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We are asked to analyze the function \( f(x) = x^2 e^{16x} \), determine the inflection points, and examine concavity over specific intervals.

### Step 1: Find the second derivative of \( f(x) \)

1. **First derivative** \( f'(x) \):

Using the product rule \( (uv)' = u'v + uv' \), where \( u = x^2 \) and \( v = e^{16x} \),
\[
f'(x) = (2x) e^{16x} + x^2 (16e^{16x}) = e^{16x}(2x + 16x^2)
\]

2. **Second derivative** \( f''(x) \):

Again, applying the product rule to \( f'(x) \):
\[
f''(x) = \left( e^{16x}(2x + 16x^2) \right)' = e^{16x}(2x + 16x^2)' + e^{16x}(2x + 16x^2)(16)
\]
\[
= e^{16x}(2 + 32x) + 16e^{16x}(2x + 16x^2)
\]
Simplify the terms:
\[
f''(x) = e^{16x}(2 + 32x + 16(2x + 16x^2)) = e^{16x}(2 + 32x + 32x + 256x^2)
\]
\[
f''(x) = e^{16x}(2 + 64x + 256x^2)
\]

### Step 2: Solve \( f''(x) = 0 \) for inflection points

To find inflection points, we set \( f''(x) = 0 \):
\[
e^{16x}(2 + 64x + 256x^2) = 0
\]
Since \( e^{16x} \neq 0 \), solve:
\[
2 + 64x + 256x^2 = 0
\]
This is a quadratic equation. Divide by 2:
\[
1 + 32x + 128x^2 = 0
\]
Solve using the quadratic formula:
\[
x = \frac{-32 \pm \sqrt{32^2 - 4(128)(1)}}{2(128)} = \frac{-32 \pm \sqrt{1024 - 512}}{256} = \frac{-32 \pm \sqrt{512}}{256}
\]
\[
x = \frac{-32 \pm 16\sqrt{2}}{256}
\]
\[
x = \frac{-1 \pm \frac{\sqrt{2}}{8}}{8}
\]
Thus, the inflection points are:
\[
C = \frac{-1

Consider the function f(x) = x^2 * e^(16x). f(x) has two inflection values at x = C and x = D with C ≤ D. Determine the values of C and D, and specify concavity on the given intervals.

Analyze the function f(x) = x^2 * e^(16x) to find its inflection points and determine concavity over given intervals. This problem includes the use of product rule, quadratic formula, and second derivative test. Suitable for students in Grade 12 or early university calculus courses.

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