https://r2.erweima.ai/mathbot/image/17271802500389sccyblk.jpg

To solve the problem, let's break it down step by step.

### Problem Statement:
We are given the function \( y = \left( x^3 - 3x + m \right)^2 \) and we need to find the value of the parameter \( m \) such that the minimum value of the function \( y \) on the interval \([-1, 1]\) is equal to 1.

### Steps to Solve:

1. **Understand the Function**:
   \[
   y = \left( x^3 - 3x + m \right)^2
   \]
   The function inside the square, \( f(x) = x^3 - 3x + m \), needs to be analyzed to find the conditions under which its minimum value is such that \( y = 1 \).

2. **Condition for Minimum Value**:
   Since \( y = (f(x))^2 \), the minimum value of \( y \) will be \( 1 \) if the minimum value of \( f(x) \) on \([-1, 1]\) is \( \pm 1 \). This is because \( (1)^2 = 1 \) and \( (-1)^2 = 1 \).

3. **First Derivative**:
   To find the minimum value of \( f(x) = x^3 - 3x + m \), compute the derivative \( f'(x) \) and set it equal to 0 to find the critical points.
   \[
   f'(x) = 3x^2 - 3
   \]
   Setting \( f'(x) = 0 \) gives:
   \[
   3(x^2 - 1) = 0
   \]
   \[
   x^2 = 1 \quad \Rightarrow \quad x = \pm 1
   \]

4. **Evaluate at Critical Points**:
   We evaluate \( f(x) \) at the critical points and endpoints of the interval \([-1, 1]\).
   - At \( x = 1 \):
     \[
     f(1) = 1^3 - 3(1) + m = 1 - 3 + m = m - 2
     \]
   - At \( x = -1 \):
     \[
     f(-1) = (-1)^3 - 3(-1) + m = -1 + 3 + m = m + 2
     \]
   - At \( x = 0 \):
     \[
     f(0) = 0^3 - 3(0) + m = m
     \]

   Now, for the minimum value to be \( \pm 1 \):
   - For \( f(1) = -1 \): \( m - 2 = -1 \Rightarrow m = 1 \)
   - For \( f(1) = 1 \): \( m - 2 = 1 \Rightarrow m = 3 \)
   - For \( f(-1) = -1 \): \( m + 2 = -1 \Rightarrow m = -3 \)
   - For \( f(-1) = 1 \): \( m + 2 = 1 \Rightarrow m = -1 \)

### Conclusion:
The values of \( m \) that satisfy the condition are \( m = 1, 3, -3, -1 \). The sum of all these values is:
\[
1 + 3 + (-3) + (-1) = 0
\]

So, the sum of all values of \( m \) is **0**.

Would you like to dive into any of these steps in more detail or have any specific questions?

### Related Questions:
1. How do we determine the global minimum of a function on a closed interval?
2. What is the significance of critical points in determining extrema?
3. Can the minimum value of the square of a function ever be negative?
4. What role does the derivative play in finding local maxima and minima?
5. How do different values of \( m \) affect the shape and position of the cubic function?

### Tip:
When working with polynomial functions, always check both the critical points and the endpoints of the interval for possible extrema.

Tính tổng tất cả các giá trị của tham số m sao cho giá trị nhỏ nhất của hàm số y = (x^3 - 3x + m)^2 trên đoạn [-1;1] bằng 1.

Solve the problem of finding values of m such that the minimum value of the function y = (x^3 - 3x + m)^2 on the interval [-1;1] equals 1. Learn about cubic functions, critical points, and the Extrema Theorem on closed intervals. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation