f′(x)=12x2−48=0⇒x2=4⇒x=2 (in [1, 3]).
Evaluate at critical point and endpoints:
f(1)=4−48+105=61.
f(2)=32−96+105=41.
f(3)=108−144+105=69.
Minimum value = 41 at x = 2.
The minimum value of f(x) = 4x^3 - 48x + 105 in the interval [1,3] is :
CUET UG 2023 — Mathematics Applied-Mathematics
2023mcqmedium
The minimum value of f(x)=4x3−48x+105 in the interval [1,3] is :
Official previous-year question
Held on 7 Jun 2023 · Verified 13 Jul 2026.
Options
- A
61
- B
41
- C
32
- D
69
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