Question

If a function \( f(x) \) is continuous in the closed interval \( [a, b] \) and the first derivative \( f'(x) \) exists in the open interval \( (a, b) \), then according to the Lagrange's mean value theorem:
\[ \frac{f(b) - f(a)}{b - a} = f'(c) \] If \( a = 0 \), \( b = 1.5 \), and \( f(x) = x(x - 1)(x - 2) \), then the value(s) of \( c \) in \( [a, b] \) is/are:

• A. \( 0.50 \)
• B. \( 0.75 \)
• C. \( 1.00 \)
• D. \( 1.50 \)

Hint:

The Lagrange Mean Value Theorem is often used to find a point where the derivative of a function equals the average rate of change over an interval. Solving the resulting equation often leads to finding the specific values of \( c \) in the interval.

Answer 1

The Correct Option is A, D

Given that \( f(x) = x(x - 1)(x - 2) \), we first compute the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}[x(x - 1)(x - 2)] \] By using the product rule and simplifying, we get:
\[ f'(x) = 3x^2 - 6x + 2 \] Now, applying the Mean Value Theorem:
\[ \frac{f(1.5) - f(0)}{1.5 - 0} = f'(c) \] First, calculate \( f(0) \) and \( f(1.5) \):
\[ f(0) = 0(0 - 1)(0 - 2) = 0, \quad f(1.5) = 1.5(1.5 - 1)(1.5 - 2) = 1.5(0.5)(-0.5) = -0.375 \] Now, use the values in the Mean Value Theorem equation:
\[ \frac{-0.375 - 0}{1.5 - 0} = f'(c) \quad \Rightarrow \quad \frac{-0.375}{1.5} = f'(c) \quad \Rightarrow \quad f'(c) = -0.25 \] Now, solve for \( c \) by setting \( f'(c) = -0.25 \) and solving the quadratic equation:
\[ 3c^2 - 6c + 2 = -0.25 \quad \Rightarrow \quad 3c^2 - 6c + 2.25 = 0 \] Using the quadratic formula:
\[ c = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2.25)}}{2(3)} = \frac{6 \pm \sqrt{36 - 27}}{6} = \frac{6 \pm \sqrt{9}}{6} = \frac{6 \pm 3}{6} \] Thus, \( c = 1 \) or \( c = 0.5 \), both of which lie within the interval \( [0, 1.5] \). Therefore, the correct answers are \( c = 0.5 \) and \( c = 1.5 \), so the correct answer is (A), (D).