use the intermediate value theorem to show that the equation tan(x)=1-x has a root in the interval (0,pi/4)

To use the Intermediate Value Theorem (IVT) to show that the equation \(\tan(x) = 1 - x\) has a root in the interval \((0, \frac{\pi}{4})\), we need to follow these steps:

### Step-by-Step Solution:

1. **Define the function:**
   Let's define a function \(f(x)\) such that:
   \[
   f(x) = \tan(x) - (1 - x)
   \]
   We want to find a root of this function in the interval \((0, \frac{\pi}{4})\), i.e., we want to show that there exists some \(c \in (0, \frac{\pi}{4})\) such that \(f(c) = 0\).

2. **Check the continuity of the function:**
   The function \(\tan(x)\) is continuous on the interval \((0, \frac{\pi}{4})\) because \(\tan(x)\) is continuous for \(x \in (0, \frac{\pi}{2})\) except at the points where it has vertical asymptotes, which are not within this interval. The function \(1 - x\) is a polynomial and is continuous everywhere. Therefore, \(f(x) = \tan(x) - (1 - x)\) is continuous on the interval \((0, \frac{\pi}{4})\).

3. **Evaluate the function at the endpoints of the interval:**

   - At \(x = 0\):
     \[
     f(0) = \tan(0) - (1 - 0) = 0 - 1 = -1
     \]

   - At \(x = \frac{\pi}{4}\):
     \[
     f\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) - \left(1 - \frac{\pi}{4}\right) = 1 - \left(1 - \frac{\pi}{4}\right) = \frac{\pi}{4}
     \]

   We observe that:
   \[
   f(0) = -1 < 0 \quad \text{and} \quad f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} > 0
   \]

4. **Apply the Intermediate Value Theorem:**
   The Intermediate Value Theorem states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one \(c \in (a, b)\) such that \(f(c) = 0\).

   Since \(f(x)\) is continuous on \((0, \frac{\pi}{4})\), and \(f(0) < 0\) while \(f(\frac{\pi}{4}) > 0\), by the Intermediate Value Theorem, there exists a \(c \in (0, \frac{\pi}{4})\) such that \(f(c) = 0\).

### Conclusion:

The equation \(\tan(x) = 1 - x\) has at least one root in the interval \((0, \frac{\pi}{4})\) by the Intermediate Value Theorem.

Would you like further details or have any questions?

### Related Questions:

1. What is the Intermediate Value Theorem, and how is it generally applied?
2. Can the function \(f(x) = \tan(x) - (1 - x)\) have more than one root in \((0, \frac{\pi}{4})\)?
3. How do you find the exact value of the root using numerical methods such as the bisection method?
4. How does the Intermediate Value Theorem differ from the Mean Value Theorem?
5. What are some real-life applications of the Intermediate Value Theorem?

### Tip:

When using the Intermediate Value Theorem, always ensure the function is continuous on the interval and that it changes signs at the endpoints.

Learn how to use the Intermediate Value Theorem to prove that the equation tan(x) = 1 - x has a root in the interval (0, pi/4). This problem involves trigonometric equations and application of the Intermediate Value Theorem in calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation