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Answer:
1. f^-1(x) = 4/(x+2) -2
2. f^-1(x) = (-(x+3)/2)^(1/5)
Step-by-step explanation:
1. As with all "inverse function" problems, solve for y:
x = f(y)
x +2 = 4/(y +2) . . . . add 2
y +2 = 4/(x +2) . . . . . multiply by (y+2)/(x+2)
y = 4/(x+2) -2 . . . . . subtract 2
We see that this function is its own inverse. The attached graph shows it is symmetrical about the line y=x.
f^-1(x) = 4/(x+2) -2
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2. x = f(y)
x +3 = -2y^5 . . . . add 3
-(x +3)/2 = y^5 . . . . . divide by 2
(-(x +3)/2)^(1/5) = y . . . . take the 5th root
f^-1(x) = (-(x +3)/2)^(1/5)
In typeset form, that is ...
[tex]\displaystyle f^{-1}(x)=\sqrt[5]{\frac{-(x+3)}{2}}\\\\\text{or}\\\\f^{-1}(x)=-\frac{1}{2}\sqrt[5]{16x +48}[/tex]
This last version is with the denominator "rationalized" and the contents of the radical "simplified." It may be a preferred form.
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The graphs show the function and inverse are symmetrical about the line y=x, as they should be.
