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Inverse Functions

  1. Oct 16, 2010 #1
    Just want to check that i am doing this question correctly.

    f(x) = 2x+5 h(x) = 1/x , x [tex]\neq[/tex]0

    Find the inverse of fh(x)

    So first i found the function fh(x)


    then let y = 2*1/x+5 , x [tex]\neq[/tex]0

    now this is the bit i cant rememeber how to do, when i try and make x the subject do i need to multiply the 2 on the RHS as well as the y on the LHS?

    if i multiply the 2 then i end up with f-1(x)=2x+5/x
    If i dont i end up with f-1(x) = x+7
  2. jcsd
  3. Oct 16, 2010 #2
    First of all, you do mean f(h(x)), and not fh(x), which to me looks like f(x)*h(x). Under that assumption...

    f(h(x)) = 2/x + 5.

    So, you have y = 2/x + 5..

    To find the inverse, you usually just switch x and y, and solve appropriately.

    I'll start with the first step: x = 2/y + 5

    Can you carry it through from here?
  4. Oct 16, 2010 #3
    I can now thanks :)

    I see what you have done but im not sure why 1/x becomes 2/x in this circumstance. Im just trying to understand the mechanics behind it so i can be fully aware. If h(x) was 1/x + 5 would it still be 2/x +5 or 2/x + 10?

    your correct in your assumption, my teacher is poor and makes us write it fh(x) instead of f(h(x))
  5. Oct 16, 2010 #4
    Whereever you saw x, you needed to replace with 1/x. So all you really have is instead of 2*x + 5, you have 2*(1/x) + 5.

    If h(x) = 1/x + 5, and f(x) = 2x + 5 then you will actually have f(h(x)) = 2(1/x + 5) + 5 = 2/x + 15.
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