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Find f(x) from f'(x)

  1. Sep 5, 2015 #1

    adjacent

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    1. The problem statement, all variables and given/known data
    A function ##f##, defined for ##x \in \mathbb{R}, x > 0## is such that:
    ##f'(x)=x^2-2+\frac{1}{x^2}##
    a Find the value of ##f''(x)## at ##x=4##
    b Given that ##f(3)=0## find ##f(x)##

    2. Relevant equations

    3. The attempt at a solution
    a

    ##f''(x) = 2x-2x^{-3}##
    ##f''(4)=2(4)-2(4^{-3})=\frac{225}{32}##
    b
    I haven't studied integration yet. So I have no idea how to do this....How am I supposed to do this?
     
  2. jcsd
  3. Sep 5, 2015 #2

    Titan97

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    You have to use integration. ##f(x)=\int f'(x) dx##
    The value of f(3) is given so that you can calculate the constant of integration.
     
  4. Sep 5, 2015 #3
    Let's denote by equality up to a constant by the sign [itex]\backsimeq[/itex], then $$f(x)\backsimeq\int f'(x)\,\mathrm{d}x=\int\left\{x^2-2+\frac{1}{x^2}\right\} \,\mathrm{d}x,$$ then use [itex]f(3)=0[/itex] to find the meant constant after calculating the integral -- of course.

    Crash Course on Basic Integration:

    And to help you with the integration, use the following rules: (use them as given for now, you'll learn as to why they hold later on in your courses)

    Let [itex]f(x),g(x)[/itex] and [itex]h(x)[/itex] be integrable functions (basically, functions which you can integrate) then: $$\int\big\{f(x)+g(x)+h(x)\big\}\,\mathrm dx=\int f(x)\,\mathrm dx+\int g(x)\,\mathrm dx+\int h(x)\,\mathrm dx.$$ Use it in this case where [itex]f:x\mapsto x^2[/itex], [itex]g:x\mapsto-2[/itex] and [itex]h:x\mapsto\frac1{x^2}[/itex].

    Integration is sometimes pictured as the inverse of differentiation. When you integrate a certain function [itex]\varphi[/itex] you ask yourself: "which function which when differentiated gives me [itex]\varphi[/itex]?" To integrate a constant function of the form [itex]f:x\mapsto a[/itex], then it is clear that the family of functions which when differentiated gives us [itex]a[/itex] are those of the form [itex]x\mapsto ax+C[/itex] -- as you can check. So in general $$\int a\,\mathrm dx=ax+C.$$
    As for functions of the form [itex]x\mapsto x^n[/itex] where [itex]n\in\mathbf Z\setminus\{-1\}[/itex], then the general rule is (why? Try to integrate the right-hand side) $$\int x^n\,\mathrm dx=\frac{x^{n+1}}{n+1}+C.$$
    These rules will be necessary for you to figure out the integral of [itex]x\mapsto x^2-2+\frac{1}{x^2}[/itex].
     
    Last edited: Sep 5, 2015
  5. Sep 5, 2015 #4

    bhobba

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    By definition the anti-derivative is any function that when differentiated gives the function you want the anti-derivative of. To find an anti-derivative of your function note an anti-derivative of x^n is 1/n+1 x^n+1 as can be checked by differentiating it.

    Now you need to know something that is fairly intuitive but actually requires some advanced math to prove (you will learn the proof when you study analysis) that all anti-derivatives have the form F(x) + C where C is a constant and F(x) is any anti-derivative. You can see by differentiating it you get f(x) that F(x) is the anti-derivative of. Such is called the indefinite integral.

    To answer your question figure out the indefinite integral and determine C using f(3) =0.

    Personally I don't think questions like that should be asked until you have studied integration.

    Thanks
    Bill
     
    Last edited: Sep 5, 2015
  6. Sep 5, 2015 #5

    epenguin

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    You don't have to be told about integration to solve this problem. You don't have to know it is integration.

    You show you know how to differentiate xn, x to a power, and a function made of a sum of only such things.

    The result of differentiating x to a power involves x to a different power.

    So given x to some power what would you have to differentiate to get it? Not so difficult surely.
     
    Last edited: Sep 5, 2015
  7. Sep 5, 2015 #6

    adjacent

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    Thanks everyone, I have figured it out now :)
     
  8. Sep 5, 2015 #7

    PeroK

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    If you know the derivative of a function and the function value at a given point, then you have uniquely defined function. A simple "proof" is to consider you have two such functions: ##f## and ##g## with ##f'(x) = g'(x)## for all ##x## and ##f(a) = g(a)## for a given ##a##.

    Now consider the function ##h = f-g##.

    ##h(a) = 0## and ##h'(x) = 0## for all ##x##, hence, clearly, ##h(x) = 0## for all ##x## and therefore ##f = g##

    This shows that if you can guess a function from its derivative and a given value, then you know that's the only possible such function.

    It's good to think this through "intuitively", in my opinion, rather than rely on a new technique like integration to crack the problem. You could even claim to be solving a "differential" equation here, just by using basic calculus and a bit of intelligent guesswork.
     
  9. Sep 5, 2015 #8

    epenguin

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    Don't tell us you have got the answer, tell us the answer you have got.
     
  10. Sep 5, 2015 #9

    bhobba

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    I think once the OP studies analysis exactly how clear that is will be called into question.

    IMHO best to avoid clearly etc at this level lest when you study analysis you don't get what its about. It may seem like a small point but when I did it we had people repeating it for the third time who still didn't get that what's clear to your intuition can lead to problems and that is the whole reason for analysis.

    To the OP - its OK to appeal to intuition at this stage - but intuition can lead you astray. Before the careful treatment of such things you had silly results like 1 - 1 + 1 -1 +1 .... = 1/2. Avoiding such requires a careful axiomatic development and was one of the great achievements of the 19th century. It's the branch of math called analysis - colloquially called doing your epsilonics for reasons that will be clear when you do it - which I of course hope you do. The university I studied at under constant pressure from students who hated it (they thought it was mind games) removed it as a graduation requirement which I personally thought was a pity.

    Thanks
    Bill
     
    Last edited: Sep 5, 2015
  11. Sep 6, 2015 #10

    adjacent

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    Not sure if I should solve this intuitively or using Integration now (Learned just today) ;)
    I think of indefinite integration as the opposite of differentiation .. So I guess doing it intuitively would also be just that.
    Integrating ##f'(x)## gives ##f(x) = \frac{x^3}{3}-2x-\frac{1}{x}+C##
    Given that ##f(3)=0##, I found C to be ##2 \frac{2}{3}##. So ##f(x)=\frac{x^3}{3}-2x-\frac{1}{x} + 2 \frac{2}{3}##
    Thanks all once again :)
     
  12. Sep 6, 2015 #11
    You made a mistake while calculating the constant:
    $$\begin{align}f(3)=0&\iff\frac{3^3}{3}-2\cdot3-\frac13+C=0\\&\iff\frac83+C=0\\&\iff C=-\frac83.\end{align}$$
     
    Last edited: Sep 6, 2015
  13. Sep 6, 2015 #12

    epenguin

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    An answer seems to have emerged but following up #7 let me repeat that for question b most of what has been said is not wrong, but irrelevant.

    Once given that the student knows rules for e.g. given f(x) to obtain f'(x), then given f'(x) he should be able to obtain f(x) from the same rules. For this:

    • He does not need intuition
    • Does not need these doctrines about integration
    • Does not need to know the concept nor the word 'integration'
    • Does not need to know anything about differention either
    • Does not need to know what f'(x) means or what anyone says it means
    • Also because it does not need to mean anything
    • Only needs to know, as he has shown he does, a subset of the rules for obtaining f' given f
    • In particular the rule for getting f'(axn) and f'(x + y)
    • Does not need to know what axn means, nor even what + means, they do not have to mean anything.
    That is, this question is not about calculus, it is algebraic, purely about consistent symbolic manipulation.
    I could almost say pure logic but an abstract algebraist would probably say you'd be unconsciously using some laws you've never questioned of arithmetic or algebra, of maybe of addition, subtraction, well at least of a multiplication and division. :oldwink::oldwink::oldwink:

    To emphasize the nature of the question you don't need to use the notation f' , you could just as well use any formula that has f and other symbols in it, e.g a rule like φ(f, a, n, x) = (f, na, n-1, x) or something even more minimal, and question something like: given (a, n, x) or (2, 2, 3) what is φ?
     
    Last edited: Sep 6, 2015
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