Solve the problem.
-$$f ( x ) = ( x - 10 ) ^ { 2 } + 2 ( x \geq 10 )$$
A) $f ^ { - 1 } ( x ) = \sqrt { x + 10 } - 2 , x \geq - 10$
B) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } + 2 } - 10 , \mathrm { x } \geq - 2$
C) $f ^ { - 1 } ( x ) = 2 x ^ { 2 } + 10 , x \geq 0$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } - 2 } + 10 , \mathrm { x } \geq 2$

Question

Quizplus · Accepted Answer

To find the inverse of the function, we set y = f(x) and solve for x. 
$$y = (x-10)^2 + 2$$
$$y-2 = (x-10)^2$$
$$\sqrt{y-2}=x-10$$
$$x=\sqrt{y-2}+10$$
Thus, the inverse function is $$f^{-1}(x) = \sqrt{x-2}+10, x \geq 2$$ which matches choice D.

Solve the Problem f(x)=(x−10)2+2(x≥10)f ( x ) = ( x - 10 ) ^ { 2 } + 2 ( x \geq 10 )f(x)=(x−10)2+2(x≥10)

Solve the Problem $f ( x ) = ( x - 10 ) ^ { 2 } + 2 ( x \geq 10 )$