Question

The value of $C$ in $(0, 2)$ satisfying the mean value theorem for the function $f(x) = x(x - 1)^2$, $x \in [0, 2]$ is equal to:}

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$\\

Answer 1

The Correct Option is B

The mean value theorem states that there exists a $C \in (a, b)$ such that: \[ f'(C) = \frac{f(b) - f(a)}{b - a}. \] Here, $f(x) = x(x - 1)^2$, $a = 0$, $b = 2$: \[ f(0) = 0, \quad f(2) = 2(2 - 1)^2 = 2. \] \[ \frac{f(b) - f(a)}{b - a} = \frac{2 - 0}{2 - 0} = 1. \] Differentiate $f(x)$: \[ f'(x) = (x - 1)^2 + 2x(x - 1). \] Solve $f'(C) = 1$: \[ (C - 1)^2 + 2C(C - 1) = 1. \] Simplify and solve for $C$ to find $C = \frac{4}{3}$.

Answer 2

1. Understand the problem:

We need to find the value of \( c \) in the interval \( (0, 2) \) that satisfies the Mean Value Theorem (MVT) for the function \( f(x) = x(x-1)^2 \) on the interval \( [0, 2] \).

2. Verify MVT conditions:

The function \( f(x) = x(x-1)^2 \) is a polynomial, so it is continuous on \( [0, 2] \) and differentiable on \( (0, 2) \). Thus, MVT applies.

3. Compute \( f(0) \) and \( f(2) \):

\[ f(0) = 0(0-1)^2 = 0 \] \[ f(2) = 2(2-1)^2 = 2 \]

4. Compute the average rate of change:

\[ \frac{f(2) - f(0)}{2 - 0} = \frac{2 - 0}{2} = 1 \]

5. Find the derivative \( f'(x) \):

First, expand \( f(x) \):

\[ f(x) = x(x^2 - 2x + 1) = x^3 - 2x^2 + x \]

Now, differentiate:

\[ f'(x) = 3x^2 - 4x + 1 \]

6. Solve \( f'(c) = 1 \) for \( c \in (0, 2) \):

\[ 3c^2 - 4c + 1 = 1 \implies 3c^2 - 4c = 0 \implies c(3c - 4) = 0 \]

Solutions: \( c = 0 \) or \( c = \frac{4}{3} \).

Since \( c \in (0, 2) \), the valid solution is \( c = \frac{4}{3} \).

7. Match the result to the options:

The value \( c = \frac{4}{3} \) corresponds to option (B).

Correct Answer: (B) \( \frac{4}{3} \)