Question

\(f(x)\) is a cubic polynomial and the curve represented by \(y=f(x)\) passes through the origin. The function \(y=f(x)\) is an increasing function in \[ (-\infty,0)\cup(1,\infty) \] and a decreasing function in \[ (0,1). \] If \[ f'(2)=6, \] then the \(c\) of Lagrange's Mean Value Theorem on the interval \([0,2]\) satisfies the equation

• A. \(3x^2-x-1=0\)
• B. \(3x^2+5x-1=0\)
• C. \(3x^2-3x-1=0\)
• D. \(3x^2+2x-1=0\)

Hint:

If a cubic function is increasing and decreasing on specified intervals, first determine the zeros of \[ \boxed{f'(x)} \] from the sign changes. Then apply Lagrange's Mean Value Theorem using \[ \boxed{f'(c)=\dfrac{f(b)-f(a)}{b-a}.} \]

Answer 1

The Correct Option is C

Step 1: Find the derivative. Since \[ f(x) \] is increasing in \[ (-\infty,0)\cup(1,\infty) \] and decreasing in \[ (0,1), \] its derivative has zeros at \[ x=0 \quad\text{and}\quad x=1. \] Hence, \[ f'(x)=k\,x(x-1), \] where \[ k>0. \] Given, \[ f'(2)=6, \] so \[ 2k=6, \] which gives \[ k=3. \] Thus, \[ f'(x)=3x(x-1). \]

Step 2: Find \(f(x)\). Integrating, \[ f(x)=x^3-\frac32x^2+C. \] Since the curve passes through the origin, \[ f(0)=0, \] so \[ C=0. \] Hence, \[ f(x)=x^3-\frac32x^2. \]

Step 3: Apply Lagrange's Mean Value Theorem. By LMVT, \[ f'(c) = \frac{f(2)-f(0)}{2-0}. \] Now, \[ f(2) = 8-6 = 2. \] Therefore, \[ \frac{f(2)-f(0)}2 = 1. \] Hence, \[ 3c(c-1)=1, \] or \[ 3c^2-3c-1=0. \] Therefore, \[ \boxed{3c^2-3c-1=0.} \] Thus, \[ \boxed{(C)} \] is the correct answer.