The maximum value of f(x) = ≤ft(displaystyle1/x)^x is

The correct answer is Option (3) → \(e^{1/e}\)

Let \(y = \left(\frac{1}{x}\right)^x = x^{-x}\), x > 0. Since the variable is in the exponent, logarithmic differentiation is used:

    \(\log y = -x\log x\)

    \(\displaystyle \frac{1}{y}\frac{dy}{dx} = -\left(\log x + x\cdot\frac{1}{x}\right) = -(\log x + 1)\)

    \(\displaystyle \frac{dy}{dx} = -x^{-x}(\log x + 1)\)

Setting \(\frac{dy}{dx} = 0\):

    \(\log x = -1,\quad x = \frac{1}{e}\)

For \(x \lt \frac{1}{e}\), \(\log x + 1 \lt 0\), so \(\frac{dy}{dx} \gt 0\); for \(x \gt \frac{1}{e}\), \(\frac{dy}{dx} \lt 0\). The derivative changes from + to −, so \(x = \frac{1}{e}\) gives the maximum:

    \(\displaystyle f\left(\frac{1}{e}\right) = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e}\)

Why the other options are incorrect

• Option (1) \(e^{-1/e}\) — Incorrect. This is \(\left(\frac{1}{e}\right)^{1/e}\), the minimum value of \(x^x\), the reciprocal of the given function.

• Option (2) \(\left(\frac{1}{e}\right)^e\) — Incorrect. It equals \(e^{-e}\), which is less than 1, while the maximum \(e^{1/e}\) is greater than 1.

• Option (4) \(\left(\frac{1}{e}\right)^{-e}\) — Incorrect. It equals \(e^{e}\), not \(e^{1/e}\); it is larger than the maximum, so f never reaches it.