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What will be the point of maximum of the function 2x^3 + 3x^2 – 36x + 10?(a) -1(b) -2(c) -3(d) -4I had been asked this question in unit test.I would like to ask this question from Calculus Application in portion Application of Calculus of Mathematics – Class 12

Answer»

Right option is (c) -3

For explanation: Let y = 2x^3 + 3x^2 – 36x + 10……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(-3) + 6 = -30, which is < 0.


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