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)]
first note that these functions are one to one and that from:
to get the inverse temporarily change this equation to
now exchange x for y and solve for y
now as for
so
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".