ a) To find $f^{-1}(x)$, we substitute $x$ with $f^{-1}x$ and $f(x)$ with $x$: $$x=2f^{-1}(x)+3$$ Then solve for $f^{-1}(x)$: $$f^{-1}(x)=\frac{x-3}{2}$$ b) The graphs of $f(x)$ and $f^{-1}(x)$ are enclosed below. c) - Evaluate $df/dx$ at $x=-1$: $$\frac{df}{dx}=2\times1+0=2$$ At $x=-1$, $df/dx=2$ - Evaluate $df^{-1}/dx$ at $x=f(-1)=2(-1)+3=1$: $$\frac{df^{-1}}{dx}=\frac{1}{2}(1-0)=\frac{1}{2}$$ At $x=1$, $df^{-1}/dx=1/2$ It can be seen that at these points, $df^{-1}/dx=1/(df/dx)$