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Old November 5th, 2009, 02:55 AM
HallsofIvy HallsofIvy is offline
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Quote:
Originally Posted by Thatoneguy12345 View Post
So the first one is g(x)=(x/2)+1

Find g^-1 [g(2)]

Second is h(x)=(x/3)+1
Find h[h^-1 (x)]
You don't need to do all the calculation bigwave did (the first is correct and nicely done- for the second he found h^-1(x), not h[h^-1(x)]) and it really doesn't matter how g and h are defined (as long as they have inverses).

By the definition of inverse, g^-1[g(2)]= 2 and h[h^-1(x)]= x!


If f is any function having an inverse and a, b is any numbers in the domains of f and f^-1 respectively, f^-1(f(a))= a and f(f^-1(a))= a. That's the definition of "inverse function".
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