find the maximum point of the curve

Question

y=x^2e^-1/2x

Answer 1

y'=2xe^-(1/2)x-(x^2e^-(1/2)x)/2=(e^-(1/2)x)(2x-x^2/2). When y'=0 there is a turning point, so 2x-x^2/2=0; 4x-x^2=0 so x(4-x)=0. x=0 and 4 are the turning points. When x=0, y=0; when x=4, y=16/e^2=2.165 approx. y"=(e^-(1/2)x)(2-x)-(e^(1/2)x)(2x-x^2/2)/2. y"<0 when x=4 because 2-x<0 and 2x-x^2/2=0. The turning point at x=4 is therefore a maximum. At x=0 y">0, a minimum.