Sketch the graph of the given function. Use technology to approximate the intercepts, coordinates of extrema, and points of inflection to one decimal place. Check your sketch using technology. HINT [See Example 1.] f(x) = 4x3 + 6x2 − 24x + 1

(c) Indicate any points of inflection. (If an answer does not exist, enter DNE.)

(x, y) =

To sketch the graph of the function \( f(x) = 4x^3 + 6x^2 - 24x + 1 \), we need to follow these steps:

### 1. Find the Intercepts
- **x-intercepts**: Set \( f(x) = 0 \) and solve for \( x \).
  \[
  4x^3 + 6x^2 - 24x + 1 = 0
  \]
  We'll solve this equation to find approximate values for the x-intercepts.

- **y-intercept**: This occurs when \( x = 0 \), so:
  \[
  f(0) = 4(0)^3 + 6(0)^2 - 24(0) + 1 = 1
  \]
  The y-intercept is at \( (0, 1) \).

### 2. Find the Extrema
- **First Derivative**: To find critical points, we compute the first derivative \( f'(x) \).
  \[
  f'(x) = 12x^2 + 12x - 24
  \]
  Set \( f'(x) = 0 \) and solve for \( x \) to find critical points:
  \[
  12x^2 + 12x - 24 = 0
  \]
  Simplifying:
  \[
  x^2 + x - 2 = 0
  \]
  Solve this quadratic equation using the quadratic formula:
  \[
  x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2}
  \]
  So the critical points are:
  \[
  x = 1 \quad \text{and} \quad x = -2
  \]
  We evaluate \( f(x) \) at these points to find the y-coordinates of the extrema.

### 3. Find Points of Inflection
- **Second Derivative**: To find points of inflection, we compute the second derivative \( f''(x) \).
  \[
  f''(x) = 24x + 12
  \]
  Set \( f''(x) = 0 \) and solve for \( x \):
  \[
  24x + 12 = 0 \implies x = -\frac{1}{2}
  \]
  So there is a possible point of inflection at \( x = -\frac{1}{2} \). We compute \( f(-\frac{1}{2}) \) to find the corresponding y-coordinate.

### Let's compute all these values precisely using technology.### Results:

1. **Extrema:**
   - Critical points occur at \( x = -2 \) and \( x = 1 \).
     - At \( x = -2 \), \( f(-2) = 41 \) (local maximum).
     - At \( x = 1 \), \( f(1) = -13 \) (local minimum).

2. **Inflection Point:**
   - There is a point of inflection at \( x = -0.5 \), with \( f(-0.5) = 14 \).

3. **Intercepts:**
   - **x-intercepts** occur approximately at:
     - \( x = -3.33 \)
     - \( x = 0.04 \)
     - \( x = 1.78 \)
   - **y-intercept** is at \( (0, 1) \).

These are the approximate key features of the graph of \( f(x) = 4x^3 + 6x^2 - 24x + 1 \).

Would you like to explore any of these details further or ask about the graphical sketch?

### Five related questions:
1. How do we verify the concavity changes at the inflection point?
2. What is the significance of critical points in determining the graph's shape?
3. How do the intercepts help in sketching the graph?
4. Can you explain the role of the first and second derivatives in finding extrema?
5. How would adding or subtracting constants affect this cubic function?

**Tip**: Inflection points occur where the concavity of the graph changes, often indicated by the second derivative equaling zero.

Sketch the graph of the given function. Use technology to approximate the intercepts, coordinates of extrema, and points of inflection to one decimal place. Check your sketch using technology. HINT [See Example 1.] f(x) = 4x^3 + 6x^2 − 24x + 1

(c) Indicate any points of inflection. (If an answer does not exist, enter DNE.)

(x, y) =

Explore how to sketch the graph of the cubic function f(x) = 4x^3 + 6x^2 − 24x + 1 by finding the intercepts, extrema, and points of inflection. Learn about critical points and inflection points with detailed step-by-step solutions. Suitable for Grades 10-12.

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