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

Question

Quizplus · Accepted Answer

To find the inverse function, $f^{-1}(x)$, we swap $x$ and $y$ in the original function and solve for $y$. Starting with $y = 7x^2 - 8$, we swap to get $x = 7y^2 - 8$, and solve for $y$. Adding 8 to both sides gives $x + 8 = 7y^2$. Dividing by 7 gives $\frac{x + 8}{7} = y^2$. Taking the square root of both sides, $y = \sqrt{\frac{x + 8}{7}}$, which matches choice C. The restriction $x \geq 0$ ensures the function is one-to-one and that its inverse is a function.

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 $f ( x ) = 7 x ^ { 2 } - 8 , x \geq 0$