https://file.aiquickdraw.com/mathbot/file/17301481603653c7i5kys.jpg

Let's solve each part step-by-step for the function \( f(x) = x^4 - 2x^2 \).

### Part (a): Calculate \( f'(x) \)

To find the derivative \( f'(x) \), we differentiate \( f(x) = x^4 - 2x^2 \) with respect to \( x \):

\[
f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^2) = 4x^3 - 4x
\]

So, \( f'(x) = 4x^3 - 4x \).

### Part (b): Determine the coordinates of any stationary points

Stationary points occur when \( f'(x) = 0 \). Setting \( 4x^3 - 4x = 0 \):

\[
4x(x^2 - 1) = 0
\]

This gives:

1. \( 4x = 0 \)  ⟹ \( x = 0 \)
2. \( x^2 - 1 = 0 \) ⟹ \( x = \pm 1 \)

So, the \( x \)-coordinates of the stationary points are \( x = 0 \), \( x = 1 \), and \( x = -1 \).

To find the corresponding \( y \)-coordinates, we substitute these \( x \)-values into \( f(x) \):

- For \( x = 0 \): \( f(0) = 0^4 - 2 \cdot 0^2 = 0 \)
- For \( x = 1 \): \( f(1) = 1^4 - 2 \cdot 1^2 = 1 - 2 = -1 \)
- For \( x = -1 \): \( f(-1) = (-1)^4 - 2 \cdot (-1)^2 = 1 - 2 = -1 \)

The stationary points are:
- \( (0, 0) \)
- \( (1, -1) \)
- \( (-1, -1) \)

### Part (c): Determine the nature of any stationary point

To determine the nature of each stationary point, we check the second derivative \( f''(x) \).

1. **Second derivative**: Differentiate \( f'(x) = 4x^3 - 4x \) again:

   \[
   f''(x) = 12x^2 - 4
   \]

2. **Nature of stationary points**:

   - For \( x = 0 \):
     \[
     f''(0) = 12 \cdot 0^2 - 4 = -4
     \]
     Since \( f''(0) < 0 \), \( (0, 0) \) is a **local maximum**.

   - For \( x = 1 \):
     \[
     f''(1) = 12 \cdot 1^2 - 4 = 12 - 4 = 8
     \]
     Since \( f''(1) > 0 \), \( (1, -1) \) is a **local minimum**.

   - For \( x = -1 \):
     \[
     f''(-1) = 12 \cdot (-1)^2 - 4 = 12 - 4 = 8
     \]
     Since \( f''(-1) > 0 \), \( (-1, -1) \) is a **local minimum**.

### Part (d): Find where the graph intersects or touches the axes

1. **y-axis**: The graph intersects the \( y \)-axis when \( x = 0 \):
   \[
   f(0) = 0
   \]
   So, the intersection with the \( y \)-axis is at \( (0, 0) \).

2. **x-axis**: The graph intersects the \( x \)-axis when \( f(x) = 0 \):
   \[
   f(x) = x^4 - 2x^2 = 0
   \]
   Factor this expression:
   \[
   x^2(x^2 - 2) = 0
   \]
   This gives:
   - \( x^2 = 0 \) ⟹ \( x = 0 \)
   - \( x^2 = 2 \) ⟹ \( x = \pm \sqrt{2} \)

   So, the points where the graph intersects the \( x \)-axis are \( (0, 0) \), \( (\sqrt{2}, 0) \), and \( (-\sqrt{2}, 0) \).

### Part (e): Sketch the graph of \( f(x) \)

To sketch \( f(x) = x^4 - 2x^2 \):
1. Mark the stationary points: \( (0, 0) \), \( (1, -1) \), and \( (-1, -1) \).
2. Mark the intercepts with the axes: \( (0, 0) \), \( (\sqrt{2}, 0) \), and \( (-\sqrt{2}, 0) \).
3. Note that \( (0, 0) \) is a maximum, while \( (1, -1) \) and \( (-1, -1) \) are minima.

Would you like a visual sketch for further clarity?

---

### Further Questions to Explore
1. What is the behavior of \( f(x) \) as \( x \to \pm \infty \)?
2. How does changing the coefficient of \( x^2 \) affect the stationary points?
3. Can \( f(x) = x^4 - 2x^2 \) have inflection points?
4. How would the graph change if \( f(x) = x^4 + 2x^2 \)?
5. Can you find the roots of \( f(x) = 0 \) by another method?

### Tip
When analyzing stationary points, always consider the second derivative to classify maxima, minima, or points of inflection.

For the function f(x) = x^4 - 2x^2, calculate f'(x), determine the coordinates and nature of any stationary points, find where the graph intersects the y-axis and x-axis, and sketch the graph of f(x).

Explore the solution to finding the derivative of f(x) = x^4 - 2x^2, locating stationary points, classifying their nature using the second derivative test, and identifying graph intersections with the x-axis and y-axis. This comprehensive problem is suitable for 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