Solve the problem.
-$$f ( x ) = \sqrt { x + 5 }$$
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \mathrm { x } ^ { 2 } - 5 , \mathrm { x } \geq 0$
B) $f ^ { - 1 } ( x ) = - x ^ { 2 } + 5 , x \geq 0$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } - 5 } , \mathrm { x } \geq 5$
D) $f ^ { - 1 } ( x ) = x ^ { 2 } - 25 , x \geq 0$

Question

Quizplus · Accepted Answer

To find the inverse function $f^{-1}(x)$, swap $x$ and $y$ in the original function $y = \sqrt{x + 5}$ and solve for $y$. This gives $x = \sqrt{y + 5}$, squaring both sides yields $x^2 = y + 5$, so $y = x^2 - 5$, with $x \geq 0$ because the square root function only outputs non-negative numbers.

Solve the Problem f(x)=x+5f ( x ) = \sqrt { x + 5 }f(x)=x+5​

Solve the Problem $f ( x ) = \sqrt { x + 5 }$