Engineering Maths - Calculus

19. The maxima and minima of the function f(x) = 2x3 - 15x2 + 36x + 10 occur respectively at

  • Option : C
  • Explanation : f(x)=2x3-15x2+36x+10 f '(x)=6x2-30x+36 and f ''(x)=12x-30 For maxima and minima, f '(x)=0 6x2-30x+36 = 0 or x2- 5x + 6 = 0 or X = 3,2
    Putting these values of x in equation (ii), we get
    f '(3) = 36 - 30 = + 6 positive
    Hence minimum value = 3
    f ' (2) = 24 - 30 = - 6 negative, hence maximum value = 2
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