$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 }$
A) $f ^ { - 1 } ( x ) = x + \sqrt { 13 } , 0 \leq x \leq \sqrt { 13 }$
B) $f ^ { - 1 } ( x ) = \left( 13 - x ^ { 2 } \right) ^ { 2 } , 0 \leq x \leq \sqrt { 13 }$
C) $f ^ { - 1 } ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 }$
D) $f ^ { - 1 } ( x ) = \sqrt { x ^ { 2 } - 13 } , 0 \leq x \leq \sqrt { 13 }$
E) $f ^ { - 1 } ( x ) = \frac { 1 } { \sqrt { 13 - x ^ { 2 } } } , 0 \leq x \leq \sqrt { 13 }$

Question

$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 }$ 
A) $f ^ { - 1 } ( x ) = x + \sqrt { 13 } , 0 \leq x \leq \sqrt { 13 }$
B) $f ^ { - 1 } ( x ) = \left( 13 - x ^ { 2 } \right) ^ { 2 } , 0 \leq x \leq \sqrt { 13 }$
C) $f ^ { - 1 } ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 }$
D) $f ^ { - 1 } ( x ) = \sqrt { x ^ { 2 } - 13 } , 0 \leq x \leq \sqrt { 13 }$
E) $f ^ { - 1 } ( x ) = \frac { 1 } { \sqrt { 13 - x ^ { 2 } } } , 0 \leq x \leq \sqrt { 13 }$

Quizplus · Accepted Answer

To find $f^{-1}(x)$, we need to solve for $x$ in terms of $f(x)$. 
$$f(x) = \sqrt{13-x^2}$$
Square both sides: 
$$f(x)^2 = 13-x^2$$
Solve for $x$: 
$$x = \pm \sqrt{13-f(x)^2}$$
Since $0 \leq x \leq \sqrt{13}$, we take the positive square root: 
$$x = \sqrt{13-f(x)^2}$$
This gives us $f^{-1}(x) = \sqrt{x^2-13}$, which is choice D.

Find f−1(x) if f(x)=13−x2,0≤x≤13\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 } Find f−1(x) if f(x)=13−x2​,0≤x≤13​

$\text { Find } f ^ { - 1 } ( x ) \text { if } f ( x ) = \sqrt { 13 - x ^ { 2 } } , 0 \leq x \leq \sqrt { 13 }$