Question

Given f(x)=-1/7√16-x^2 find f^-1(x)

Answer 1

Answer:

$f^{-1}(x)=\pm \sqrt{49x^2-16}$

Not a function.

Step-by-step explanation:

The given function is;

$f(x)=-\frac{1}{7}\sqrt{16-x^2}$

Let $y=-\frac{1}{7}\sqrt{16-x^2}$

Interchange x and y;

$x=-\frac{1}{7}\sqrt{16-y^2}$

Solve for y;

$-7x=\sqrt{16-y^2}$

Square both sides

$(-7x)^2=(\sqrt{16-y^2})^2$

$49x^2^2=16-y^2$

$49x^2^2-16=y^2$

$y=\pm \sqrt{49x^2-16}$

The inverse is

$f^{-1}(x)=\pm \sqrt{49x^2-16}$

$f^{-1}(x)=\pm \sqrt{49x^2-16}$ is not a function because one x-value maps onto to different y-values.