Answer:
A
Step-by-step explanation:
Given quadratic function:
[tex]y=3x^2 - 9, \qquad x \leq 0[/tex]
The domain of the given function is restricted to values of x less than or equal to zero. Therefore:
As 3x² ≥ 0, then range of the given function is restricted to values of y greater than or equal to -9.
[tex]\hrulefill[/tex]
To find the inverse of the given function, first interchange the x and y variables:
[tex]x = 3y^2 - 9[/tex]
Now, solve the equation for y:
[tex]\begin{aligned}x& = 3y^2 - 9\\\\x+9&=3y^2\\\\\dfrac{x+9}{3}&=y^2\\y&=\pm \sqrt{\dfrac{x+9}{3}}\end{aligned}[/tex]
The range of the inverse function is the domain of the original function.
As the domain of the original function is restricted to x ≤ 0, then the range of the inverse function is restricted to y ≤ 0.
Therefore, the inverse function is the negative square root:
[tex]f^{-1}(x)=-\sqrt{\dfrac{x+9}{3}}[/tex]
The domain of the inverse function is the range of the original function.
As the range of the original function is restricted to y ≥ -9, then the domain of the inverse function is restricted to x ≥ -9.
[tex]\boxed{f^{-1}(x)=-\sqrt{\dfrac{x+9}{3}}\qquad x \geq -9}[/tex]
So the correct statement is:
- A) The inverse of y is equal to negative square root of the quantity x plus 9 over 3 end quantity such that x is greater than or equal to negative 9.
