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Math Help - [f(x+h) - f(x)]/h Help!

  1. #1
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    [f(x+h) - f(x)]/h Help!

    Hey guys, my math prof gave me a problem i'm not sure exactly how to solve.

    f(x) = (x+1)^2
    lim h->0

    I'm guessing the pattern is: [f(x+h) - f(x)]/h

    so you would get this: [(x+1+h)^2 - (x+1)^2]/h

    and the next step is: [(x+1)^2 + 2(x+1)h + h^2 - (x+1)^2]/h

    i dont get how to go from the first step to the second step =/

    (i tried to use the math tags but they didnt come out right, sorry for not formatting them correctly)
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  2. #2
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    Here is the way LaTeX works at this site: [tex]\dfrac{f(x+h)-f(x)}{h} [/tex] gives  \dfrac{f(x+h)-f(x)}{h}


    I suggest using f(x)=x^2+2x+1
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  3. #3
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    i keep trying to use it but i get a syntax error. Here's a picture of what i dont understand. How do you get from the first line to the second line?

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    MHF Contributor harish21's Avatar
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    (x+1+h)^2

    Consider this term as (a+b)^2 ; where a = x+1 and b = h

    (a+b)^2 = a^2+2ab+b^2

    so,

    (x+1+h)^2=(x+1)^2+2(x+1)h+h^2
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  5. #5
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    thanks! i understand how they got from the second to the third line, but how do you get from the third line to the fourth line?
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  6. #6
    MHF Contributor harish21's Avatar
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    \dfrac{2(x+1)h+h^2}{h} = \dfrac{h[2(x+1)+h]}{h}

    the h on the numerator and the h on the denominator cancel each other out and you are left with

    \displaystyle{\lim_{h \to 0} (2(x+1)+h)}
    Last edited by harish21; October 28th 2010 at 09:40 AM.
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  7. #7
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    Using ‘\displaystyle’ makes the limit display correctly.
    [tex] \displaystyle\lim_{h \to 0} (2(x+1)+h) [/tex] gives  \displaystyle\lim_{h \to 0} (2(x+1)+h)
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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Plato View Post
    Using ‘\displaystyle’ makes the limit display correctly.
    [tex] \displaystyle\lim_{h \to 0} (2(x+1)+h) [/tex] gives  \displaystyle\lim_{h \to 0} (2(x+1)+h)
    Did the Editing.

    Thanks for the input.
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  9. #9
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    thanks guys! this forum is such a great help
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  10. #10
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    Here is how I always do it:

    \displaystyle f'(x) = \lim_{h\to{0}}\frac{(1+x+h)^2-(1+x)^2}{h}

    Let 1+x+h = a and 1+x = b,

    then h = a-(1+x) = a-b.

    Also h\to{0} \Leftrightarrow {a-b}\to{0} \Rightarrow a\to{b}

    Thus \displaystyle f'(x) = \lim_{a\to{b}}\frac{a^2-b^2}{a-b} = \lim_{a\to{b}}\frac{(a-b)(a+b)}{a-b}[/Math]<br /> <br />
 [Math]\Rightarrow \displaystyle f'(x) = \lim_{a\to{b}}(a+b) = 2b = 2(1+x).
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