How do you find the local extrema for $(e^{x}) (x^{2})$ $(e^x)(x^2)$ ?

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Answer 1 · 2017-02-20T16:58:08

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

Suppose that, $f (x) = x^{2} e^{x} .$ $f(x)=x^2e^x.$

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.!

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