Question

Let \[ f(x)=x^3-\frac32x^2, \] which of the following is not correct?

• A. In \((0,2)\), \(3x^2-3x-1=0\) has at least one solution.
• B. \(f''(x)=0\) has a solution in \((0,2)\).
• C. \(f(x)\) is increasing in \((-\infty,0)\) and \((1,\infty)\).
• D. \(f(x)\) is decreasing in \((0,\infty)\).

Hint:

To determine increasing or decreasing intervals, \[ \boxed{ \begin{aligned} f'(x)>0 &\Rightarrow \text{Increasing}, f'(x)<0 &\Rightarrow \text{Decreasing}. \end{aligned} } \] Always analyze the sign of the first derivative on each interval.

Answer 1

The Correct Option is D

Step 1: Find the first derivative. Given, \[ f(x)=x^3-\frac32x^2. \] Differentiating, \[ f'(x)=3x^2-3x \] \[ =3x(x-1). \]

Step 2: Determine the intervals of increase and decrease. For \[ x<0, \] both \(x\) and \((x-1)\) are negative. Hence, \[ f'(x)>0. \] For \[ 0<x<1, \] \[ f'(x)<0. \] For \[ x>1, \] \[ f'(x)>0. \] Therefore, \[ f(x) \] is increasing in \[ (-\infty,0) \] and \[ (1,\infty), \] and decreasing only in \[ (0,1). \] Thus, statement (D) is false.

Step 3: Verify the remaining statements. Second derivative, \[ f''(x)=6x-3. \] Setting \[ f''(x)=0, \] we obtain \[ 6x-3=0 \] \[ x=\frac12, \] which lies in \[ (0,2). \] Hence, statement (B) is true. Also, \[ 3x^2-3x-1=0 \] has roots \[ x=\frac{3\pm\sqrt{21}}6. \] One root lies in \[ (0,2), \] so statement (A) is also true. Therefore, \[ \boxed{\text{Statement (D) is not correct}.} \] Thus, \[ \boxed{(D)} \] is the correct answer.