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GATE 2016 ME (ME1) – Question 36

Engineering Mathematics · Calculus: Partial derivatives, Taylor series, maxima and minima, Fourier series · 2 marks · Numerical answer

For $f(x)=2x^3-3x^2$ on $[-1,2]$, the global minimum is ______.

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Correct answer: -5.1 to -4.9

Explanation

For f=2x³−3x², f-prime=6x(x−1), so stationary points are0 and1. Evaluate these and the endpoints−1 and2: f values are0,−1,−5 and4 respectively. The smallest is **−5**, attained at x=−1. Checking only stationary points would miss the global minimum.