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) $f ^ { - 1 } ( x ) = 36 + x ^ { 2 } , 0 \leq x \leq 6$

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The Answer is A

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 $f ( x ) = \sqrt { 36 - x ^ { 2 } } , 0 \leq x \leq 6$