GATE 2026 ES – Question 13
The global minimum of $f(x)=3x^4+4x^3-12x^2+6$ for real $x$ is
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Correct answer: (C) −26
Explanation
Differentiate: $f^{\prime}(x)=12x^3+12x^2-24x=12x(x+2)(x-1)$. Critical points are $-2,0,1$,with values $-26,6,1$. The positive quartic leading term ensures $f(x)\to+\infty$ at both ends, so the least critical value is the global minimum: **−26,C**.