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April 16, 2014

April 16, 2014

Posted by **noel** on Thursday, May 24, 2012 at 10:00pm.

a. f(x) = 1/x-2 + 3

b. f(x) = 2(x – 4)^2 + 5

- math -
**Steve**, Friday, May 25, 2012 at 12:27amy = 1/(x-2) + 3

y-3 = 1/(x-2)

x-2 = 1/(y-3)

x = 1/(y-3) + 2

so, f^{-1}(x) = 1/(x-3) + 2

y = 2(x-4)^2 + 5

y-5 = 2(x-4)^2

(y-5)/2 = (x-4)^2

now, we can use ± root, so let's choose the + root:

√[(y-5)/2] = x-4

√[(y-5)/2]+4 = x

so, f^{-1}(x) = √[(x-5)/2] + 4

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