Consider f(x) = x^2e^-x. Find the intervals on which f is increasing and decreas

Question

Consider f(x) = x^2e^-x. Find the intervals on which f is increasing and decreasing, and find the local extrema of f.

Explanation / Answer

you surely know that f ' = x e^(-x) [ 2 - x ] and thus relative min when x = 0 { f = 0 } and relative max at x = 2 { f = 4 e^(-2) } f ' ' = e^(-x) [ 2 - x ] - x e^(-x) [ 2 - x ] - x e^(-x) = e^(-x) { 2 - x - x [ 2 - x] - x } x² - 4 x + 2 = [ x-2]² - 2---> x = 2 - v2 is where f ' is changing the fastest df/dx = (x^2)(e^x) + (2x)(e^x) = (x^2 + 2x)(e^x) f is increasing when df/dx > 0 Or (x^2 + 2x)(e^x) > 0 Dividing by e^x, x^2 + 2x > 0 x(x + 2) > 0------------------(1) The critical points for the above in equality are x = 0, x = -2 At x = 0 and at x = -2, the LHS of (1) becoms 0. But we want greater than 0. So x = - 2, x = 0 are not valid. Consider the following cases: - 1. x < -2 x is negative x+2 is negative x(x+2) is positive. Condition (1) is satisfied. 2. - 2 0 x is positive x+2 is positive x(x+2) is positive Condition (1) is satisfied. So condition (1) is satisfied in following intervals: - x < - 2 and x > 0 Ans: (x < - 2) U (x > 0) In interval notation, (-infinity, - 2) U (0, infinity)

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