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Given, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?(a) 1(b) 2(c) 3(d) 4I had been asked this question in exam.This question is from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer»

Correct answer is (c) 3

Best explanation: We have, f(x) = x^3 – 12x^2 + 45X + 8……….(1)

Differentiating both SIDES of (1) with respect to x we

f’(x) = 3x^2 – 24x + 45

3x^2 – 24x + 45 = 0

Or x^2 – 8X + 15 = 0

Or (x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however SMALL, then,

f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.

f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.

Clearly, f’(x) changes sign from positive on the left to negative on the right of the POINT x = 3.

So, f(x) has maximum at 3.


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