Math CentralQuandaries & Queries


Question from jess, a student:

if g(x)=2x+3 find g^-1(g(4))

how do I solve I can't figure out what to do with the 4

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):
g(x) = 2x + 3
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.

Stephen La Rocque.

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