Answer:
[tex]\textsf{Inverse:}\quad f^{-1}(x)=-9+\sqrt{x-6}[/tex]
[tex]\textsf{Domain:}\quad x \geq 6[/tex]
[tex]\textsf{Range:}\quad f^{-1}(x) \geq -9[/tex]
Step-by-step explanation:
The given function is:
[tex]f(x)=(x+9)^2+6[/tex]
Since its domain is restricted to x ≥ -9, its range is also restricted.
To find the minimum value of its range, substitute x = -9 into the function:
[tex]f(-9)=(-9+9)^2+6\\\\f(-9)=(0)^2+6\\\\f(-9)=0+6\\\\f(-9)=6[/tex]
Therefore, the range of the function is f(x) ≥ 6.
To find the inverse f⁻¹(x), begin by replacing f(x) with y:
[tex]y=(x+9)^2+6[/tex]
Swap the x and y:
[tex]x=(y+9)^2+6[/tex]
Now, rearrange to isolate y:
[tex]\begin{aligned}x&=(y+9)^2+6\\\\x-6&=(y+9)^2\\\\\pm\sqrt{x-6}&=y+9\\\\y&=-9\pm\sqrt{x-6}\end{aligned}[/tex]
Replace y with f⁻¹(x):
[tex]f^{-1}(x)=-9\pm \sqrt{x-6}[/tex]
When solving for the inverse, taking the square root introduces both the positive and negative square roots, so we have two possible inverse functions. Since the domain and range of the original function are restricted, we can use these values to determine which function is the inverse.
The domain of the inverse of a function is the same as the range of the original function. Given the range of the original function is f(x) ≥ 6, then the domain of the inverse function is restricted to x ≥ 6.
The range of the inverse of a function is the same as the domain of the original function. Given the domain of the original function is x ≥ -9, then the range of the inverse function is restricted to f⁻¹(x) ≥ -9.
[tex]\textsf{When}\;x\geq 6 \implies f^{-1}(x)=-9-\sqrt{x-6} \leq -9[/tex]
[tex]\textsf{When}\;x\geq 6 \implies f^{-1}(x)=-9+\sqrt{x-6} \geq -9[/tex]
As the range is restricted to f⁻¹(x) ≥ -9, we choose the positive square root in the inverse function to maintain consistency with the range. Therefore, the inverse function is:
[tex]f^{-1}(x)=-9+\sqrt{x-6}[/tex]
Its domain is x ≥ 6 and its range is f⁻¹(x) ≥ -9.
