The minimum value of \(f(x)=4x^3-48x+105\) in the interval [1, 3] is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 41
On a closed interval the absolute minimum is the least of the values at the critical points inside the interval and at the end points.
\(f'(x) = 12x^2 - 48\)
\(12x^2 - 48 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\)
Only x = 2 lies in [1, 3]. The values:
\(f(1) = 4 - 48 + 105 = 61\)
\(f(2) = 32 - 96 + 105 = 41\)
\(f(3) = 108 - 144 + 105 = 69\)
The least of these is 41, at x = 2.
Why the other options are incorrect
Option (1) 61 — Incorrect. 61 is the value at the end point x = 1; f is smaller at x = 2.
Option (3) 32 — Incorrect. f takes the values 61, 41 and 69 at x = 1, 2, 3, and f(2) = 41 is the least, so 32 is never reached on [1, 3].
Option (4) 69 — Incorrect. 69 = f(3) is the maximum value on [1, 3], not the minimum.