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)

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

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

### Step 2: Check continuity of the function
The function \( f(x) = e^x + 7x - 8 \) is continuous because both \( e^x \) and \( 7x \) 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 \( f(x) \) takes values of opposite signs at two points \( a \) and \( b \), then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \).

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

1. **Evaluate \( f(0) \):**
   \[
   f(0) = e^0 + 7(0) - 8 = 1 - 8 = -7
   \]
   So, \( f(0) = -7 \).

2. **Evaluate \( f(1) \):**
   \[
   f(1) = e^1 + 7(1) - 8 = e + 7 - 8 = e - 1 \approx 2.718 - 1 = 1.718
   \]
   So, \( f(1) \approx 1.718 \).

### Step 4: Conclusion using IVT
We have:
- \( f(0) = -7 \) (a negative value)
- \( f(1) \approx 1.718 \) (a positive value)

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

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 \( e^x \) 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.

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)

Explore how the Intermediate Value Theorem guarantees the existence of a root for the equation e^x = 8 - 7x within the interval (0, 1). This problem covers exponential functions and calculus concepts, providing a detailed step-by-step solution. Suitable for Grades 11-12.

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