Use the Intermediate Value Theorem to show that there is a solution to the given equation in the
specified interval:
i. [4 points] ln(x) = 4 −
√
x, for x ∈ (4,9).
1
ii. [4 points] e
x = 3 − 2x, for x ∈ (0,1)

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

### Part i: \( \ln(x) = 4 - \sqrt{x}, \text{ for } x \in (4,9) \)

We can rewrite the equation as a function:

\[
f(x) = \ln(x) - (4 - \sqrt{x})
\]
We aim to find if there exists \( c \in (4,9) \) such that \( f(c) = 0 \), by applying the IVT.

#### Step 1: Check Continuity
- Both \( \ln(x) \) and \( \sqrt{x} \) are continuous for \( x \in (4,9) \).
- Therefore, \( f(x) = \ln(x) - (4 - \sqrt{x}) \) is also continuous on \( (4,9) \).

#### Step 2: Evaluate \( f(x) \) at the endpoints of the interval

1. **At \( x = 4 \)**:
   \[
   f(4) = \ln(4) - (4 - \sqrt{4}) = \ln(4) - (4 - 2) = \ln(4) - 2
   \]
   Using \( \ln(4) \approx 1.386 \), we get:
   \[
   f(4) \approx 1.386 - 2 = -0.614
   \]

2. **At \( x = 9 \)**:
   \[
   f(9) = \ln(9) - (4 - \sqrt{9}) = \ln(9) - (4 - 3) = \ln(9) - 1
   \]
   Using \( \ln(9) \approx 2.197 \), we get:
   \[
   f(9) \approx 2.197 - 1 = 1.197
   \]

#### Step 3: Apply the IVT
- \( f(4) \approx -0.614 \) and \( f(9) \approx 1.197 \), so \( f(4) < 0 \) and \( f(9) > 0 \).
- Since \( f(x) \) is continuous and changes sign between \( x = 4 \) and \( x = 9 \), the IVT guarantees that there is at least one solution \( c \in (4,9) \) such that \( f(c) = 0 \).

### Conclusion for Part i:
There exists a solution to the equation \( \ln(x) = 4 - \sqrt{x} \) in the interval \( (4,9) \).

---

### Part ii: \( e^x = 3 - 2x, \text{ for } x \in (0,1) \)

We rewrite the equation as a function:

\[
g(x) = e^x - (3 - 2x)
\]
We aim to show there is a solution \( c \in (0,1) \) such that \( g(c) = 0 \), using the IVT.

#### Step 1: Check Continuity
- Both \( e^x \) and \( 3 - 2x \) are continuous for all \( x \in (0,1) \).
- Therefore, \( g(x) = e^x - (3 - 2x) \) is continuous on \( (0,1) \).

#### Step 2: Evaluate \( g(x) \) at the endpoints of the interval

1. **At \( x = 0 \)**:
   \[
   g(0) = e^0 - (3 - 2(0)) = 1 - 3 = -2
   \]

2. **At \( x = 1 \)**:
   \[
   g(1) = e^1 - (3 - 2(1)) = e - (3 - 2) = e - 1
   \]
   Using \( e \approx 2.718 \), we get:
   \[
   g(1) \approx 2.718 - 1 = 1.718
   \]

#### Step 3: Apply the IVT
- \( g(0) = -2 \) and \( g(1) \approx 1.718 \), so \( g(0) < 0 \) and \( g(1) > 0 \).
- Since \( g(x) \) is continuous and changes sign between \( x = 0 \) and \( x = 1 \), the IVT guarantees that there is at least one solution \( c \in (0,1) \) such that \( g(c) = 0 \).

### Conclusion for Part ii:
There exists a solution to the equation \( e^x = 3 - 2x \) in the interval \( (0,1) \).

---

Do you want more details on any step, or further explanation?

Here are some related questions to deepen your understanding:

1. How can the Intermediate Value Theorem be applied to functions with discontinuities?
2. What are some common cases where the Intermediate Value Theorem does not apply?
3. How can we estimate the value of the solution in each case using numerical methods?
4. Can the Intermediate Value Theorem guarantee more than one solution in an interval?
5. How does the continuity of a function affect the applicability of the IVT?

**Tip**: The IVT is useful not only for proving the existence of a solution but also for narrowing down intervals where a root might lie, which can then be solved more precisely using other methods like Newton's method or the bisection method.

Use the Intermediate Value Theorem to show that there is a solution to the given equation in the specified interval:

 i. ln(x) = 4 − √x, for x ∈ (4,9).
 ii. e^x = 3 − 2x, for x ∈ (0,1).

Learn how to apply the Intermediate Value Theorem to prove the existence of solutions for ln(x) = 4 - √x in the interval (4,9) and e^x = 3 - 2x in the interval (0,1). This solution involves continuous functions and step-by-step evaluations of endpoints. Suitable for Grades 11-12 and college students.

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