Explanation:
The second derivative is already given, but here's how you can find it using product rule:
f(x) = x²e²ˣ
f'(x) = x² (2e²ˣ) + (2x) e²ˣ
f'(x) = 2e²ˣ (x² + x)
f"(x) = 2e²ˣ (2x + 1) + 4e²ˣ (x² + x)
f"(x) = 2e²ˣ (2x² + 4x + 1)
Now, we evaluate f"(x). e²ˣ is always positive. So let's check when the other factor is 0:
2x² + 4x + 1 = 0
x = [ -4 ± √(16 − 4(2)(1)) ] / 2(2)
x = [ -4 ± √(16 − 8) ] / 4
x = (-4 ± 2√2) / 4
x = (-2 ± √2) / 2
There are two places when f"(x) is 0. So we can split the domain into 3 intervals:
x < (-2 − √2) / 2
(-2 − √2) / 2 < x < (-2 + √2) / 2
x > (-2 + √2) / 2
Evaluate the sign of f"(x) in each interval.
x < (-2 − √2) / 2 → f"(x) > 0
(-2 − √2) / 2 < x < (-2 + √2) / 2 → f"(x) < 0
x > (-2 + √2) / 2 → f"(x) > 0
So f(x) is concave down in the interval (-2 − √2) / 2 < x < (-2 + √2) / 2, and concave up everywhere else.