To find the maximum value of the function f(x) = x2 e2x for x ≥ 0, we start by calculating its derivative, f'(x).
Using the product rule, where u = x2 and v = e2x, we have:
- u’ = 2x
- v’ = 2e2x
Now, applying the product rule:
f'(x) = u'v + uv' = (2x)e2x + (x2)(2e2x)
This simplifies to:
f'(x) = e2x (2x + 2x2) = 2x e2x (x + 1)
The function reaches its critical points when f'(x) = 0. Setting the derivative equal to zero:
2x e2x (x + 1) = 0
This equation gives us:
- x = 0
- x + 1 = 0 (not applicable since x ≥ 0)
Now, we need to evaluate the function at the critical point and the endpoint of the interval. Evaluating at x = 0:
f(0) = 02 e0 = 0
The function f(x) as x → ∞ tends to ∞ since both x2 and e2x grow without bound.
To determine the maximum, we observe that while f(0) = 0, as x increases positively, f(x) increases indefinitely. Therefore, there is no finite maximum value as x approaches infinity.
In conclusion, the function f(x) = x2 e2x does not have a maximum finite value in the interval [0, ∞), but it tends to infinity as x increases.