The function f is one-to-one. Find its inverse.
-$$f ( x ) = 7 x ^ { 2 } - 8 , x \geq 0$$
A) $f ^ { - 1 } ( x ) = \sqrt { \frac { 7 } { x + 8 } }$

B) $f ^ { - 1 } ( x ) = \frac { 7 } { \sqrt { x } + 8 }$

C) $f ^ { - 1 } ( x ) = \sqrt { \frac { x + 8 } { 7 } }$

D) $f ^ { - 1 } ( x ) = - \sqrt { \frac { x + 8 } { 7 } }$

Question

The function f is one-to-one. Find its inverse.
-$$f ( x ) = 7 x ^ { 2 } - 8 , x \geq 0$$
A) $f ^ { - 1 } ( x ) = \sqrt { \frac { 7 } { x + 8 } }$

B) $f ^ { - 1 } ( x ) = \frac { 7 } { \sqrt { x } + 8 }$

C) $f ^ { - 1 } ( x ) = \sqrt { \frac { x + 8 } { 7 } }$

D) $f ^ { - 1 } ( x ) = - \sqrt { \frac { x + 8 } { 7 } }$

Quizplus · Accepted Answer

To find the inverse function, $f^{-1}(x)$, we first replace $f(x)$ with $y$, giving us $y = 7x^2 - 8$. Then, we solve for $x$ in terms of $y$. Swapping $x$ and $y$ gives us $x = 7y^2 - 8$, but since we're solving for $y$ in terms of $x$, we actually start with $y = 7x^2 - 8$, then swap $x$ and $y$ to get $x = 7y^2 - 8$. Solving for $y$ gives us $y = \sqrt{\frac{x + 8}{7}}$, which corresponds to choice C. The restriction $x \geq 0$ ensures the function is one-to-one, allowing for an inverse to exist.

The Function F Is One-To-One f(x)=7x2−8,x≥0f ( x ) = 7 x ^ { 2 } - 8 , x \geq 0f(x)=7x2−8,x≥0

The Function F Is One-To-One $f ( x ) = 7 x ^ { 2 } - 8 , x \geq 0$