# Thread: Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

1. ## Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

2. Originally Posted by gnarlycarly227
Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

So, if I understand correctly. You are trying to determine whether the inverse function of f(x) is function. First, you must find the inverse function. To do this, typically one "switches" the x and y values and solves for y. So, let's do that.

$y = \frac{1}{3} \sqrt{16-x^2}$
flip x and y

$x=\frac{1}{3} \sqrt{16-y^2}$
solve for y, first multiply both side by 3

$3x=\sqrt {16-y^2}$
square both sides to solve to rid of the radical

$(3x)^2=(\sqrt {16-y^2})^2$
$9x^2 =16-y^2$
now solve for y^2

$-y^2 = 9x^2 - 16$
$y^2=16-9x^2$
take the root to isolate y

$y=\pm \sqrt {16-9x^2}$

since for any x, y (your inverse function) can have 2 values the inverse is not a function