How do you find the local extrema for #(e^x)(x^2)#?

1 Answer
Feb 20, 2017

Answer:

#" Local Mninma is 0, Local Maxima is 4/e^2."#

Explanation:

Suppose that, #f(x)=x^2e^x.#

For Local Maxima, #f'(x)=0, &, f''(x)<0.#

For Local Minima, #f'(x)=0, &, f''(x)>0.#

#f(x)=x^2e^x :. f'(x)=x^2e^x+2xe^x=x(x+2)e^x.#

# f'(x)=(x^2+2x)e^x, :. f''(x)=(x^2+2x)e^x+(2x+2)e^x.#

#:. f''(x)=(x^2+4x+2)e^x.#

Now, #f'(x)=0 rArr x(x+2)e^x=0 rArr x=0, or, x=-2.#

#f''(0)=2>0, rArr" f has a local Minima at x=0, and, it is f(0)=0."#

Similarly, #f''(-2)=-2e^-2<0 :. f(-2)=4/e^2" is local Maxima."#

Enjoy Maths.!