as before, f(x) = x e^x
f'(x) = e^x (1+x)
f'(x)=0 when 1+x = 0
So, x = -1
x e^x = -1/e
What is the minimum value of f(x) = xe^x?
-1/e
–e
–1
0
Is it 0 or -1/e?
1 answer
-1/e
–e
–1
0
Is it 0 or -1/e?
1 answer