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Help with these two exam questions

  1. Oct 11, 2007 #1
    I had these two questions in an exam and do not know how to find the answers...

    find (f-1)'(3) for f(x) = sqrt (x^3 + x^2 + x + 6)


    the int of x/(x + 4) dx

    If anyone can get me started on these and show which steps I have to take, that would be great...
  2. jcsd
  3. Oct 11, 2007 #2
    I've got an idea about the first 1, but not 100% on it. I think that the derivative of a function is 1/the derivative of the inverse function, so just calculate the df/dx for x = 3, then the answer is the reciprocal of the number you get.

    For the second question, ever thought about writing it as [itex] \int 1 dx - \int \frac{4}{x+4} dx [/itex] ?
  4. Oct 11, 2007 #3
    Well, to begin with, how did you transform the integral into that? I triend doing y = (x +4), but it didn't work.
  5. Oct 11, 2007 #4
    can you clarify what you meant for 1... You mean I should take the derivitative of that functiona and plug in 3 for x???
  6. Oct 11, 2007 #5


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    I assume that by (f-1) you mean the derivative of 1/f(x) (because if it were the inverse, the exercise would become quite messy and ugly). You can do this in two ways (three, actually):
    a) Write [tex]1/\sqrt{\cdots} = (\cdots)^{-1/2}[/tex] and use the familiar rule.
    b) Use the quotient rule
    c) Use a rule for [tex]\frac{d(1/f(x))}{dx}[/tex] if you know it (otherwise you can derive it using b)
  7. Oct 11, 2007 #6


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    I don't really understand what is going on in 1 but 2 looks simple

    [tex]\int \frac{x}{x+4} dx[/tex]

    take [tex]\frac{x}{x+4} = \frac{x+4-4}{x+4} = \frac{x+4}{x+4} -\frac{4}{x+4} = 1-\frac{4}{x+4}[/tex]

    so then

    [tex]\int \frac{x}{x+4} dx = \int(1- \frac{4}{x+4}) dx[/tex]
  8. Oct 11, 2007 #7
    I'm not entirely sure about the first part, but I think that if you find the derivative for f, then put 3 into the resulting equation, and then simply do (your answer)^-1 you will get the derivative of the function f^-1 at x = 3.

    The second part, all I did was break the integral up. Is the following not true?
    [tex] 1 - \frac{4}{x+4} = \frac{x}{x+4} [/tex]
    If it is, then the integral of one (with respect to x), must be the integral of the other (w.r.t x). I just transformed the integral by inspection, there isn't anything special going on there.
  9. Oct 11, 2007 #8
    yer. I'm assuming he means the inverse of the function. It's not exactly clear.
  10. Oct 11, 2007 #9


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    That is not true.
    As a counterexample, consider
    [tex]f(x) = x; g(x) = f(x)^{-1} = \frac{1}{x}[/tex].
    Clearly, [itex]f'(x) = 1[/itex], so by your argument we should have [itex]g'(x) = 1 / f'(x) = 1 / 1 = 1[/itex] for all x.
    But actually,
    [tex]g'(x) = - \frac{1}{x^2}[/tex]
    which is in general not equal to 1 (in fact, if x is real, it will never hold).

    You really need the quotient rule or "power rule" (I don't know what it's called - if it has a name - I mean the rule dx^n/dx = n x^(n-1)) to find the derivative of 1/f(x).
  11. Oct 11, 2007 #10

    I said that I wasn't sure about it - Thats my disclaimer :P
  12. Oct 11, 2007 #11
    I meant the derivative of the inverse of the function at 3... Thanks for the help with the second question - I can't believe they did that +4 -4 thing!!!!

    I would still like help with 1, though. The answer is supposed to be 1.
    Last edited: Oct 11, 2007
  13. Oct 11, 2007 #12

    Does that help at all?
  14. Oct 11, 2007 #13
    honestly, no.
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