The function f is one-to-one. Find its inverse.
-$$f ( x ) = x ^ { 2 } + 1 , x \geq 0$$
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } } - 1 , \mathrm { x } \geq 0$
B) $f ^ { - 1 } ( x ) = \sqrt { x - 1 } , x \geq 1$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } + 1 } , \mathrm { x } \geq - 1$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } } - 1 , \mathrm { x }

Question

The function f is one-to-one. Find its inverse.
-$$f ( x ) = x ^ { 2 } + 1 , x \geq 0$$
A) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } } - 1 , \mathrm { x } \geq 0$
B) $f ^ { - 1 } ( x ) = \sqrt { x - 1 } , x \geq 1$
C) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } + 1 } , \mathrm { x } \geq - 1$
D) $\mathrm { f } ^ { - 1 } ( \mathrm { x } ) = \sqrt { \mathrm { x } } - 1 , \mathrm { x } < 0$

Quizplus · Accepted Answer

To find the inverse function, swap $x$ and $y$ and solve for $y$: $x = y^2 + 1$. Solving for $y$ gives $y = \sqrt{x - 1}$, valid for $x \geq 1$ because the original function is defined for $x \geq 0$, and $x^2 + 1 \geq 1$.

The Function F Is One-To-One f(x)=x2+1,x≥0f ( x ) = x ^ { 2 } + 1 , x \geq 0f(x)=x2+1,x≥0

The Function F Is One-To-One $f ( x ) = x ^ { 2 } + 1 , x \geq 0$