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Math Help - Differentiation mixed with Line Equation

  1. #1
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    Differentiation mixed with Line Equation

    the problems goes like this.

    f(x) = x^5 + 5e^x , find f'(5)

    So I did that and found f'(x) = 5x^4 + 5e^x. Finding f'(5) = 3867

    Use this to find the equation of the tangent line to the curve y = x^5 + 5e^x at the point (a,f(a)) when a=5 . The equation of this tangent line can be written in the form y=mx+b where m is: 3867

    I found out that part easily, by putting 5 into f'(x)

    But the last part of the question is:

    and where b is: ?

    I have no idea. I tried putting 5 into y = mx+b and adding in the slope but that gives me .2 which is an incorrect answer. Been trying to figure this one out forever!

    Please help!!

    Thanks,

    -MoMo
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  2. #2
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    To find b in y=mx+b

    make y = f(5) and x = 5

     f(x) = x^5 + 5e^x

     f(5) = 5^5 + 5e^5 = \dots

    after you have those values together with your gradient solve for b.

    To find b in f(5)=m\times5+b
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  3. #3
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    well so f(5) = 5^5 + 5e^5 = 3867

    m = 3867 as well

    therefore if I do it

    f(5) = m * 5 + b

    I'm getting 3867 = 3867 * 5 + b

    which gives me b = .2

    But that is incorrect? Am I missing something? or perhaps my interpretation is wrong..
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  4. #4
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    Quote Originally Posted by mowgli View Post

    I'm getting 3867 = 3867 * 5 + b

    which gives me b = .2
    This is not correct. Look closer.
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  5. #5
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    i know i will feel really stupid when i see it but I just can't.

    f(5) = 3867 and I already discovered that m = 3867 by inputing 5 into f'(x)

    I dont see how I can rearrange y = mx+b any other way to discover b.

    I feel so frustrated, because I thought it was simple but I dont know why i can't put my head around this.
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  6. #6
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     3867 = 3867 \times 5 + b

     3867 = 19335 + b

     3867 - 19335 = b

     b= -15468
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  7. #7
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    LOL OMG!!! i've been dividing 3867 by 19355!! Epic Fail T_T

    Thank you sooo much!
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