Engineering Maths - Calculus
11. Minimum point of the function
is at
- Option : A
- Explanation : f'(x)= x2-1 For maximum and minimum value, f'(x) = 0 ∴ x2 -1= 0 x = ± 1 Again f(x) = 2x At x = 1, f '(x) = 2 > 0 f '(x) = -2 < 0 Hence minimum at x = 1
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