Home › JEE Advanced › Mathematics › Limits and Continuity › For y = x³ - 6x² + 9x, find critical points (whe…
For y = x³ - 6x² + 9x, find critical points (where y' = 0):
Ax = 1, 3 (where y' = 3x² - 12x + 9 = 0)
Bx = -1, -3
Cx = 6
Dx = 0 only
Answer & Solution
Correct answer: A. x = 1, 3 (where y' = 3x² - 12x + 9 = 0)
y' = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3). Critical points: x = 1, 3. Second derivative test: y'' = 6x - 12. At x = 1: y'' = -6 (max). At x = 3: y'' = 6 (min).
Related questions
Two things this chain introduces beside continuity are differentiability and:Composites of continuous functions, by Theorem 2, are:A function that fails the limit test at a point is:To differentiate an implicit relation you use the:The function f(x) = |x| at x = 0 is continuous but not:For f(x) = 2x + 3 at x = 1, the limit and f(1) are both:An alternative to repeated product rule for many factors is:The chain rule can be extended to composites of: