Math Problem Statement

Use the Intermediate Value Theorem to show that the equation e^x = sin(x) has infinitely many solutions.

Solution

The goal is to show that the equation ex=sin(x)e^x = \sin(x) has infinitely many solutions using the Intermediate Value Theorem (IVT). Let's break it down step by step.

Step 1: Understanding the Functions

  1. The function exe^x is continuous and strictly increasing for all real xx. As xx \to -\infty, ex0e^x \to 0, and as xx \to \infty, exe^x \to \infty.
  2. The function sin(x)\sin(x) is continuous and oscillates between -1 and 1 for all real xx, repeating its values periodically.

Step 2: Intermediate Value Theorem (IVT)

The IVT states that if a continuous function f(x)f(x) takes values f(a)f(a) and f(b)f(b) at two points aa and bb, and if a value NN lies between f(a)f(a) and f(b)f(b), then there exists at least one cc in the interval (a,b)(a, b) such that f(c)=Nf(c) = N.

Step 3: Application of IVT

  1. Define the function f(x)=exsin(x)f(x) = e^x - \sin(x). This function is continuous because both exe^x and sin(x)\sin(x) are continuous.
  2. We need to show that f(x)=0f(x) = 0 (which corresponds to ex=sin(x)e^x = \sin(x)) has infinitely many solutions.

Step 4: Choosing Intervals to Apply IVT

  • Since exe^x \to \infty as xx \to \infty and sin(x)\sin(x) is bounded between -1 and 1, the function f(x)=exsin(x)f(x) = e^x - \sin(x) grows without bound as xx \to \infty.
  • At xx \to -\infty, ex0e^x \to 0, while sin(x)\sin(x) still oscillates between -1 and 1, so f(x)=exsin(x)f(x) = e^x - \sin(x) will oscillate between values close to 1-1 and values close to 11.

For large negative xx, exe^x approaches zero, so f(x)sin(x)f(x) \approx -\sin(x). In particular:

  • At some points, sin(x)=1\sin(x) = 1, giving f(x)1f(x) \approx -1.
  • At some points, sin(x)=1\sin(x) = -1, giving f(x)1f(x) \approx 1.

Thus, for large negative values of xx, f(x)f(x) takes values on both sides of zero.

Step 5: Conclusion from IVT

Since f(x)f(x) oscillates and takes both positive and negative values as xx \to -\infty, and f(x)f(x) \to \infty as xx \to \infty, the IVT guarantees that there are infinitely many points where f(x)=0f(x) = 0, meaning ex=sin(x)e^x = \sin(x) has infinitely many solutions.

This shows that the equation ex=sin(x)e^x = \sin(x) has infinitely many solutions, without the need for a graph.


Do you have any questions or would you like further clarification?

Here are 5 related questions:

  1. How does the behavior of the sine function differ from that of exe^x as xx increases or decreases?
  2. Can we apply the IVT to any continuous function, or are there conditions that must be met?
  3. How do periodic functions influence the application of the IVT?
  4. Could a similar technique be used to find solutions to ex=cos(x)e^x = \cos(x)?
  5. How would you find approximate values for some of the solutions of ex=sin(x)e^x = \sin(x)?

Tip: When using the Intermediate Value Theorem, ensure that the function is continuous over the interval you're examining.

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Math Problem Analysis

Mathematical Concepts

Exponential Function
Sine Function
Intermediate Value Theorem

Formulas

f(x) = e^x - sin(x)

Theorems

Intermediate Value Theorem

Suitable Grade Level

University Calculus