Find the inverse function of $f ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$ .

A) $f ^ { - 1 } ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$
B) $f ^ { - 1 } ( x ) = \sqrt { x ^ { 2 } - 36 } , 0 \leq x \leq 6$
C) $f ^ { - 1 } ( x ) = 36 - x ^ { 2 } , 0 \leq x \leq 6$
D) $f ^ { - 1 } ( x ) = \sqrt { 36 + x ^ { 2 } } , 0 \leq x \leq 6$
E) $f ^ { - 1 } ( x ) = 36 + x ^ { 2 } , 0 \leq x \leq 6$

Question

Find the inverse function of $f ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$ .

A) $f ^ { - 1 } ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$
B) $f ^ { - 1 } ( x ) = \sqrt { x ^ { 2 } - 36 } , 0 \leq x \leq 6$
C) $f ^ { - 1 } ( x ) = 36 - x ^ { 2 } , 0 \leq x \leq 6$
D) $f ^ { - 1 } ( x ) = \sqrt { 36 + x ^ { 2 } } , 0 \leq x \leq 6$
E) $f ^ { - 1 } ( x ) = 36 + x ^ { 2 } , 0 \leq x \leq 6$

Quizplus · Accepted Answer

To find the inverse function, swap $x$ and $y$ and solve for $y$. Starting with $y = \sqrt{36 - x^2}$, swapping gives $x = \sqrt{36 - y^2}$. Solving for $y$ yields $y = \sqrt{36 - x^2}$, which is the same as the original function, but with the roles of $x$ and $y$ switched, fitting the definition of an inverse function within the given domain $0 \leq x \leq 6$.

Find the Inverse Function Of f(x)=36−x2,0≤x≤6f ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6f(x)=36−x2​,0≤x≤6

Find the Inverse Function Of $f ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$