$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = 6 x ^ { 2 } , x \geq 0 \text {. }$
A) $f ^ { - 1 } ( x ) = \sqrt { \frac { 1 } { 6 x } }$
B) $f ^ { - 1 } ( x ) = \frac { 1 } { 6 x ^ { 2 } }$
C) $f ^ { - 1 } ( x ) = \sqrt { \frac { 6 } { x } }$
D) $f ^ { - 1 } ( x ) = \frac { 1 } { 6 \sqrt { x } }$
E)$f ^ { - 1 } ( x ) = \sqrt { \frac { x } { 6 } }$

Question

$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = 6 x ^ { 2 } , x \geq 0 \text {. }$ 
A) $f ^ { - 1 } ( x ) = \sqrt { \frac { 1 } { 6 x } }$
B) $f ^ { - 1 } ( x ) = \frac { 1 } { 6 x ^ { 2 } }$
C) $f ^ { - 1 } ( x ) = \sqrt { \frac { 6 } { x } }$
D) $f ^ { - 1 } ( x ) = \frac { 1 } { 6 \sqrt { x } }$
E)$f ^ { - 1 } ( x ) = \sqrt { \frac { x } { 6 } }$

Quizplus · Accepted Answer

We will use the definition of inverse functions:

If $y=f(x)$, then $x=f^{-1}(y)$

We want to find $f^{-1}(x)$ such that $f(f^{-1}(x))=x$ for all $x\geq 0$.

Let's substitute $f^{-1}(x)$ into the original function for $x$:

$f(f^{-1}(x))=6(f^{-1}(x))^2$

We want this to equal $x$, so:

$6(f^{-1}(x))^2=x$

Solving for $f^{-1}(x)$, we have:

$f^{-1}(x)=\sqrt{\frac{x}{6}}$

And since this holds for all $x\geq 0$, the answer is (C).

Find f−1(x) if f(x)=6x2,x≥0. \text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = 6 x ^ { 2 } , x \geq 0 \text {. } Find f−1(x) if f(x)=6x2,x≥0.

$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = 6 x ^ { 2 } , x \geq 0 \text {. }$