Hi Jess.

Short solution:
g^-1(g(x)) means to feed x into the g(x) formula, then take that answer and feed it backwards through it again !

So simply, g^-1(g(4)) = 4, because the g^-1( ) cancels the g( ). This is valid because g(x) is defined as a one-to-one function (in this case, a line).

Long solution:
g(x) = 2x + 3
so g(4) = 2(4) + 3 = 8 + 3 = 11.

This means that g^-1(g(4)) = g^-1(11).

Now g^-1(x) means to do things backwards, so I replace g(x) with x and x with g^-1(x):
So
g(x) = 2x + 3
becomes
x = 2(g^-1(x)) + 3

Next I solve for g^-1(x):
x - 3 = 2(g^-1(x))
g^-1(x) = (x - 3) / 2

So if we want g^-1(11), then x = 11 for this step and we just fill it in:

g^-1(11) = (11 - 3) / 2 = 8 / 2 = 4.

So the answer is 4, just like it was in the quick solution.

You have to be careful that the function you are reversing is one-to-one when using the quick solution, whereas the long solution will show you a problem.

For example, consider h(x) = x². What is h^-1(h(-4)) ?

The short answer tells us that h^-1( h(x) ) = x, so the answer would be -4, right? Not so fast!

The long answer says h(-4) = (-4)^2 = 16 so h^-1(h(-4)) = h^-1(16). To find h^-1(x) we proceed:
x = ( h^-1(x) )²
± sqrt(x) = h^-1(x)
h^-1(x) = ± sqrt(x).

So with x now equal to 16, we have h^-1(16) = ± 4 rather than -4. So we'd have an incomplete answer. The real answer to the question is what this long solution method gave us!

This is because h(x) = x² is not a one-to-one function - there are two different x values that generate the same result, 16.

So use the quick way if you are sure (and can satisfy yourself and your teacher) that it is a one-to-one function (all line equations are one-to-one functions, for example), but if in doubt, use the long way.

Cheers,
Stephen La Rocque.