Question

Find the intervals in which the following function \( f(x) \) is strictly increasing or strictly decreasing: \[ f(x) = 20 - 9x + 6x^2 - x^3. \]

Hint:

The sign of \( f'(x) \) determines whether \( f(x) \) is increasing (\( f'(x)>0 \)) or decreasing (\( f'(x)<0 \)).

Answer 1

Step 1: Compute the first derivative \( f'(x) \): \[ f'(x) = \frac{d}{dx}\left(20 - 9x + 6x^2 - x^3\right) = -9 + 12x - 3x^2. \] 
Step 2: Factorize \( f'(x) \): \[ f'(x) = -3(3 - 4x + x^2) = -3(x - 3)(x - 1). \] 
Step 3: Solve \( f'(x) = 0 \): \[ (x - 3)(x - 1) = 0 \quad \Rightarrow \quad x = 1, \, x = 3. \] The critical points are \( x = 1 \) and \( x = 3 \). 
Step 4: Test the sign of \( f'(x) \) in each interval \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \): 
- For \( x \in (-\infty, 1) \): Choose \( x = 0 \): \[ f'(0) = -3(0 - 3)(0 - 1) = -3(-3)(-1) = -9 \, ({negative, decreasing}). \] - For \( x \in (1, 3) \): Choose \( x = 2 \): \[ f'(2) = -3(2 - 3)(2 - 1) = -3(-1)(1) = 3 \, ({positive, increasing}). \] - For \( x \in (3, \infty) \): Choose \( x = 4 \): \[ f'(4) = -3(4 - 3)(4 - 1) = -3(1)(3) = -9 \, ({negative, decreasing}). \] 
Step 5: Conclusion: - \( f(x) \) is strictly decreasing in \( (-\infty, 1) \cup (3, \infty) \). - \( f(x) \) is strictly increasing in \( (1, 3) \).