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Math Help - Tangent Line

  1. #1
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    Tangent Line

    Find an equation of the line that is tangent to the graph of f for the given value of x.

    f(x)=(x^2-3)^5(2x-1)^3 x=2

    I got the derivative 2(1-2x)^2(x^2-3)^4(13x^2-5x-9)

    Then I plugged in 2 for x and I got 594

    I'm not sure if this is right and where to go next..thanks
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  2. #2
    MHF Contributor Amer's Avatar
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    f(x)=(x^2-3)^5(2x-1)^3...  x=2

    f'(x)=5(2x)(x^2-3)^4(2x-1)^3+3(2)(2x-1)^2(x^2-3)^5

    f'(2)=5(4)(4-3)^4(3)^3+6(4-1)^2(4-3)^5=20(27)+54=540+54=594

    now

    f(x)=(x^2-3)^5(2x-1)^3...f(2)=27

    the equation of the tangent

    594=\left(\frac{(y-27)}{(x-2)}\right)
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