Let's solve the problem step by step: Given: \( f'(x) = (x+11)^{15} \cdot (x+5)^4 \cdot (x-2)^{14} \) We need to determine where \( f \) is **increasing** or **decreasing**. --- ## **Step 1: Find the critical points** Set \( f'(x) = 0 \): \[ (x+11)^{15} = 0 \implies x = -11 \] \[ (x+5)^4 = 0 \implies x = -5 \] \[ (x-2)^{14} = 0 \implies x = 2 \] So, critical points: \( x = -11, -5, 2 \) --- ## **Step 2: Test the sign of \( f'(x) \) in each interval** Intervals: \( (-\infty, -11) \), \((-11, -5)\), \((-5, 2)\), \((2, \infty)\) ### **Parity of exponents:** - \( (x+11)^{15} \) : Odd power, changes sign at \( x = -11 \) - \( (x+5)^4 \) : Even power, always non-negative, does **not** change sign at \( x = -5 \) - \( (x-2)^{14} \) : Even power, always non-negative, does **not** change sign at \( x = 2 \) Let's pick test points in each interval: --- ### **Interval 1: \( (-\infty, -11) \), test \( x = -12 \)** - \( x+11 = -1 \Rightarrow (-1)^{15} = -1 \) - \( x+5 = -7 \Rightarrow (-7)^4 = 2401 \) - \( x-2 = -14 \Rightarrow (-14)^{14} > 0 \) \[ f'(-12) = (-) \cdot (+) \cdot (+) = - \] So \( f \) is **decreasing** on \( (-\infty, -11) \) --- ### **Interval 2: \( (-11, -5) \), test \( x = -10 \)** - \( x+11 = 1 \Rightarrow (1)^{15} = 1 \) - \( x+5 = -5 \Rightarrow (-5)^4 = 625 \) - \( x-2 = -12 \Rightarrow (-12)^{14} > 0 \) \[ f'(-10) = (+) \cdot (+) \cdot (+) = + \] So \( f \) is **increasing** on \( (-11, -5) \) --- ### **Interval 3: \( (-5, 2) \), test \( x = 0 \)** - \( x+11 = 11 \Rightarrow (11)^{15} > 0 \) - \( x+5 = 5 \Rightarrow (5)^4 > 0 \) - \( x-2 = -2 \Rightarrow (-2)^{14} > 0 \) \[ f'(0) = (+) \cdot (+) \cdot (+) = + \] So \( f \) is **increasing** on \( (-5, 2) \) --- ### **Interval 4: \( (2, \infty) \), test \( x = 3 \)** - \( x+11 = 14 \Rightarrow (14)^{15} > 0 \) - \( x+5 = 8 \Rightarrow (8)^4 > 0 \) - \( x-2 = 1 \Rightarrow (1)^{14} > 0 \) \[ f'(3) = (+) \cdot (+) \cdot (+) = + \] So \( f \) is **increasing** on \( (2, \infty) \) --- ## **Step 3: Check at critical points** - At \( x = -11 \), \( (x+11)^{15} \) changes sign (root of **odd** multiplicity), so sign changes. - At \( x = -5 \), \( (x+5)^4 \) has **even** multiplicity, so **no sign change**. - At \( x = 2 \), \( (x-2)^{14} \) has **even** multiplicity, so **no sign change**. --- ## **Step 4: Final Answer** ### **Maximal intervals where \( f \) is increasing:** \[ \boxed{(-11, -5) \cup (-5, 2) \cup (2, \infty)} \] ### **Maximal intervals where \( f \) is decreasing:** \[ \boxed{(-\infty, -11)} \] --- **Summary Table:** | Interval | \( f'(x) \) sign | \( f(x) \) behavior | |---------------------|:----------------:|:-------------------:| | \( (-\infty, -11) \) | – | Decreasing | | \( (-11, -5) \) | + | Increasing | | \( (-5, 2) \) | + | Increasing | | \( (2, \infty) \) | + | Increasing | **Answer:** - \( f \) is **decreasing** on \( (-\infty, -11) \) - \( f \) is **increasing** on \( (-11, -5) \), \( (-5, 2) \), and \( (2, \infty) \)