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Math Help - Chain rule/Tangent Line

  1. #1
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    Chain rule/Tangent Line

    Hi,
    Would anyone be able to show me how to go about obtaining the average of all values of x where the tangent line is horizontal? The first part (a) of the question asked me to find an equation for the tangent line to the graph of f at x=1. I determined that y=-2/81 and the slope to be -34/243. My equation turned out to be y=-34/243x + 40/243. As far as part b goes, I have no clue what to do. Any help would be greatly appreciated. I tried including a picture of what I've done so far (hope you can see it). Thanks! Ana


    f(x) = 2x/(2-5x)4

    x=1 (this is the given information)


    B) Find the average of all values of where the tangent line is horizontal. If there are no such values, enter -1000.

    Chain rule/Tangent Line-ptdc0018.jpg
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  2. #2
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    Re: Chain rule/Tangent Line

    Quote Originally Posted by phenol View Post
    Hi,
    Would anyone be able to show me how to go about obtaining the average of all values of x where the tangent line is horizontal? The first part (a) of the question asked me to find an equation for the tangent line to the graph of f at x=1. I determined that y=-2/81 and the slope to be -34/243. My equation turned out to be y=-34/243x + 40/243. As far as part b goes, I have no clue what to do. Any help would be greatly appreciated. I tried including a picture of what I've done so far (hope you can see it). Thanks! Ana


    f(x) = 2x/(2-5x)4

    x=1 (this is the given information)


    B) Find the average of all values of where the tangent line is horizontal. If there are no such values, enter -1000.

    Click image for larger version. 

Name:	PTDC0018.jpg 
Views:	11 
Size:	150.3 KB 
ID:	24963
    Your tangent line equation is ok if you did the arithmetic correctly.

    Simplify your derivative and set it equal to 0, then solve for all possible value(s) of x. Note that there is only one value of x where the tangent line is horizontal ... which would be equal to the average of that single value.
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  3. #3
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    Re: Chain rule/Tangent Line

    Quote Originally Posted by phenol View Post
    Hi,
    Would anyone be able to show me how to go about obtaining the average of all values of x where the tangent line is horizontal? The first part (a) of the question asked me to find an equation for the tangent line to the graph of f at x=1. I determined that y=-2/81
    This is incorrect. y= +2/81. The numerator, 2x= 2, is positive and the denominator is positive because it is a 4th power.

    and the slope to be -34/243.
    This is completely incorrect. First, the derivative is positive. Second, the numerator is wrong. You do have the correct denomiator.

    My equation turned out to be y=-34/243x + 40/243. As far as part b goes, I have no clue what to do. Any help would be greatly appreciated. I tried including a picture of what I've done so far (hope you can see it). Thanks! Ana
    The tangent will be horizontal when the derivative is 0. For what values of x is that true? Average all those x values.


    f(x) = 2x/(2-5x)4

    x=1 (this is the given information)


    B) Find the average of all values of where the tangent line is horizontal. If there are no such values, enter -1000.

    Click image for larger version. 

Name:	PTDC0018.jpg 
Views:	11 
Size:	150.3 KB 
ID:	24963[/QUOTE]
    Thanks from phenol
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  4. #4
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    Re: Chain rule/Tangent Line

    I figured out part b, I came out with -2/15. Thanks for your help! Ana
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