https://r2.erweima.ai/mathbot/image/1727042834680yq5zl091.jpg

The given function is:

\[
y = -x^4 + 2x^3 - 2x - 1
\]

To find the **local extrema** (cực trị), we need to follow these steps:

### Step 1: Find the first derivative \( y' \)
The derivative of the function with respect to \( x \) is:

\[
y' = \frac{d}{dx} (-x^4 + 2x^3 - 2x - 1) = -4x^3 + 6x^2 - 2
\]

### Step 2: Set the first derivative to zero to find critical points
We solve \( y' = 0 \):

\[
-4x^3 + 6x^2 - 2 = 0
\]

First, divide the equation by \(-2\) to simplify:

\[
2x^3 - 3x^2 + 1 = 0
\]

Now, solve this cubic equation. We try rational root testing, and \( x = 1 \) is a root. Let's perform synthetic division of \( 2x^3 - 3x^2 + 1 \) by \( (x - 1) \).

#### Synthetic division:
The quotient is \( 2x^2 - x - 1 \). So we have:

\[
2x^3 - 3x^2 + 1 = (x - 1)(2x^2 - x - 1)
\]

Now, solve \( 2x^2 - x - 1 = 0 \) using the quadratic formula:

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}
\]

Thus, the solutions are:

\[
x = \frac{1 + 3}{4} = 1 \quad \text{and} \quad x = \frac{1 - 3}{4} = -\frac{1}{2}
\]

### Step 3: Second derivative test
Now we perform the second derivative test to determine if these points are maxima or minima.

The second derivative \( y'' \) is:

\[
y'' = \frac{d}{dx}(-4x^3 + 6x^2 - 2) = -12x^2 + 12x
\]

#### For \( x = 1 \):
\[
y''(1) = -12(1)^2 + 12(1) = -12 + 12 = 0
\]
The second derivative test is inconclusive at \( x = 1 \).

#### For \( x = -\frac{1}{2} \):
\[
y''\left(-\frac{1}{2}\right) = -12\left(-\frac{1}{2}\right)^2 + 12\left(-\frac{1}{2}\right) = -12\left(\frac{1}{4}\right) + 12\left(-\frac{1}{2}\right) = -3 - 6 = -9
\]
Since \( y''\left(-\frac{1}{2}\right) < 0 \), the function has a local maximum at \( x = -\frac{1}{2} \).

### Step 4: Check the nature of \( x = 1 \) by analyzing intervals
Since the second derivative test is inconclusive at \( x = 1 \), we can analyze the sign of the first derivative around \( x = 1 \).

- For \( x = 0.5 \) (to the left of \( x = 1 \)): 
  \[
  y'(0.5) = -4(0.5)^3 + 6(0.5)^2 - 2 = -4(0.125) + 6(0.25) - 2 = -0.5 + 1.5 - 2 = -1
  \]
  So \( y' < 0 \), indicating the function is decreasing.

- For \( x = 1.5 \) (to the right of \( x = 1 \)): 
  \[
  y'(1.5) = -4(1.5)^3 + 6(1.5)^2 - 2 = -4(3.375) + 6(2.25) - 2 = -13.5 + 13.5 - 2 = -2
  \]
  So \( y' < 0 \), indicating the function is also decreasing to the right.

Thus, \( x = 1 \) is a point of inflection.

### Conclusion:
- Local maximum at \( x = -\frac{1}{2} \)
- Point of inflection at \( x = 1 \)

Would you like more details on any part of the process?

---

### Related questions:
1. How do you solve cubic equations in general?
2. What does the second derivative test tell us about extrema?
3. How do you use synthetic division to simplify cubic polynomials?
4. Can you explain the concept of points of inflection in functions?
5. What is the role of the first derivative in finding critical points?

### Tip:
Always check the nature of critical points using both the first and second derivatives to ensure correct classification.

Tìm cực trị của hàm số y = -x^4 + 2x^3 - 2x - 1

A detailed guide to finding the local extrema of the function y = -x^4 + 2x^3 - 2x - 1. This calculus problem involves using the first and second derivative tests to determine critical points, solving a cubic equation, and applying the quadratic formula. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation