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Math Help - horizontal tangent

  1. #1
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    horizontal tangent

    7. Find all values of x for the given function where the tangent line is horizontal.

    f(x) = root( x^2 + 18x + 86)

    9. Find derivative

    y = ( 6e^x) / ( 2e^x + 1)
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  2. #2
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    Quote Originally Posted by ArmiAldi View Post
    7. Find all values of x for the given function where the tangent line is horizontal.

    f(x) = root( x^2 + 18x + 86)

    9. Find derivative

    y = ( 6e^x) / ( 2e^x + 1)
    f(x) = \sqrt{ x^2 + 18x + 86} = (x^2+18x+86)^{\frac12} . Now calculate the first derivative:

    f'(x)= \frac12 \cdot (x^2+18x+86)^{-\frac12} \cdot (2x+18) = \frac{2x+18}{{\frac12}\cdot \sqrt{x^2+18x+86}}

    f'(x) = 0 if the numerator equals zero and the denominator is not zero:

    2x+18=0~\implies~x=-9

    f(x)=\frac{6e^x}{2e^x + 1} . Use quotient rule:

    f'(x) = \frac{(2e^x + 1) \cdot 6e^x - 6e^x \cdot 2e^x}{(2e^x+1)^2} = \frac{6e^x}{(2e^x+1)^2}
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  3. #3
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    Quote Originally Posted by earboth;111724[snip
    f(x)=\frac{6e^x}{2e^x + 1} . Use quotient rule:

    f'(x) = \frac{(2e^x + 1) \cdot 6e^x - 6e^x \cdot 2e^x}{(2e^x+1)^2} = \frac{6e^x}{(2e^x+1)^2}
    Alternatively, note that

    f(x)=\frac{6e^x}{2e^x + 1} = \frac{(6e^x + 3) - 3}{2e^x + 1} = 3 - \frac{3}{2e^x + 1}.

    The derivative of the first term is obviously zero. To get the derivative of the second term, you can use either the chain or quotient rule.
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