Inverse Functions — Question 10

Question 10 - Solution

Domain. The square root and denominator require

$$\boxed{D=[-4,1)\cup(1,\infty)}.$$

One-to-one. For $x>-4$, $x\ne1$,

$$f'(x)=\frac{-(x+9)}{2\sqrt{x+4}(x-1)^2}<0.$$

The left branch decreases from $0$ to $-\infty$ and has range $(-\infty,0]$.

The right branch decreases from $+\infty$ to $0$ and has range $(0,\infty)$.

These ranges are disjoint, so $f$ is one-to-one on its entire domain and its range is $\mathbb R$.

Inverse. Write $y=\sqrt{x+4}/(x-1)$. Squaring gives

$$y^2x^2-(2y^2+1)x+y^2-4=0.$$

For $y\ne0$, the quadratic formula gives

$$x=\frac{2y^2+1\pm\sqrt{1+20y^2}}{2y^2}.$$

The original, unsquared equation requires $x<1$ when $y<0$ and $x>1$ when $y>0$.

Also $y=0$ gives $x=-4$. Thus

$$\boxed{f^{-1}(y)=\begin{cases} \dfrac{2y^2+1-\sqrt{1+20y^2}}{2y^2},&y<0,\\[6pt] -4,&y=0,\\[3pt] \dfrac{2y^2+1+\sqrt{1+20y^2}}{2y^2},&y>0. \end{cases}}$$

The sign choice restores the original equation after squaring. The inverse has domain $\mathbb R$ and range $D$.