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Math Help - find the exact values for the horizontal tangent

  1. #1
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    find the exact values for the horizontal tangent

    i just would like someone to check my work on this one i think ive done it correctly

    f(x)= x^3+x^2-x-17
    f' = 3x^2-2x-1

    3x^2-2x-1=0

    (3x+1)(x-1)=0

    x= 1, -1/3

    1^3+1^2-1-17=-16
    (-1/3)^3+(-1/3)^2-(-1/3)-17=-17.3


    (1, -16), (-1/3,-17.3)
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  2. #2
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    Quote Originally Posted by dat1611 View Post
    i just would like someone to check my work on this one i think ive done it correctly

    f(x)= x^3+x^2-x-17
    f' = 3x^2-2x-1
    Already this is incorrect f(x)=  x^3+x^2-x-17 \implies f'(x)=  3x^2+2x-1
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  3. #3
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    ok that was a stupid mistake so now i get

    (-1,-16), (1/3,-17.2)
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  4. #4
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    correct!
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