f′(x)=6x2+6x−12=6(x2+x−2)=6(x+2)(x−1)
For strict increase, f′(x)>0.
(x+2)(x−1)>0 when x<−2 or x>1.
So f is strictly increasing on (−∞,−2)∪(1,∞).
CUET UG 2022 — Mathematics Calculus
The interval in which the function f(x)=2x3+3x2−12x+1 is strictly increasing is -
Held on 4 Aug 2022 · Verified 13 Jul 2026.
[−2,1]
(−∞,1]
(−∞,−2)∪(1,∞)
(−∞,−1]∪[2,∞)
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