What will be nature of the f(x) = 10 – 9x + 6x^2 – x^3 for x > 3?

(a) Decreases

(b) Increases

(c) Cannot be determined for x > 3

(d) A constant function

The question was posed to me during an online exam.

I’d like to ask this question from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12

NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options

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The correct option is (a) Decreases

For explanation: f(x) = 10 – 9x + 6x^2 – x^3

Thus, f’(x) = – 9 + 12x –3 x^2

= -3(x^2 – 4x + 3)

Or f’(x) = -3(x – 1)(x – 3) ……….(1)

If x > 3, then x – 1 > 0 and x – 3 > 0

Hence, (x – 1)(x – 3) > 0

Thus, from (1) it readily follows that, f’(x) < 0, when x > 3

So, f(x) decreases for values of x > 3.