Consider the function f(x)=x2e4xf(x)=x2e4x.

f(x)f(x) has two inflection points at *x* = *C* and *x* = *D* with *C < D*
where *C* is
and *D* is

Finally for each of the following intervals, tell whether f(x)f(x) is concave up or concave down.
(−∞,C)(-∞,C): Select an answer Concave Up Concave Down
(C,D)(C,D): Select an answer Concave Up Concave Down
(D,∞)(D,∞) Select an answer Concave Up Concave Down

To find the inflection points of the function \( f(x) = x^2 e^{4x} \), we need to determine the second derivative and analyze its sign.

1. **First derivative**:
   \[
   f'(x) = 2x e^{4x} + x^2 \cdot 4 e^{4x} = e^{4x} (2x + 4x^2)
   \]

2. **Second derivative**:
   Using the product rule:
   \[
   f''(x) = \frac{d}{dx}(e^{4x}(2x + 4x^2)) = e^{4x}(2 + 8x) + 4e^{4x}(2x + 4x^2)
   \]
   This simplifies to:
   \[
   f''(x) = e^{4x}(2 + 8x + 8x + 16x^2) = e^{4x}(16x^2 + 16x + 2)
   \]
   Setting \( f''(x) = 0 \):
   \[
   16x^2 + 16x + 2 = 0
   \]
   Dividing by 2:
   \[
   8x^2 + 8x + 1 = 0
   \]
   Using the quadratic formula:
   \[
   x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 8 \cdot 1}}{2 \cdot 8} = \frac{-8 \pm \sqrt{64 - 32}}{16} = \frac{-8 \pm \sqrt{32}}{16} = \frac{-8 \pm 4\sqrt{2}}{16} = \frac{-2 \pm \sqrt{2}}{4}
   \]

   Therefore, the inflection points are:
   \[
   C = \frac{-2 - \sqrt{2}}{4}, \quad D = \frac{-2 + \sqrt{2}}{4}
   \]

3. **Concavity**:
   - For \( (-\infty, C) \): Choose a test point (e.g., \( x = -1 \)):
     \[
     f''(-1) = e^{-4}(16(-1)^2 + 16(-1) + 2) = e^{-4}(16 - 16 + 2) = 2 > 0 \quad \text{(Concave Up)}
     \]
   - For \( (C, D) \): Choose a test point (e.g., \( x = 0 \)):
     \[
     f''(0) = e^{0}(16(0)^2 + 16(0) + 2) = 2 > 0 \quad \text{(Concave Up)}
     \]
   - For \( (D, \infty) \): Choose a test point (e.g., \( x = 1 \)):
     \[
     f''(1) = e^{4}(16(1)^2 + 16(1) + 2) = e^{4}(16 + 16 + 2) = 34e^{4} > 0 \quad \text{(Concave Up)}
     \]

### Summary:
- Inflection points:
  - \( C = \frac{-2 - \sqrt{2}}{4} \)
  - \( D = \frac{-2 + \sqrt{2}}{4} \)

- Concavity:
  - \( (-\infty, C) \): Concave Up
  - \( (C, D) \): Concave Up
  - \( (D, \infty) \): Concave Up

Do you need further details or have any questions? Here are some related questions:
1. How do you interpret inflection points in a graph?
2. What does concavity imply about the behavior of a function?
3. How would the concavity change if the function were modified?
4. Can you provide an example of a function with different concavity?
5. What role does the exponential function play in this context? 

**Tip:** Always verify the sign of the second derivative when determining concavity.

Consider the function f(x) = x^2 e^{4x}. f(x) has two inflection points at x = C and x = D with C < D where C is and D is. Finally for each of the following intervals, tell whether f(x) is concave up or concave down. (-∞, C): Select an answer Concave Up Concave Down (C, D): Select an answer Concave Up Concave Down (D, ∞): Select an answer Concave Up Concave Down

Learn how to find the inflection points and determine concavity for the function f(x) = x^2 e^{4x}. This problem includes calculating derivatives, using the quadratic formula to solve for inflection points, and analyzing concavity in different intervals. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation