Answer:
[tex]f^{-1}(x)=\boxed{-x^2+14x-47} \quad \textsf{for\;the\;domain}\;\; \boxed{[\:7, \infty)}[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=\sqrt{2-x}+7, \quad x \leq 2[/tex]
The given function has a restricted domain.
Therefore, the range of the given function is also restricted to [7, ∞).
The inverse of a function is its reflection in the line y = x.
To find the inverse of a function, first swap f(x) for y:
[tex]\implies y=\sqrt{2-x}+7[/tex]
Rearrange the equation to isolate x:
[tex]\implies y-7=\sqrt{2-x}[/tex]
[tex]\implies (y-7)^2=2-x[/tex]
[tex]\implies x=2-(y-7)^2[/tex]
Swap the x for f⁻¹(x) and the y for x:
[tex]\implies f^{-1}(x)=2-(x-7)^2[/tex]
Expand (if necessary):
[tex]\implies f^{-1}(x)=2-(x^2-14x+49)[/tex]
[tex]\implies f^{-1}(x)=-x^2+14x-47[/tex]
The domain of the inverse function is the range of the function.
Therefore, since the range of the function is [7, ∞), then the domain of the inverse function is [7, ∞).
Learn more about inverse functions here:
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