Verify lmvt for f(x) = 1 + 2x + 3x^2 + 4x^3 on [0,1]

To verify the **Lagrange Mean Value Theorem (LMVT)** for the function \( f(x) = 1 + 2x + 3x^2 + 4x^3 \) on the interval \([0, 1]\), we follow these steps:

### Step 1: Check the conditions for LMVT
The **LMVT** requires:
1. **Continuity** on the closed interval \([a, b]\): \( f(x) \) is a polynomial function, and all polynomial functions are continuous on any interval. Thus, \( f(x) \) is continuous on \([0, 1]\).
2. **Differentiability** on the open interval \((a, b)\): Since \( f(x) \) is a polynomial, it is differentiable everywhere, including on the open interval \((0, 1)\).

Thus, the conditions for LMVT are satisfied.

### Step 2: Apply LMVT formula
According to the **LMVT**, there exists a point \( c \in (0, 1) \) such that:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

For the given interval \([0, 1]\), \( a = 0 \) and \( b = 1 \). So, we need to compute:

\[
f'(c) = \frac{f(1) - f(0)}{1 - 0} = f(1) - f(0)
\]

### Step 3: Calculate \( f(0) \) and \( f(1) \)
Substitute \( x = 0 \) and \( x = 1 \) into the function:

\[
f(0) = 1 + 2(0) + 3(0)^2 + 4(0)^3 = 1
\]
\[
f(1) = 1 + 2(1) + 3(1)^2 + 4(1)^3 = 1 + 2 + 3 + 4 = 10
\]

Now, compute:

\[
f'(c) = f(1) - f(0) = 10 - 1 = 9
\]

### Step 4: Differentiate \( f(x) \)
Now, find \( f'(x) \), the derivative of \( f(x) = 1 + 2x + 3x^2 + 4x^3 \):

\[
f'(x) = 0 + 2 + 6x + 12x^2 = 2 + 6x + 12x^2
\]

### Step 5: Find \( c \)
Set \( f'(c) = 9 \) and solve for \( c \):

\[
2 + 6c + 12c^2 = 9
\]

Simplify the equation:

\[
6c + 12c^2 = 9 - 2
\]
\[
6c + 12c^2 = 7
\]
\[
2c + 4c^2 = \frac{7}{3}
\]

This is a quadratic equation. Now, we solve for \( c \).

\[
4c^2 + 2c - \frac{7}{3} = 0
\]

Multiplying the whole equation by 3 to eliminate fractions:

\[
12c^2 + 6c - 7 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
c = \frac{-6 \pm \sqrt{6^2 - 4(12)(-7)}}{2(12)}
\]
\[
c = \frac{-6 \pm \sqrt{36 + 336}}{24}
\]
\[
c = \frac{-6 \pm \sqrt{372}}{24}
\]
\[
c = \frac{-6 \pm 6\sqrt{31}}{24}
\]
\[
c = \frac{-1 \pm \sqrt{31}}{4}
\]

Since \( c \) must lie in the interval \((0, 1)\), we take the positive root:

\[
c = \frac{-1 + \sqrt{31}}{4} \approx 0.7808
\]

Thus, \( c \approx 0.7808 \).

### Conclusion:
The value \( c \approx 0.7808 \) satisfies the Lagrange Mean Value Theorem for the given function \( f(x) = 1 + 2x + 3x^2 + 4x^3 \) on the interval \([0, 1]\).

Would you like more details on any step or further explanation? Here are five related questions:

1. How do you apply the quadratic formula to solve equations?
2. Can LMVT be used for non-polynomial functions?
3. Why are polynomial functions always differentiable?
4. What is the geometric interpretation of LMVT?
5. How is LMVT related to Rolle's Theorem?

**Tip:** Verifying LMVT often involves solving quadratic equations, which can give multiple solutions. Always check if the solutions lie within the given interval.

Verify LMVT for f(x) = 1 + 2x + 3x^2 + 4x^3 on [0,1]

Verify the Lagrange Mean Value Theorem (LMVT) for the function f(x) = 1 + 2x + 3x^2 + 4x^3 on the interval [0, 1]. This solution covers key concepts in calculus, including differentiation and solving quadratic equations. Suitable for college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation