Please help me answer this calculus questions.

Question

Determine the maximum and minimum values of the curve y = xe^x + 3

Answer 1

Given:

f(x) = x exp( x ) + 3

Find:

Mins = ?

Maxs = ?

Solution:

f'(x) = x exp( x ) + exp( x ) = 0

exp( x ) * ( x + 1 ) = 0

exp( x ) = 0 or x + 1 = 0

Technically, as x → -∞, exp( x ) → 0

So there is a "maximum" as x → -∞ of +3

And the regular minimum would occur at x = -1

Having a minimum of - 1/e ≈ -0.36788

I hope this helps! Message me in the comments if you have any questions, comments, or concerns!

Answer 2

lim_x→∞ xe^x + 3 = ∞ so the function has no absolute max on the reals.

y^' = e^x + xe^x (by product rule) and

e^x(1 + x) = 0 ; x = -1 a local min as y^' goes from negative to positive there.

y(-1) = - 1/e + 3 ≈ 2.632 the absolute minimum value