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Math Help - For what x value does the tangent of the graph of f(x) = xe^2x have a slope = 0?

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    For what x value does the tangent of the graph of f(x) = xe^2x have a slope = 0?

    I said

    f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0


    Is this right?
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  2. #2
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    Quote Originally Posted by satx View Post
    I said

    f'(x) = m = 0 = 2xe^2x ---> x = 0/(2xe^2x) = 0


    Is this right?
    Your method is right although you need to use the product rule when finding f'(x)

    f(x) = xe^{2x}

    u = x \: \rightarrow \: u' = 1

    v = e^{2x} \: \rightarrow \: v' = 2e^{2x}

    f'(x) = u'v + v'u
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  3. #3
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    Wouldn't that equal 2?
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    Quote Originally Posted by satx View Post
    Wouldn't that equal 2?
    i believe you have, f'(x)=2e^{2x}x+e^{2x}
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  5. #5
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    to solve f'(x)=0
    f'(x)=2e^{2x}x+e^{2x}=e^{2x}(2x+1)=0\Leftrightarro  w 2x+1=0
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    Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???
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    Quote Originally Posted by Raoh View Post
    to solve f'(x)=0
    f'(x)=2e^{2x}x+e^{2x}=e^{2x}(2x+1)=0\Leftrightarro  w 2x+1=0
    So e^0 is 0? I'm so confused
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    Quote Originally Posted by satx View Post
    Ok then that would equal one then, if x = 0, no? (e^2(0))(0) + e^2(0)) = 0 + 1 = 1???
    i think you're trying to solve f'(x)=0
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  9. #9
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    Quote Originally Posted by Raoh View Post
    i think you're trying to solve f'(x)=0
    Well isn't the slope equal to the derivative?
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    Quote Originally Posted by satx View Post
    So e^0 is 0? I'm so confused
    well e^0=1,what i wrote means f'(-\frac{1}{2})=0 and the x you're searching for is -\frac{1}{2}.
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  11. #11
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    Quote Originally Posted by satx View Post
    Well isn't the slope equal to the derivative?
    yes
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