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Math-Calculus (check please)

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FOr the function f(x)=x^2e^x

Find the critical points, increasing decreasing, first and second derivative and concavity

I am having trouble finding the concavity >_<

First derivative: xe^x(x+2)
Second : e^x(x^2+4x+2)

Critical points from first derivate:
x=0,-2

Increasing: (-infinity,0) (-2, infinity)
Decreasing: (0,-2)

I can't figure out how to do the concavity part of it.

  • Math-Calculus (check please) -

    The "concavity" of the function comes from the second derivative, where there is a critical point.

    If the second derivative is positive at a point where the first derivative is zero, the function is "concave upward" there. If it is negative, it is concave downward.

  • Math-Calculus (check please) -

    Increasing: (-infinity,0) (-2, infinity)
    Decreasing: (0,-2)

    This does not make sense. Coming from -oo you get to x = -2 before you get to x = 0

    Now if I put in a large - number, say -100 for x
    I get (-100)^2 e^-100= 10^4 *3.72*10^-44
    which sure looks like 0
    If I put in x = -2 I get
    (-2)^2 e^-2 = .54
    so it increases from x = -oo to x = 0

    Now at x = 0, we know f(x) is zero, so it is decreasing from x = -2 to x = 0
    We can be sure that it then heads up to the right so increasing from x = 0 to x --> +oo

    Well, I suppose concave means curving upward. That would happen where the second derivative is +
    Now where is the second derivative zero?
    x^2+4x + 2 = 0
    x = [ -4 +/- sqrt(16 -8) ] /2
    = -2 +/- .5 sqrt 8
    = -2 +/- sqrt 2
    so the second derivative may change sign at -3.41 and at -.586
    in other words to the left and right of where the function is level at (-2,.54)

    To the left of -3.41 ,say x = -5, the second derivative is +, so curving up, concave from -oo to -3.41
    at x = -2, the second derivative is negative, so curves down from -3.41 to - .586
    Then at x = 0 the second derivative is + again so it starts curving up at x = -.586 and continues curving up from then on

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