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Math Help - Difference Quotient, fifth power

  1. #1
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    Difference Quotient, fifth power

    Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

    The exact wording is:

    Let f(x) = 2x^5-3x^2+1

    What is the difference quotient f' of f?


    I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Shonard View Post
    Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

    The exact wording is:

    Let f(x) = 2x^5-3x^2+1

    What is the difference quotient f' of f?


    I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?
    The difference quotient can be defined (geometrically) as the slope of a secant line through the graph of a function f(x). In other words

    The difference quotient: \frac{\Delta{y}}{\Delta{x}}.

    But your problem is asking you to find f'(x)=\lim_{\Delta{x}\to0}\frac{\Delta{y}}{\Delta{  x}}.

    Do you know how to do this?
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  3. #3
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    Quote Originally Posted by Shonard View Post
    Hello, I have a question regarding difference quotients. I haven't been able to find a single example of a difference quotient to an exponent higher than 3, so I was hoping someone here could give me a hand.

    The exact wording is:

    Let f(x) = 2x^5-3x^2+1

    What is the difference quotient f' of f?


    I solved it to the best of my knowledge, but the answer is rather long which causes me to doubt it. Any thoughts?
    Do you know the Binomial Theorem? It will help you to expand (x + h)^5.

    (a + b)^n = \sum_{r = 0}^n {n\choose{r}} a^{n - r}b^r, provided n is a positive integer.


    So that would mean (x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5.


    Do you think now you can evaluate \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}?
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  4. #4
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    I know how to solve the difference quotient through the formula..

    f(x+h)-f(x)/h

    the fifth power is making it tricky though
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  5. #5
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    you are correct. the answer would be rather lengthy.
    f(x)=2x^5-3x^2+1
    f(x+h)=2(x+h)^5-3(x+h)^2+1

    f'(x)=lim h->0 f(x+h)-f(x)/h

    u will have to expand (x+h)^5 using binomial theorem which makes the problem lengthy. at last h would get cancelled. and u will get f'(x)=10x^4-6x
    which is nothing but the differentiation of f(x).
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  6. #6
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    Read my post.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Shonard View Post
    I know how to solve the difference quotient through the formula..

    f(x+h)-f(x)/h

    the fifth power is making it tricky though
    As prove it mentioned, if you insist upon differentiating this with the definition, then the binomial theorem will help you. Or, perhaps, Pascal's triangle...
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Do you know the Binomial Theorem? It will help you to expand (x + h)^5.

    (a + b)^n = \sum_{r = 0}^n {n\choose{r}} a^{n - r}b^r, provided n is a positive integer.


    So that would mean (x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5.


    Do you think now you can evaluate \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}?
    Ok I attempted to evaluate using the expansion of (x+h)^5 and got:

    10x^4 + 20x^3h + 20x^2h^2 + 10xh^3 + 2h^4 - 2x - h

    does this seem correct?
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  9. #9
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    f(x) = 2x^5 - 3x^2 + 1.


    f(x + h) = 2(x + h)^5 - 3(x + h)^2 + 1

     = 2(x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5) - 3(x^2 + 2xh + h^2) + 1

     = 2x^5 + 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 3x^2 - 6xh - 3h^2 + 1.


    f(x + h) - f(x) = 2x^5 + 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 3x^2 - 6xh - 3h^2 + 1
     - (2x^5 - 3x^2 + 1)

     = 10x^4h + 20x^3h^2 + 20x^2h^3 + 10xh^4 + 2h^5 - 6xh - 3h^2.


    \frac{f(x + h) - f(x)}{h} = 10x^4 + 20x^3h + 20x^2h^2 + 10xh^3 + 2h^4 - 6x - 3h.


    Now let h \to 0 and we find

    f'(x) = 10x^4 - 6x.
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  10. #10
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    Ah, I see my mistake thank you.
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