Question

The function $f\left(x\right)=\frac{5}{x+2}$is one to one. Find $f^{-1}$. Find the domain and the range of $f$and the domain and the range of $f^{-1}$.

Short answer

$f^{-1}=\frac{5-2x}{x}$

The domain of $f$is defined for all values of $x$except for $x=-2$and its range is, all the real numbers.

The domain of$f^{-1}$is defined for all values of $x$except for$x=0$and the range is all the real numbers.

Step-by-step solution

Step 1 Given function is,

$f\left(x\right)=\frac{5}{x+2}$.

Find $f^{-1}$.

Solve the given function for$x$.

$$\begin{gathered} f\left(x\right)=\frac{5}{x+2} \\ f\left(x\right)\left(x+2\right)=5 \\ xf\left(x\right)+2f\left(x\right)=5 \\ xf\left(x\right)=5-2f\left(x\right) \\ x=\frac{5-2f\left(x\right)}{f\left(x\right)} \end{gathered}$$

Step 2 Now substitute fx=x,x=f-1in the obtained equation.

$f^{-1}=\frac{5-2x}{x}$

The domain of $f\left(x\right)$is defined for all values of $x$except for $x=-2$and the given function can take up any of the values in the range of real numbers . So, the range of $f$is all the real numbers.

The domain of $f^{-1}$is defined for all the values of $x$except for $x=0$and the given function can take up any of the values in the range of real numbers . So, the range of$f^{-1}$is all the real numbers.