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What will be the minimum value of the function 2x^3 + 3x^2 – 36x + 10?(a) -31(b) 31(c) -34(d) 34I got this question at a job interview.Question is taken from Calculus Application topic in section Application of Calculus of Mathematics – Class 12

Answer»

The correct choice is (c) -34

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(2) + 6 = 30 > 0

Putting x = 2 in (1) we get its minimum value as,

2x^3 + 3x^2 – 36x + 10 = 2(2)^3 + 3(2)^2 – 36(2) + 10

= -34


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