Given #f(x)= x^2/(x+2)# how do you find #f(x-2)#?

1 Answer

Substitute #x-2# everywhere there is an #x# and you'll get to #f(x-2) = (x-2)^2/x#


When working function problems, it's all in the substitution!

We are starting with #f(x) = x^2/(x+2)#

So each time we're given an "x", we're going to square it, then divide itself (after we add 2 to it first). It's easier to see that if we said #x=1# that we'd do the following:

#f(1) = 1^2/(1+2) = 1/3#

So - if we substitute a number into this function, we can come up with a single answer (substituting 1 generates an answer of 1/3).

What happens if we alter the rule? That is what your question is doing - instead of just dropping in any given number (i.e. "x"), we're instead going to subtract 2 from it first, then see what the answer is. What then is the general rule for substituting in #x-2#?

Let's see - we substitute just like above:

#f(x-2) = (x-2)^2/((x-2)+2)#

See? Everywhere there was an x, there is now x-2. Let's simplify this expression:

#f(x-2) = (x-2)^2/x#

And I don't think we can do much more than that. If we expand out the numerator, there will be terms without an x, so there isn't a clean way to get the x out from the denominator without it being a mess - it'd look like #x-4-4/x# and that's just not simple at all!