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Reverse Composition to find composition of a derivative

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]
    I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})
    [/tex]
    [tex]
    Find: g'(x-1)
    [/tex]

    2. Relevant equations
    In order to find [itex]g'(x-1)[/itex] I know the following steps have to be taken:
    [tex]
    f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)
    [/tex]

    3. The attempt at a solution
    Composing ##\sqrt{x}## into ##f(x)## isn't hard, given that I have ##f(x)##. Finding the derivative and then composing ##x-1## into that can also be done without much difficulty.
    The only problem that I find myself coming across is how exactly to move from ##f(x+1)## to ##f(x)##. I don't remember any sort of formal equation of something like this, so I started just playing around with it.
    I defined ##h(x)=x+1## so that my original equation becomes ##f(h(x))= \sqrt{x^2-2x}##.
    Then I tried to logic out what had to be done to ##h(x)## to become ##f(h(x))##. First I decided it needed to be squared. $$h(x)^2 = (x+1)^2 = x^2+2x+1$$It also needs to be taken to a one-half power, but doing that would negate the square, and so something else needs to be happening under the radical.
    So I tried $$\sqrt{h(x)^2-1}=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x+1-1}=\sqrt{x^2+2x}$$
    Unfortunately, that does not equal ##f(x+1)##.
    I happen to be stuck here and cannot figure out how to move forward.
     
  2. jcsd
  3. May 2, 2013 #2
    You seem to be over-thinking on the problem. The first task as you said is to find f(x).

    You have f(x+1). Replace x with something which can give you f(x).
     
  4. May 2, 2013 #3
    I don't understand. Replace x with some arbitrary coefficient?
     
  5. May 2, 2013 #4
    No no. Replace x with something like x+k, which turns x+1 to x.
     
  6. May 2, 2013 #5
    So...it would be something along the lines of f(x+k)=√x2-2x? I'm not making the connection.
     
  7. May 2, 2013 #6
    Replace x with x+k, you get ##f(x+k+1)=\sqrt{(x+k)^2-2(x+k)}##. What should be k if LHS is to be ##f(x)##?
     
  8. May 2, 2013 #7
    I'll admit I was still a little confused by what you meant, so I decided to try out a few numbers first. At last I tried k=-1 and noticed that inside of the function, f(x+k+1) = f(x) if k=-1, are things like this true for all composition functions and the like?
     
  9. May 2, 2013 #8

    Mark44

    Staff: Mentor

    This is obviously a calculus problem, so belongs in the Calculus & Beyond section. I am moving it there.
     
  10. May 2, 2013 #9
    Yes, k=-1 is the right choice. So what is ##f(x)## now?

    Yes, they are.
     
  11. May 2, 2013 #10
    I apologize for that, since the question was focusing on the algebra of the problem I thought it was best suited for the precalculus section.

    $$f(x)=\sqrt{x^2-4x+3}$$
     
  12. May 2, 2013 #11
    That's correct. You have f(x) now, do the remaining part.
     
  13. May 2, 2013 #12
    Then it follows that
    $$f(\sqrt{x})=\sqrt{x-4\sqrt{x}+3}=g(x)$$
    $$g'(x)=\frac{\sqrt{x}-2}{2\sqrt{x^2-4\sqrt{x^3}+3x}}$$
    $$g'(x-1)=\frac{\sqrt{x-1}-2}{2\sqrt{(x-1)(x+2)-4\sqrt{(x-1)^3}}}$$

    Those last two functions need to be simplified still further . The problem I am having, however, is that each time I go about this I seem to end up going in circles. I've been thinking about trying to factor using fractional exponents, but I'm not sure if that would get me somewhere useful. Any advice?
     
  14. May 2, 2013 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't see any terribly useful ways to simplify those. Why do you think you have to simplify?
    You could factor a (x-1) out of the terms in the radical in the denominator of the second form but that really doesn't look all that much simpler.
     
    Last edited: May 2, 2013
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