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Given, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its minimum?(a) 1(b) 7(c) 3(d) 5I have been asked this question in quiz.My query is from Calculus Application topic in division Application of Calculus of Mathematics – Class 12

Answer»

The correct answer is (d) 5

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.


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