Question

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

Answer 1

`" 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.!