Question

If \(f(x)=x^3-3x^2+1\), then the interval in which the function is decreasing is:

• A. \((-\infty,0)\)
• B. \((0,2)\)
• C. \((2,\infty)\)
• D. \((-\infty,\infty)\)

Hint:

If: \[ f'(x)>0 \] then function is increasing. If: \[ f'(x)<0 \] then function is decreasing.

Answer 1

The Correct Option is B

Concept: A function decreases in the interval where: \[ f'(x)<0 \] Therefore, we first differentiate the function and then analyze the sign of the derivative.

Step 1: Differentiating the function. Given: \[ f(x)=x^3-3x^2+1 \] Differentiating: \[ f'(x)=3x^2-6x \] Taking common factor: \[ f'(x)=3x(x-2) \]

Step 2: Finding the critical points. Set derivative equal to zero: \[ 3x(x-2)=0 \] Therefore: \[ x=0,\ 2 \] These points divide the number line into intervals: \[ (-\infty,0),\ (0,2),\ (2,\infty) \]

Step 3: Checking the sign of derivative. For \(x<0\), take \(x=-1\): \[ f'(-1)=3(-1)(-3)>0 \] Hence function is increasing. For \(0<x<2\), take \(x=1\): \[ f'(1)=3(1)(-1)<0 \] Hence function is decreasing. For \(x>2\), take \(x=3\): \[ f'(3)=3(3)(1)>0 \] Hence function is increasing. Therefore, the function decreases in: \[ (0,2) \]