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Math Help - help on a calculus problem

  1. #1
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    help on a calculus problem

    find the change in elevation when X=2 for the path modeled by the function:
    f(x)=0.04(8x-x^2), where x and f(x) are measured in miles...how can this problem be estimated using precalculus and how can it be solved with calculus?

    any help would be appreciated thanks
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  2. #2
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    Quote Originally Posted by PMoNEY23 View Post
    find the change in elevation when X=2 for the path modeled by the function:
    f(x)=0.04(8x-x^2), where x and f(x) are measured in miles...how can this problem be estimated using precalculus and how can it be solved with calculus?

    any help would be appreciated thanks
    A rate of change is the slope of a curve at a point... which is the same as the slope of the line tangent to the curve at that point.

    So you could draw a tangent to the curve at the point x=2 and measure its slope using \frac{\textrm{Rise}}{\textrm{Run}}.

    You could also choose another point very close to 2, say 2.001, and work out the slope of the line that would cross the two points, using \frac{\textrm{Rise}}{\textrm{Run}}=\frac{y_2-y_1}{x_2-x_1}. This will give you a good approximation.

    Using calculus, if a function f(x)=ax^n, its derivative (rate of change at any point) is given by f'(x)=nax^{n-1}. Substitute x=2 into the derivative and this will be the slope of the curve at that point.
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  3. #3
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    how exactly can i find the derivative for that function? Can you plz explain in steps
    Thanks
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  4. #4
    Super Member 11rdc11's Avatar
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    f(x)=0.04(8x-x^2)

    Using the power rule that prove it mentioned

    f'(x)=nax^{n-1}

    with a being the coefficent and n the power

    f(x)= .32x - .04x^2

    f'(x) = 32 - .08x
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