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)