Given the function f defined by f(x) = x^2e^(2x) for x ≥ 0, what is the maximum value of f(x)?

To find the maximum value of the function f(x) = x2e2x for x ≥ 0, we first need to take its derivative and find the critical points.

1. **Derivative Calculation**: Using the product rule, we differentiate f(x):
Let u = x2 and v = e2x. Then, the derivatives are u’ = 2x and v’ = 2e2x. Applying the product rule:

f'(x) = u’v + uv’ = 2xe2x + x2(2e2x) = e2x(2x + 2x2) = 2xe2x(x + 1).

2. **Finding Critical Points**: Set the derivative equal to zero:

2xe2x(x + 1) = 0.

This gives us two cases:

  • 2x = 0 which implies x = 0
  • x + 1 = 0 which gives a negative solution and is not considered since x ≥ 0.

Thus, the only critical point is x = 0.

3. **Evaluate the Function**: Now we need to evaluate f(0):
f(0) = 02e0 = 0.

4. **Check Behavior as x Approaches Infinity**: As x increases towards infinity, e2x grows very rapidly and dominates the x2 term:

Thus, as x → ∞, f(x) → ∞. Therefore, there is no maximum value at a finite point.

In conclusion, the function does not have a maximum value at finite x, but trends towards infinity as x increases.

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