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Math Help - Finding horizontal tangent lines

  1. #1
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    Finding horizontal tangent lines

    1. find each value of x at which f(x) =x(2x-8)^3 has a horizontal tangent line.

    y'=(2x-8)^3 + 6x(2x-8)^2

    set = to 0
    (2x-8)^3 + 6x(2x-8)^2=0

    factoring out
    = [(2x-8)^2][(2x-8)+(6x)]

    = (2x-8)=0
    x=4
    = (6x)=0
    x=0
    Key says the answer is x=1, and x=4.
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  2. #2
    MHF Contributor
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    Talking

    Quote Originally Posted by hazecraze View Post
    = [(2x-8)^2][(2x-8)+(6x)]

    = (2x-8)=0
    How did you arrive at this equality? How did you go from "2x - 8 = 0" to "2x = 0 and 6x = 0"?

    Instead, start with the factorization:

    (2x\, -\, 8)^2 (8x\, -\, 8)

    Set this equal to zero, and solve both factors, using the methods you learned back in algebra.
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  3. #3
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    How did I not see that? Got it now though.
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