Simple Inverse Function problems!

Thatoneguy12345 · #1 $Old$ November 4th, 2009, 07:55 PM

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)]

bigwave · #2 $Old$ November 5th, 2009, 12:13 AM

first note that these functions are one to one and that from:

$g(x) = \frac{x}{2} \ + \ 1$

$g(2) = \frac{2}{2} \ + \ 1 \ = 2$

to get the inverse temporarily change this equation to

$y = \frac{x}{2} \ + 1$

now exchange x for y and solve for y

$x = \frac{y}{2} + 1$

$y = 2x - 2 \ or \ g^{-1}(x) = 2x - 2$

$then\ g^{-1}[g(2)] = 2(2) - 2 \ = \ 2$

now as for $h(x) = \frac{x}{3} + 1$

$h^{-1}(x) = 3x -3$

so $h[h^{-1}(x)] = \frac{3x-3}{3} - 3 = x-4$

HallsofIvy · #3 $Old$ November 5th, 2009, 03:55 AM

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".