# Given, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?

Category: QuestionsGiven, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?
Editor">Editor Staff asked 11 months ago

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) 4

This question is from Calculus Application in section Application of Calculus of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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.