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Math Help - Derivative help

  1. #1
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    Derivative help

    I am stuck on a few questions, can someone please help.

    Let
    f(x)=2*x^2+5*x - 3
    and let x0= 5.

    A. The average rate of change of f between x= 5 and x= 5.14 equals ____


    B. The average rate of change of f between x= 5 and x= 5.014 equals ___


    C. The average rate of change of f between x= 5 and x= 5.0014 equals ____


    D. The instantaneous rate of change of f at x= 5 equals ____
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  2. #2
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    Quote Originally Posted by lilwayne View Post
    I am stuck on a few questions, can someone please help.

    Let
    f(x)=2*x^2+5*x - 3
    and let x0= 5.

    A. The average rate of change of f between x= 5 and x= 5.14 equals ____


    B. The average rate of change of f between x= 5 and x= 5.014 equals ___


    C. The average rate of change of f between x= 5 and x= 5.0014 equals ____


    D. The instantaneous rate of change of f at x= 5 equals ____
    You really should post your efforts, thoughts or where you are stuck...

    Hint: the average rate of change is the slope of the straight line joining the two points,

    hence it is \displaystyle\frac{f(x_2)-f(x_1)}{x_2-x_1}

    The instantaneous rate of change is given by the slope of the tangent at a particular f(x).
    The derivative is the slope of the tangent evaluated at the given x.
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  3. #3
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    that's what i don't get, what do i use the formula for, like the f(x2) comes from where? If i figure out how to do the first one, i could probably do the rest by my self.
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  4. #4
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    Quote Originally Posted by lilwayne View Post
    I am stuck on a few questions, can someone please help.

    Let
    f(x)=2*x^2+5*x - 3
    and let x0= 5.

    A. The average rate of change of f between x= 5 and x= 5.14 equals ____


    B. The average rate of change of f between x= 5 and x= 5.014 equals ___


    C. The average rate of change of f between x= 5 and x= 5.0014 equals ____


    D. The instantaneous rate of change of f at x= 5 equals ____
    Here's how it works...

    (A)

    x_2>x_1\Rightarrow\ x_1=5,\;\;\;x_2=5.14

    f\left(x_1\right) is the result you get when you place the value x_1 into f(x)

    in other words, just replace x with the value x_1 and calculate the result.

    Do the same for x_2

    Then you can calculate the average rate of change.
    It's the same as calculating the slope of a line between two given points

    \displaystyle\ slope=\frac{y_2-y_1}{x_2-x_1}

    Hence, the solution for (A) is

    \displaystyle\frac{\left[2(5.14)^2+5(5.14)-3\right]-\left[2(5)^2+5(5)-3\right]}{5.14-5}

    Use your calculator for that.

    (B) and (C) are evaluated in the same way.

    How are you a calculating derivatives for part (D) ?
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  5. #5
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    i calculated 46.71 and punched that in, but it said incorrect :S
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  6. #6
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    oops calculated wrong, got it right now lol thanks
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