Find the absolute extrema of the function \(f(x)= xe^{- x^2/18}\) on the interval \([-2,4]\).

Ans : f(x)=xe^(-x^2/8) differentiating by applying product rule, f'(x)=x(e^(-x^2/8))((-2x)/8)+e^(-x^2/8) f'(x)=e^(-x^2/8)(-x^2/4+1) f'(x)=-1/4e^(-x^2/8)(x^2-4) Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers. Now to find the critical numbers, solve for x for f'(x)=0. -1/4e^(-x^2/8)(x^2-4)=0 x^2-4=0 , x=+-2 e^(-x^2/8)=0 ,

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.