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Math Help - Rate of change?

  1. #1
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    Rate of change?

    Hey guys, here's another one.. I can't believe I don't remember how to do any of this..Thanks.
    http://oi44.tinypic.com/33dbec4.jpg
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  2. #2
    MHF Contributor Amer's Avatar
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    Re: Rate of change?

    f(x) = \frac{x}{x^2+2}

    determine \frac{f(x+h)-f(x)}{h} in the simplest form

    f(x+h) = \frac{x+h}{(x+h)^2+2}

    and f(x) = \frac{x}{x^2+2}
    so

    \frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{x}{x^2+2}}{h}

    try to cancel the h in the denominator
    Last edited by Amer; December 7th 2011 at 07:50 PM. Reason: fixing f(x)
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  3. #3
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    Re: Rate of change?

    If f(x)= \frac{x}{x^2+2} then f(x+h)= \frac{x+h}{(x+h)^2+2}

    Now go get em!
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  4. #4
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    Re: Rate of change?

    You guys are the best. Thanks so much! I'll post the answers when I get them
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  5. #5
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    Re: Rate of change?

    Quote Originally Posted by Amer View Post
    f(x) = \frac{x}{x^2+2}

    determine \frac{f(x+h)-f(x)}{h} in the simplest form

    f(x+h) = \frac{x+h}{(x+h)^2+2}

    and f(h) = \frac{h}{h^2+2}
    so

    \frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{h}{h^2+2}}{h}

    try to cancel the h in the denominator
    What you have done is \displaystyle \begin{align*} \frac{f(x + h) - f(h)}{h} \end{align*} , not \displaystyle \begin{align*} \frac{f(x + h) - f(x)}{h} \end{align*} ...
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  6. #6
    MHF Contributor Amer's Avatar
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    Re: Rate of change?

    Quote Originally Posted by Prove It View Post
    What you have done is \displaystyle \begin{align*} \frac{f(x + h) - f(h)}{h} \end{align*} , not \displaystyle \begin{align*} \frac{f(x + h) - f(x)}{h} \end{align*} ...
    yeah, thanks for pointing that
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  7. #7
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    Re: Rate of change?

    Quote Originally Posted by Amer View Post
    yeah, thanks for pointing that
    Definitely. Thank you! I think I can figure it out from here
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  8. #8
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    Re: Rate of change?

    Quote Originally Posted by Amer View Post
    f(x) = \frac{x}{x^2+2}

    determine \frac{f(x+h)-f(x)}{h} in the simplest form

    f(x+h) = \frac{x+h}{(x+h)^2+2}

    and f(x) = \frac{x}{x^2+2}
    so

    \frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{x}{x^2+2}}{h}

    try to cancel the h in the denominator
    Okay.. I'm lost.

    Somehow I got to:

    (x^3 + hx^2 + 2x +2h) - ((x^2 +hx)^2 +2x)) / h

    Am I even on the right track?
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  9. #9
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    Re: Rate of change?

    What happened to the denominators?

    \frac{\frac{x+h}{(x+h)^2+ 2}- \frac{x}{x^2+ 2}}{h}
    Get common denominators:
    \frac{(x+h)(x^2+2)}{h(x^2+2)((x+h)^2+ 2)}- \frac{x((x+h)^2+ 2)}{h(x^2+ 2)((x+h)^2+ 2)}
    \frac{x^3+ hx^2+ 2x+ 2h- (x^3+ 2hx^2+ (h^2+2)x}{h(x^2+ 2)((x+h)^2+ 2)}

    Reduce the numerator. you can pretty much ignore the denominator. Once you cancel that original "h", the rest of the denominator goes to (x^2+ 2)^2.
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  10. #10
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    Re: Rate of change?

    Quote Originally Posted by HallsofIvy View Post
    What happened to the denominators?

    \frac{\frac{x+h}{(x+h)^2+ 2}- \frac{x}{x^2+ 2}}{h}
    Get common denominators:
    \frac{(x+h)(x^2+2)}{h(x^2+2)((x+h)^2+ 2)}- \frac{x((x+h)^2+ 2)}{h(x^2+ 2)((x+h)^2+ 2)}
    \frac{x^3+ hx^2+ 2x+ 2h- (x^3+ 2hx^2+ (h^2+2)x}{h(x^2+ 2)((x+h)^2+ 2)}

    Reduce the numerator. you can pretty much ignore the denominator. Once you cancel that original "h", the rest of the denominator goes to (x^2+ 2)^2.

    Uhh. Alright. I'm pretty confused on this.. So I ended up with:
    2 - x^2 - hx / x^4 + 2hx^3 + 2h^2 + h^2x^2 + 4x^2 + 4hx +4.

    I think.. I overworked it..
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