Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.
Please help me solve this problem I'm desperate!!!!!!
Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.
Please help me solve this problem I'm desperate!!!!!!
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.
$\displaystyle y = \frac{1}{3} \sqrt{16-x^2}$
flip x and y
$\displaystyle x=\frac{1}{3} \sqrt{16-y^2}$
solve for y, first multiply both side by 3
$\displaystyle 3x=\sqrt {16-y^2}$
square both sides to solve to rid of the radical
$\displaystyle (3x)^2=(\sqrt {16-y^2})^2 $
$\displaystyle 9x^2 =16-y^2 $
now solve for y^2
$\displaystyle -y^2 = 9x^2 - 16$
$\displaystyle y^2=16-9x^2$
take the root to isolate y
$\displaystyle y=\pm \sqrt {16-9x^2}$
since for any x, y (your inverse function) can have 2 values the inverse is not a function