Question

Given, f(x) = x^3 – 12x^2 + 45x + 8. What is the minimum value of f(x)?

(a) -1

(b) 0

(c) 1

(d) Value does not exist

I have been asked this question during an online exam.

Enquiry is from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer 1

Right answer is (c) 1

The 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’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.

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

Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.

So, f(x) has minimum at 5.

Putting, x = 5 in (1)

Thus, its maximum value is,

f(3) = 5^3 – 12*5^2 + 45*5 + 8 = 58.