f(x)= x^2e^-x at x= a passes through the origin Solution df/dx = (x^2)(e^x) + (2

Question

f(x)= x^2e^-x at x= a passes through the origin

Explanation / Answer

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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