Question

Let $f(x) = (x - 2)^{17} (x + 5)^{24}$. Then

• A. f does not have a critical point at x=2
• B. f has a minimum at x=2
• C. f has neither a maximum nor a minimum at x=2
• D. f has a maximum at x=2

Answer 1

The Correct Option is C

We are asked about the nature of the point \( x = 2 \) for the function: \[ f(x) = (x - 2)^{17} (x + 5)^{24} \] To determine whether it is a maximum, minimum, or neither, we analyze the derivative.

Step 1: First Derivative using Product Rule
Let: \[ f(x) = u(x) \cdot v(x) \quad \text{where } u(x) = (x - 2)^{17},\quad v(x) = (x + 5)^{24} \] Then: \[ f'(x) = u'(x)v(x) + u(x)v'(x) \]

Differentiate: \[ u'(x) = 17(x - 2)^{16},\quad v'(x) = 24(x + 5)^{23} \] So: \[ f'(x) = 17(x - 2)^{16}(x + 5)^{24} + (x - 2)^{17} \cdot 24(x + 5)^{23} \]

Factor common terms: \[ f'(x) = (x - 2)^{16}(x + 5)^{23} \left[ 17(x + 5) + 24(x - 2) \right] \]

Simplify the bracket: \[ 17(x + 5) + 24(x - 2) = 17x + 85 + 24x - 48 = 41x + 37 \] So: \[ f'(x) = (x - 2)^{16}(x + 5)^{23}(41x + 37) \]

Step 2: Check critical point at \( x = 2 \)
Set \( f'(x) = 0 \). This happens when: \[ (x - 2)^{16} = 0 \Rightarrow x = 2 \quad \text{(critical point)} \]

So \( x = 2 \) is a critical point. Now determine its nature.

Step 3: Use the first derivative test near \( x = 2 \)

• Choose \( x < 2 \), say \( x = 1.9 \): \( x - 2 < 0 \), \( (x - 2)^{16} > 0 \), since even power
• \( x + 5 > 0 \), \( 41x + 37 = 41(1.9) + 37 \approx 115.9 > 0 \)
• So \( f'(x) > 0 \) just before 2
• Now take \( x > 2 \), say \( x = 2.1 \): \( x - 2 > 0 \), all other terms also positive
• So \( f'(x) > 0 \) after 2 as well

Conclusion: \( f'(x) \) does not change sign at \( x = 2 \), and remains positive on both sides. So the graph is increasing before and after \( x = 2 \) \Rightarrow it is neither a maximum nor a minimum

Answer:

\[ \boxed{\text{f has neither a maximum nor a minimum at } x = 2} \]