If the function is one-to-one, find its inverse. If not, write "not one-to-one."
-$$f ( x ) = 6 x ^ { 2 } - 9 , x \geq 0$$
A) $f ^ { - 1 } ( x ) = \frac { 6 } { \sqrt { x } + 2 }$
B) $f ^ { - 1 } ( x ) = \sqrt { \frac { x + 9 } { 6 } } , x \geq - 9$
C) not a one-to-one
D) $f ^ { - 1 } ( x ) = \sqrt { \frac { 6 } { x + 2 } } , x \neq - 9$

Question

Quizplus · Accepted Answer

Given the restriction $x \geq 0$, the function $f(x) = 6x^2 - 9$ is one-to-one because for every x in the domain, there is a unique y in the codomain, and it passes the horizontal line test. To find the inverse, we swap $x$ and $y$ and solve for $y$: $x = 6y^2 - 9$. Solving for $y$, we get $y = \sqrt{\frac{x + 9}{6}}$, which matches choice B, with the domain $x \geq -9$ to ensure the expression under the square root is non-negative.

If the Function Is One-To-One, Find Its Inverse f(x)=6x2−9,x≥0f ( x ) = 6 x ^ { 2 } - 9 , x \geq 0f(x)=6x2−9,x≥0

If the Function Is One-To-One, Find Its Inverse $f ( x ) = 6 x ^ { 2 } - 9 , x \geq 0$