Question:

<p><strong>Step 1: Set up the problem</strong><br/>We need to compare: $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6}, 7^{1/7}$</p><p><strong>Step 2: Analyze the function $f(x) = x^{1/x}$</strong><br/>To find which of these numbers is greatest, consider $f(x) = x^{1/x}$ for $x > 0$. Taking the natural logarithm:<br/>$$\ln f(x) = \frac{1}{x} \ln x = \frac{\ln x}{x}$$</p><p><strong>Step 3: Find the critical point using derivatives</strong><br/>Differentiate with respect to $x$:<br/>$$\frac{d}{dx}[\ln f(x)] = \frac{d}{dx}\left[\frac{\ln x}{x}\right]$$<br/>$$= \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$$</p><p><strong>Step 4: Set derivative equal to zero</strong><br/>$$\frac{1 - \ln x}{x^2} = 0$$<br/>$$1 - \ln x = 0$$<br/>$$\ln x = 1$$<br/>$$x = e \approx 2.718$$</p><p><strong>Step 5: Determine the nature of the critical point</strong><br/>For $0 < x < e$: $\ln x < 1$, so the derivative is positive (function increasing)<br/>For $x > e$: $\ln x > 1$, so the derivative is negative (function decreasing)<br/>Therefore, $f(x) = x^{1/x}$ has a maximum at $x = e$.</p><p><strong>Step 6: Compare the given numbers</strong><br/>Since $e \approx 2.718$ lies between 2 and 3, and $f(x)$ achieves its maximum at $x = e$:<br/>- The point closest to $e$ among our integers is $x = 3$<br/>- Therefore, $3^{1/3}$ is the greatest among the given numbers</p><p><strong>Step 7: Verify with approximate values</strong><br/>$2^{1/2} \approx 1.414$<br/>$3^{1/3} \approx 1.442$<br/>$4^{1/4} \approx 1.414$<br/>$5^{1/5} \approx 1.380$<br/>$6^{1/6} \approx 1.348$<br/>$7^{1/7} \approx 1.320$<br/><br/>$∴$ The greatest number is $3^{1/3}$, which corresponds to option (b)</p>