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What will be the minima for the function f(x) = x^4 – 8x^3 + 22x^2 – 24x + 8?(a) -1(b) 0(c) 2(d) 3I have been asked this question during an interview.This is a very interesting question from Calculus Application in section Application of Calculus of Mathematics – Class 12

Answer»

The correct OPTION is (d) 3

For EXPLANATION: We have, x^4 – 8x^3 + 22x^2 – 24x + 8 ……….(1)

Differentiating both sides of (1) with respect to x, we get,

f’(x) = 4x^3 – 24x^2 + 44x – 24 and f”(x) = 12x^2 – 48x + 44……….(2)

At an EXTREMUM of f(x), we have f’(x) = 0

Or 4x^3 – 24x^2 + 44x – 24 = 0

Or x^2(x – 1) – 5x(x – 1) + 6(x – 1) = 0

Or (x – 1)(x^2 – 5x + 6) = 0

Or (x – 1)(x – 2)(x – 3) = 0

So, x = 1, 2, 3

Now, f”(x) = 12x^2 – 48x + 44

f”(1) = 8 > 0

f”(2) = -4 < 0

f”(3) = 8 < 0

So, f(x) has minimum at x = 1 and 3.


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