https://file.aiquickdraw.com/mathbot/file/1731924951979jovdtuti.jpg

To solve the problem described in this exercise:

We are tasked with proving that the equation \( f(x) = 0 \) has at least **seven roots** in the interval \([0, 1]\), given the properties of \( f \), a continuous function on \([0,1]\):

1. \( f(0) = f(1) = 0 \),
2. \( f(x + \frac{3}{10}) \neq f(x) \) for all \( x \in [0, \frac{7}{10}] \).

---

### **Solution Outline**

#### Step 1: Analyze the Functional Properties
1. From the second condition \( f(x + \frac{3}{10}) \neq f(x) \) for \( x \in [0, \frac{7}{10}]\), we can deduce that \( f \) is **not periodic** over this interval and takes distinct values at points spaced \( \frac{3}{10} \) apart.

2. The function \( f(x) \) must satisfy \( f(0) = 0 \) and \( f(1) = 0 \). Because \( f \) is continuous, the Intermediate Value Theorem guarantees that if \( f \) changes sign in subintervals of \([0,1]\), there will be roots.

---

#### Step 2: Divide the Interval \( [0,1] \)
Divide the interval \( [0,1] \) into overlapping subintervals of length \( \frac{3}{10} \):
- \( [0, \frac{3}{10}] \),
- \( [\frac{1}{10}, \frac{4}{10}] \),
- \( [\frac{2}{10}, \frac{5}{10}] \),
- \( [\frac{3}{10}, \frac{6}{10}] \),
- \( [\frac{4}{10}, \frac{7}{10}] \),
- \( [\frac{5}{10}, \frac{8}{10}] \),
- \( [\frac{6}{10}, 1] \).

Each of these intervals has length \( \frac{3}{10} \), and the second condition ensures that \( f(x + \frac{3}{10}) \neq f(x) \) within these intervals.

---

#### Step 3: Count Roots
- By the **Intermediate Value Theorem**, since \( f(0) = f(1) = 0 \), and \( f \) is continuous, there must be at least one root in each of these intervals.

- Since there are **seven overlapping intervals** and \( f \) cannot repeat values, \( f(x) = 0 \) must have at least **seven distinct roots** in \([0,1]\).

---

### Conclusion:
The equation \( f(x) = 0 \) has at least **seven roots** in the interval \([0,1]\).

---

Do you want further clarification on any step or additional examples?

---

#### Related Questions to Expand Understanding:
1. What does the Intermediate Value Theorem state, and how is it applied here?
2. Why does the condition \( f(x + \frac{3}{10}) \neq f(x) \) ensure distinct roots?
3. How would the solution change if the interval was \([0,2]\)?
4. Could a function with a different continuity condition (e.g., discontinuous) still satisfy this property?
5. How can you generalize this problem for intervals other than \( \frac{3}{10} \)?

#### Tip:
Always check for overlapping intervals and use symmetry or periodicity constraints to ensure all roots are distinct.

Soit f une fonction continue sur [0,1] tels que f(0) = f(1) = 0 et pour tout x ∈ [0, 7/10], f(x + 3/10) ≠ f(x). Montrer que l'équation f(x) = 0 admet au moins sept racines dans l'intervalle [0,1].

Discover how to show that a continuous function f(x) on [0,1], satisfying specific conditions, has at least seven roots. This exercise uses the Intermediate Value Theorem to analyze distinct values and overlapping intervals. Ideal for high school or pre-university 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