Functions

Inverse Derivative Rule: $g^{'} (y) = \frac{1}{f ^{'} ( x )}$ where $y = f (x)$ .
Identify $x$ for $y = 0$ : Solve $x^{3} - 6 x^{2} + 11 x - 6 = 0$ .
- By inspection, $x = 1$ gives $1 - 6 + 11 - 6 = 0$ . So $f (1) = 0$ .
Derivative of $f$ : $f^{'} (x) = 3 x^{2} - 12 x + 11$ .
Evaluate at $x = 1$ : $f^{'} (1) = 3 (1)^{2} - 12 (1) + 11 = 2$ .
Apply Rule: $g^{'} (0) = \frac{1}{f ^{'} ( 1 )} = \frac{1}{2}$ .

This hard difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of functions. It appeared in the 2025 exam.

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Solution

Given $f (x) = x^{1/ x}$ for $x > 0$ . The maximum value of $f (x)$ is: