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

To determine the interval in which the function \( f(x) = x^3 - 3x^2 + 1 \) is decreasing, we need to find the derivative of the function and analyze its sign. Here's how we do it step-by-step:

1. Find the derivative of the function \( f(x) \).

\(f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 1) = 3x^2 - 6x\)

1. Set the derivative equal to zero to find the critical points.

\(3x^2 - 6x = 0\)

1. Factor the equation \( 3x^2 - 6x = 0 \).

\(3x(x - 2) = 0\)

1. Solve for \( x \).

\(x = 0 \quad \text{or} \quad x = 2\)

1. Determine the intervals around the critical points: \( (-\infty, 0) \), \( (0, 2) \), and \( (2, \infty) \).
2. Test a point from each interval in the derivative to see where it is increasing or decreasing.

• For \( x \in (-\infty, 0) \), choose \( x = -1 \):

\(f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 \quad (\text{positive})\)

• For \( x \in (0, 2) \), choose \( x = 1 \):

\(f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 \quad (\text{negative})\)

• For \( x \in (2, \infty) \), choose \( x = 3 \):

\(f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 \quad (\text{positive})\)

From this analysis, we see that the function is decreasing in the interval \( (0, 2) \).

Conclusion: The interval in which the function \( f(x) \) is decreasing is \( (0, 2) \). Thus, the correct answer is:

\((0,2)\)