There is an extreme value when x=√2.

Then x=√2, (x 2+(a+2) x+a) e x = 0.

rule

2+(a+2)√2+a=0

The solution is a=-2.

f(x)=(x^2-2x)e^x

(2) Find the monotone interval of y=f(x)

f'(x)=(x^2-2)e^x

When x & gt√2, f' (x) > 0, the function increments once.

When -√ 2

When x

f''(x)=(x^2+2x-2)e^x

x=-√2,f''(√2)=2√2e^√2>； 0

Function to get the minimum value.

f(√2)=(2-2√2)e^(√2)