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Math Help - How to solve the following (integral calculus) problem?

  1. #1
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    Question How to solve the following (integral calculus) problem?

    The derivitive of

    f(x) = e^(kx) -√2x (x is out of the square root)

    and the graph of f(x) has a stationary point at x=1 when k is a positive real number.

    1) show the value of k=log e(√2)

    2) if the graph of f(x) is passing through (0,0) find the rule foor f(x)


    note: in the part "k=log e(√2)" e is the base


    please explain me how to solve this,

    Thanks in advance!
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  2. #2
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    Thumbs up

    2) if the graph of f(x) is passing through (0,0) find the rule foor f(x)



    1. Differentiate f(x) to get f'(x) = e^(kx) -x√2, i think is what you mean?

    Set f'(x) = 0, x=1, and then solve for k,
    Thus, e^(k) = √2
    and log e^k = k = log e(√2) = √2


    2. I'm not sure what you mean by rule floor, below i have integrated the function as that was in your description,

    f(x) = [(1/k)*(e^(kx))]-[((x^2)√2)/2] + c where k is as above
    using initial conditions, x=0, f(x)=y=0, we find c=-1/k

    Hope this helps!
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