GATE 2025 AE – Question 40
The maximum value of the function $f(x) = (x-1)(x-2)(x-3)$ in the domain $[0, 3]$ occurs at $x = $ ______ (rounded off to two decimal places).
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Correct answer: 1.41 to 1.43
Explanation
$f(x) = x^3 - 6x^2 + 11x - 6$ and $f'(x) = 3x^2 - 12x + 11 = 0$ gives $x = 2 \pm \frac{1}{\sqrt{3}} = 1.423$ or $2.577$. The value at $x = 1.423$ is $+0.385$ (a maximum) and at $x = 2.577$ it is $-0.385$. The end points give $f(0) = -6$ and $f(3) = 0$. So the maximum is at $x = 1.42$.