$1, 2^{1/2}, 3^{1/3}, 4^{1/4}, 5^{1/5}, 6^{1/6} , 7^{1/7}$.

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?

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    $\begingroup$ Let $f(x)=x^{1/x}$ with $x\in [1,7]$ and find maximum using calculus. $\endgroup$ – Taylor Jun 19 '15 at 3:21
  • $\begingroup$ Alternate solution: log tables. I can lay hands on them. $\endgroup$ – Joshua Jun 20 '15 at 18:51
  • $\begingroup$ @Taylor How does one find a maximum of a function defined on integers, using calculus? $\endgroup$ – DVD Jun 24 '15 at 0:33
  • $\begingroup$ why this kind of question has a lot of viewers and upvotes? $\endgroup$ – Luis Felipe Jul 3 '15 at 3:02
  • $\begingroup$ @LuisFelipeVillavicencioLopez don't be jealous..:) $\endgroup$ – anshabhi Jul 3 '15 at 7:20

11 Answers 11


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


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    $\begingroup$ +1 just adding that "raising both sides to the sixth power" is not a magical choice, instead it is the lowest common denominator for 2 and 3, thus turning the fractional exponents (hard to compute mentally) into integer exponents (easy or not-so-hard to compute) $\endgroup$ – Rolazaro Azeveires Jun 20 '15 at 9:41
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    $\begingroup$ An alternative way to determine that $3$ must be the solution and not $2$ is to realise $2^{1/2} = 4^{1/4}$, and 4 was already counted out by the calculus approach. $\endgroup$ – leftaroundabout Jun 20 '15 at 14:26
  • $\begingroup$ Another alternative way is to see that $e$ is approximately $2.71828$ which is close to $3$ compared to $2$. Since, there is a maxima at $x=e$, $3^{1/3}>2^{1/2}$ $\endgroup$ – Dhanush Krishna Mar 22 '16 at 16:52
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    $\begingroup$ @DhanushKrishna: That method works here, but it is "by accident." The values of $x$ closer to each other does not necessarily mean that the values of $y$ are also closer to each other. For example, consider $y=x^2$. $0.6$ is closer to $1$ than it is to $0$, but $0.6^2$ is not closer to $1^2$ than it is to $0^2$. $\endgroup$ – Rory Daulton Mar 22 '16 at 17:39
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    $\begingroup$ @DhanushKrishna: Then consider the function $f(x)=-\frac{e^x}x$. That has a maximum at $x=1$, but $f(2.5)$ is closer to $f(1)$ than $f(0.25)$ is although $0.25$ is closer to $1$ than $2.5$ is. $\endgroup$ – Rory Daulton Mar 24 '16 at 10:08

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.


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.


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:


  • 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:


  • 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:


  • 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:


  • 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:


Hence the answer is $3^{1/3}$.


$3^{\frac{1}{3}}$ 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:



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 $x>e$.


HINT: This will helps you to solve the problem purely algebraically. $$n^{\dfrac1{n}}\lt(n+1)^{\dfrac1{n+1}}\iff n^{n+1}\lt(n+1)^n\iff n\lt\left(1+\dfrac1{n}\right)^n\color{Red}{\lt 3}.$$

  • $\begingroup$ i dont understand if this holds: $n^{\dfrac1{n}}\lt(n+1)^{\dfrac1{n+1}}$ then the maximum should be the $7^{1/7}$, no? $\endgroup$ – Nikos M. Jun 24 '15 at 5:14
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    $\begingroup$ No!. If $n^{\dfrac1{n}}\lt(n+1)^{\dfrac1{n+1}},$ then $n\lt3.$ $\endgroup$ – Bumblebee Jun 24 '15 at 7:55
  • $\begingroup$ ahh, ok, but still the maximum is supposed to be on $3$ and according to the inequality it is $2$, maybe you should put $\ge$ instead of $\gt$ $\endgroup$ – Nikos M. Jun 24 '15 at 12:27
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    $\begingroup$ When $n=2$ we have $2^{\frac12}\lt 3^{\frac13}.$ $\endgroup$ – Bumblebee Jun 25 '15 at 3:19

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!


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):

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.


An alternative, brute-force way to tackle this problem is basically what Rory did to decide between $2^{1/2}$ and $3^{1/3}$: if you exponentiate all numbers by $7!$, you are sure to get all integer values

$$ \bigl\{k^{\frac{7!}{k}}|k\in\{1\ldots7\}\bigr\} $$

Unfortunately they're rather big-ish...

GHCi> [ k ^ (product [1..7] `div` k) | k<-[1..7] ]

but looking at the length of the numbers shows the third one to be biggest (because longest).

Of course, you couldn't feasibly calculate these giant powers without a calculator (not without some approximation... but then then better do the original problem directly), however if not 7 was the highest candidate but 4 or 5, it would still be doable (if laborious).

GHCi> [ k ^ (product [1..5] `div` k) | k<-[1..5] ]
  [1, 1152921504606846976, 12157665459056928801, 1152921504606846976,

Furthermore, you don't really need $n!$, only the least common multiple.

GHCi> [ k ^ (foldr lcm 1 [1..5] `div` k) | k<-[1..5] ]
  [1, 1073741824, 3486784401, 1073741824, 244140625]
  • $\begingroup$ You can do this much more feasibly if you take the options two at a time (to order each pair of solutions). $\endgroup$ – Mario Carneiro Jun 23 '15 at 20:47
  • $\begingroup$ ...as shown in barak manos' solution, right. $\endgroup$ – leftaroundabout Jun 23 '15 at 21:25

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 $3^{1/3}$.


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.


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