Let
𝑓
(
𝑥
)
=
𝑥
3
−
5
𝑥
2
+
5
𝑥
+
2.
Which of the following is a correct classification of the critical points of
𝑓
?

Question

Let 
𝑓
(
𝑥
)
=
𝑥
3
−
5
𝑥
2
+
5
𝑥
+
2.
 Which of the following is a correct classification of the critical points of 
𝑓
?

BlackTom AI · Accepted Answer

Compute f'(x) = 3x^2 − 10x + 5. Setting f'(x) = 0 finds critical points. 1. Relevant knowledge points and concepts - Critical points occur where f'(x) = 0 or f' is undefined. - The second derivative test: if f''(c) < 0 then f has a local maximum at c; if f''(c) > 0 then f has a local minimum at c. 2. Step-by-step solving f'(x) = 0 3x^2 − 10x + 5 = 0. Use the quadratic formula: x = [10 ± sqrt( (-10)^2 − 4·3·5 )] / (2·3) = [10 ± sqrt(100 − 60)] / 6 = [10 ± sqrt(40)] / 6 = [10 ± 2√10] / 6 = (5 ± √10) / 3. So the two critical points are x = (5 − √10)/3 and x = (5 + √10)/3. 3. Intermediate calculations shown - Discriminant: 100 − 60 = 40, sqrt(40) = 2√10. - Simplification: (10 ± 2√10)/6 = (5 ± √10)/3. 4. Apply second derivative test and classify Compute f''(x) = 6x − 10. Evaluate at x = (5 − √10)/3: f'' = 6*(5 − √10)/3 − 10 = 2*(5 − √10) − 10 = 10 − 2√10 − 10 = −2√10 < 0, so x = (5 − √10)/3 is a local maximum. Evaluate at x = (5 + √10)/3: f'' = 2*(5 + √10) − 10 = 10 + 2√10 − 10 = 2√10 > 0, so x = (5 + √10)/3 is a local minimum. 5. Why other classifications would be incorrect Any alternative assignment (for example swapping maximum/minimum, claiming both are minima or maxima, or saying there are no extrema) contradicts the sign of f'' at these points. The negative second derivative at the smaller root forces a maximum, and the positive second derivative at the larger root forces a minimum. 6. Conclusion and confirmation Thus f has a local maximum at x = (5 − √10)/3 and a local minimum at x = (5 + √10)/3. This matches the provided answer (written as x = 5/3 − 10/3 and x = 10/3 + 5/3 when interpreted as (5 ± √10)/3).

Let 𝑓 ( 𝑥 ) = 𝑥 3 − 5 𝑥 2 + 5 𝑥 + 2. Which of the following is a correct classification of the critical points of 𝑓 ?单项选择题

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