+0

+1
57
3

find the inverse f^-1 of f(x) = 3x^2, where x ≥ 0. Then find f^-1(12)

Mar 28, 2020

#2
+22
+1

To find the inverse, switch x and y in the equation, making it $$x=3y^2$$ and plugging in 12, you get 2 or -2.

Mar 28, 2020
#3
+4569
0

First, we solve f^-1, which can be done by setting f(y)=x...

Thus, we have 3y^2=x

$$y=\frac{\sqrt{3x}}{3}$$.

Thus, f^-1(12) is $$y=\frac{\sqrt{3*12}}{3}=\pm2.$$

.
Mar 28, 2020
edited by tertre  Mar 28, 2020