Calculus

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Consider the function f(x)=xe^2x. Find the intervals on which f(x) is concave up and the intervals where it is concave down. Show a sign graph.

Please show work in detail so I can follow for future reference. Thanks!

  • Calculus -

    The curve is concave up if f '' (x) is positive and
    concave down if f '' (x) is negative

    f ' (x) = x(2)e^(2x) + e^(2x)
    = e^(2x) (2x + 1)

    f '' (x) = (2x+1)(2)e^(2x) + 2 e^(2x)
    = 2 e^(2x) ((2x+1) + 1)
    = 2 e^(2x) (2x + 2)
    = 4 e^(2x) (x+1)
    since 4e^(2x) > 0 for all values of x, we just have to look at
    x + 1

    x+1 > 0 for x > -1
    x+1 < 0 for x < -1

    so the curve is concave up for all values of x > -1
    and concave down for all values of x < -1

    check the graph by Wolfram
    http://www.wolframalpha.com/input/?i=plot+x+e%5E%282x%29%2C+-3+%3C+x+%3C+1+%2C+-.25+%3C+y+%3C+2

    notice there is a point of inflection at x= -1

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