Which of the numbers $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$ is largest, and how to find out without calculator? $1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$.
I got this question in an Application of Derivatives test. I think log might be used here to compare the values, but even then the values are very close to each other and differ by less than 0.02, which makes it difficult to get some specific answer to this question. How to solve this by a definite method?
Source: ISI entrance exam
 A: $3^{\frac{1}{3}}$ is the largest.
In general $y=x^{\frac{1}{x}}$ reaches its maximum at $x=e$ **. In this case, technically you can't know whether y is bigger when $x=2$ or $x=3$, but I guess you can always do quick mental math for these to realize $2^3<3^2$ and thus $2^{\frac{1}{2}}<3^{\frac{1}{3}}$.
** Proof:
$d(\ln{y})=d(\frac{\ln{x}}{x})$
$\frac{dy}{y}=\frac{1-\ln{x}}{x^2}dx$
solve $\frac{dy}{dx}=0$ to get $x=e$. It is also easy to verify that $\frac{dy}{dx}>0$ when $0<x<e$; $\frac{dy}{dx}<0$ when $x>e$.
A: HINT: This will helps you to solve the problem purely algebraically. $$n^{\dfrac1{n}}\lt(n+1)^{\dfrac1{n+1}}\iff n^{n+1}\lt(n+1)^n\iff n\lt\left(1+\dfrac1{n}\right)^n\color{Red}{\lt 3}.$$
A: Not a perfect answer but I guess this graph would help too!
Graph of $$y=x^{\frac{1}{x}}$$:

Graph of $$y=\dfrac{d}{dx}(x^{\frac{1}{x}})$$:

Hope this helps!
A: This is not calculus-based answer. (The original question suggests that the OP is mostly interested in answers using calculus, but I guess alternative approaches can be interesting too.)

We can show that $$\sqrt[n]n \ge \sqrt[n+1]{n+1}$$ for $n\ge 3$. 

The above inequality is clearly equivalent to
$$n^{n+1} \ge (n+1)^n$$
(for $n\ge 3$).
Several approaches how to show this fact can be found here (and maybe in a few other posts on this site):


*

*Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$

*proving that $(n-1)^n>n^{n-1}$
Now it remains to check that $1^2<2^1$ and $2^3<3^2$, so $n=3$ is indeed the first $n$ where the inequality occurs.
A: From the logarithmic derivative, $(\ln(x)/x)'=\left(1-\ln(x)\right)/x^2$, we deduce that the function $x^{1/x}$ has a single maximum at $x=e$.
As $2^{1/2}=4^{1/4}$ any intermediate value is higher, and so is $3^{1/3}$.
A: Hint: look at $f(x) = x^{1/x}$ and decide if this function is increasing or not. Write it as $x^{1/x} = e^{(\ln x)/x}$ to differentiate it easier.
A: Since $1$ is obviously smaller than all the rest, we can skip it.
Check the rest by taking a pair of numbers, comparing them and proceeding with the larger one:



*

*Take $2^{1/2}$ and $3^{1/3}$

*Raise them both to the power of $6$ (the LCM of $2$ and $3$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(2^{1/2}\right)^{6}=2^3=8<9=3^2=\left(3^{1/3}\right)^{6}$$



*

*Take $3^{1/3}$ and $4^{1/4}$

*Raise them both to the power of $12$ (the LCM of $3$ and $4$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{12}=3^4=81>64=4^3=\left(4^{1/4}\right)^{12}$$



*

*Take $3^{1/3}$ and $5^{1/5}$

*Raise them both to the power of $15$ (the LCM of $3$ and $5$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{15}=3^5=243>125=5^3=\left(5^{1/5}\right)^{15}$$



*

*Take $3^{1/3}$ and $6^{1/6}$

*Raise them both to the power of $6$ (the LCM of $3$ and $6$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{6}=3^2=9>6=6^1=\left(6^{1/6}\right)^{6}$$



*

*Take $3^{1/3}$ and $7^{1/7}$

*Raise them both to the power of $21$ (the LCM of $3$ and $7$)

*Since they are both positive, their order will be preserved and you will get:


$$\left(3^{1/3}\right)^{21}=3^7=2187>343=7^3=\left(7^{1/7}\right)^{21}$$

Hence the answer is $3^{1/3}$.
A: You can use calculus to find where the maximum of $f(x)=x^{1/x}$ occurs. You will find that it is at $x=e$. You will also find that $f(x)$ is increasing for $0< x<e$ and decreasing for $e<x$.
Unfortunately, the choice $e^{1/e}$ was not given. The two values of $x$ that you are given that bracket $e$ are $2$ and $3$, so the correct choice is either $2^{1/2}$ or $3^{1/3}$.
We can decide between the two of them by raising both sides to the sixth power.
\begin{align}
2^{1/2} &\stackrel{?}{=} 3^{1/3} \\
\left(2^{1/2}\right)^6 &\stackrel{?}{=} \left(3^{1/3}\right)^6 \\
2^3 &\stackrel{?}{=} 3^2 \\
8 &\stackrel{?}{=} 9
\end{align}
We see that $x=3$ gives us the larger function value, so the correct answer is
$$3^{1/3}$$
A: Consider the function $f(x) = x^{1/x}$. We ask, when is this maximized? So we ask, what is $f'(x)$? To do this, we use logarithmic derivatives.
$$ \log f(x) = \log x^{1/x} = \frac{\log x}{x},$$
so
$$ \frac{f'(x)}{f(x)} = \frac{1 - \log x}{x^2},$$
or rather
$$ f'(x) = \frac{1 - \log x}{x^2} x^{1/x}.$$
Notice that this is positive when $1 > \log x$ and negative when $1 < \log x$, so the actual max is at $x = e$. This means that the max value of $f(x)$ will be at either $2^{1/2}$ or $3^{1/3}$. We might guess that since $e$ is closer to $3$, that $e^{1/3}$ will be the bigger of the two. Using whatever method of your choosing to evaluate $2^{1/2}$ and $3^{1/3}$, we can confirm that heuristic: $3^{1/3}$ is larger.
A: If you haven't been introduced to elementary calculus yet, you can take a look at this solution. Comparing integers is easier than comparing irrational numbers. Since $[1, 2, \cdots, 7] = 420$, we can take the $420$th powers of all these numbers will give us a bunch of integers to compare: $$1, 2^{210}, 3^{140}, 2^{210}, 5^{84}, 2^{70} 3^{70}, 7^{60}.$$ We see after some basic arithmetic that $3^{140}$ is the largest among these numbers. Taking the $140$th root, it is obvious that $3^{1/3}$ is the largest among the original bunch of numbers.
