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Math Help - Help!! Difference Quotients.

  1. #1
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    Post Help!! Difference Quotients.

    Find the difference quotient of f, that is, find [f(x + h) - f(x)]/h, where h cannot = 0, for the given function.

    Be sure to simplify.

    f(x) = (x+3)^2
    f(x)=3x^2-5x+4
    g(x)=x-3

    NOTE: This question comes from a pre-calculus textbook and so the explanation must be in pre-calculus terms and NOT calculus.

    Thankx.
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  2. #2
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    Quote Originally Posted by AbbbyyJackson View Post
    Find the difference quotient of f, that is, find [f(x + h) - f(x)]/h, where h cannot = 0, for the given function.

    Be sure to simplify.

    f(x) = (x+3)^2
    f(x)=3x^2-5x+4
    g(x)=x-3

    NOTE: This question comes from a pre-calculus textbook and so the explanation must be in pre-calculus terms and NOT calculus.

    Thankx.

    \frac{(x+h+3)^2-(x+3)^2}{h}

    =\frac{[(x+3)+h]^2-(x=3)^2}{h}

    =\frac{(x+3)^2+2h(x+3)+h^2-(x+3)^2}{h}

    =\frac{2h(x+3)+h^2}{h}

    =2(x+3)+h
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  3. #3
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    Quote Originally Posted by adkinsjr View Post
    \frac{(x+h+3)^2-(x+3)^2}{h}

    =\frac{[(x+3)+h]^2-(x=3)^2}{h}

    =\frac{(x+3)^2+2h(x+3)+h^2-(x+3)^2}{h}

    =\frac{2h(x+3)+h^2}{h}

    =2(x+3)+h
    Thankx.
    But Can You Explain What You Did.
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  4. #4
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    Quote Originally Posted by AbbbyyJackson View Post
    Thankx.
    But Can You Explain What You Did.
    Sure, If we have f(x)=(x+3)^2 then we simply replace the x with whatever value is in the parenthesis on the right side. For example f(2)=(2+3)^2, or f(a)=(a+3)^2. So what if I want to input x+h? What I do is replace the x in (x+3)^2 with x+h. Therefore:

    f(x+h)=[(x+h)+3)]^2

    The numerator in the difference quotient is f(x+h)-f(x). We substitute what we know:

    f(x+h)-f(x)=[(x+h)+3]^2-(x+3)^2.


    The (x+3)^2 suggest that we should rearrange the first term in the numerator.

    [(x+3)+h]^2-(x+3)^2. Notice that the first term is just a binomial. Therefore we can take it's square as (x+3)^2+2h(x+3)+h^2

    The numerator becomes:

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

    Notice the cancellation of the (x+3)^2 term:

    =2h(x+3)+h^2

    The difference quotient is now:

    \frac{2h(x+3)+h^2}{h}

    The h in the denominator cancels:

    =2(x+3)+h
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