Math Problem Statement

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. ex = 8 − 7x, (0, 1)

Solution

To use the Intermediate Value Theorem (IVT) to show that there is a root of the equation ex=87xe^x = 8 - 7x in the interval (0,1)(0, 1), follow these steps:

Step 1: Rewrite the equation

We can rewrite the equation ex=87xe^x = 8 - 7x as a function where we move all terms to one side: f(x)=ex(87x)f(x) = e^x - (8 - 7x) This simplifies to: f(x)=ex+7x8f(x) = e^x + 7x - 8 Our goal is to show that there is a root of f(x)=0f(x) = 0 in the interval (0,1)(0, 1).

Step 2: Check continuity of the function

The function f(x)=ex+7x8f(x) = e^x + 7x - 8 is continuous because both exe^x and 7x7x are continuous functions, and the sum of continuous functions is also continuous.

Step 3: Apply the Intermediate Value Theorem

The Intermediate Value Theorem states that if a continuous function f(x)f(x) takes values of opposite signs at two points aa and bb, then there exists at least one c(a,b)c \in (a, b) such that f(c)=0f(c) = 0.

Now, evaluate the function at the endpoints of the interval (0,1)(0, 1):

  1. Evaluate f(0)f(0): f(0)=e0+7(0)8=18=7f(0) = e^0 + 7(0) - 8 = 1 - 8 = -7 So, f(0)=7f(0) = -7.

  2. Evaluate f(1)f(1): f(1)=e1+7(1)8=e+78=e12.7181=1.718f(1) = e^1 + 7(1) - 8 = e + 7 - 8 = e - 1 \approx 2.718 - 1 = 1.718 So, f(1)1.718f(1) \approx 1.718.

Step 4: Conclusion using IVT

We have:

  • f(0)=7f(0) = -7 (a negative value)
  • f(1)1.718f(1) \approx 1.718 (a positive value)

Since f(x)f(x) is continuous and changes from negative to positive between x=0x = 0 and x=1x = 1, by the Intermediate Value Theorem, there must be some c(0,1)c \in (0, 1) such that f(c)=0f(c) = 0. Therefore, there is a root of the equation ex=87xe^x = 8 - 7x in the interval (0,1)(0, 1).

Would you like further details or have any questions about this solution?

Here are 5 additional questions you can explore related to the Intermediate Value Theorem:

  1. How would you find the exact value of the root using numerical methods like bisection or Newton's method?
  2. Can the Intermediate Value Theorem be applied if the function is not continuous?
  3. What are the key differences between the Intermediate Value Theorem and the Mean Value Theorem?
  4. How would you use the IVT to find roots for more complex functions?
  5. How does the behavior of exe^x influence the application of IVT to this particular problem?

Tip: The Intermediate Value Theorem only guarantees the existence of a root, not the exact value. To find the precise root, you’ll often need numerical methods like the bisection method.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Exponential Functions
Intermediate Value Theorem

Formulas

f(x) = e^x + 7x - 8
f(0) = e^0 + 7(0) - 8 = -7
f(1) = e^1 + 7(1) - 8 = e - 1 ≈ 1.718

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12