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Let f(x) = (x – 2)17 (x + 5)24. Then
  • a)
    f does not have a critical point at 2
  • b)
    f has a minimum at 2
  • c)
    f has neither minimum nor a maximum at x = 2
  • d)
    f has a maximum at 2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let f(x) = (x – 2)17 (x + 5)24. Thena)f does not have a critical...
Here f(x) = (x – 2)17 (x + 5)24 
f17 (2) ≠ 0 
⇒ Odd derivative of function is non-zero. So, f has neither a minimum nor a maximum at 2.
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Most Upvoted Answer
Let f(x) = (x – 2)17 (x + 5)24. Thena)f does not have a critical...
Here f(x) = (x – 2)17 (x + 5)24 
f17 (2) ≠ 0 
⇒ Odd derivative of function is non-zero. So, f has neither a minimum nor a maximum at 2.
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Community Answer
Let f(x) = (x – 2)17 (x + 5)24. Thena)f does not have a critical...
Understanding the Function f(x)
The function given is f(x) = (x - 2)^17 * (x + 5)^24. To analyze the behavior of f(x) at x = 2, we need to investigate its critical points and local extrema.
Critical Points
- A critical point occurs where the derivative f'(x) is zero or undefined.
- To find f'(x), we apply the product rule and analyze the behavior near x = 2.
Behavior at x = 2
- At x = 2, the term (x - 2)^17 equals zero, making f(2) = 0.
- The derivative f'(x) will also involve (x - 2) terms, leading to f'(2) = 0.
Local Extrema Analysis
- To determine the nature of the critical point at x = 2, we examine f(x) around this point.
- As x approaches 2 from the left, f(x) is negative (since (x - 2)^17 is negative).
- As x approaches 2 from the right, f(x) is still negative (since (x - 2)^17 is positive).
- Therefore, f(x) does not change sign around x = 2, indicating that it is neither a maximum nor a minimum.
Conclusion
- Since f(x) does not have a change in direction around x = 2, it confirms that:
- f has neither a minimum nor a maximum at x = 2.
Thus, the correct answer is option 'C'.
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