Can I discuss the monotonicity of the following function without using differentiation?

$$f(x) = x + \frac{9}{x}$$

Could anyone help me?

  • 4
    $\begingroup$ Well, what do you want to say? If $x$ is large then the $\frac 9x$ is negligible, so the function approaches the line $y=x$ (clearly monotonic increasing). If $x$ is small then it looks like $\frac 9x$, so monotone decreasing. $\endgroup$ – lulu Jun 16 '17 at 10:29
  • $\begingroup$ One approach is to say that it's well known that given two numbers with a constant product, the closer they are, the smaller their sum. $\endgroup$ – user49640 Jun 16 '17 at 10:29
  • 1
    $\begingroup$ We may use$ f(x)-f(y)=x+\frac{9}{x}-y-\frac{9}{y}=x-y-\frac{9(x-y)}{xy}=(x-y)\left(1-\frac{9}{xy}\right)$. By carefully considering different cases (where $1-\frac{9}{xy}$ is positive or negative), we can decide $f$ is increasing or decreasing on which part of the curve. $\endgroup$ – CY Aries Jun 16 '17 at 10:32
  • $\begingroup$ @user49640 I do not know how to apply your idea here $\endgroup$ – user426277 Jun 16 '17 at 16:07
  • $\begingroup$ @CYAries how can I do this, could you provide details please ? $\endgroup$ – user426277 Jun 16 '17 at 16:14

The function is odd, so it's enough to consider $x > 0$.

We have $$f(y) - f(x)= (y-x)\left(1 - \frac{9}{xy}\right).$$

Now, if $x < y \leq 3$, then $xy < 9$, and this shows that $f(y) - f(x) < 0$. So $f$ is strictly decreasing on $(0,3]$.

But if $3 \leq x < y$, then $xy > 9$, so $f(y) - f(x) > 0$. Hence $f$ is strictly increasing on $[3,+\infty)$.

  • $\begingroup$ Ok, but what if $x < 3 < y$? $\endgroup$ – Chris Jun 16 '17 at 17:24
  • $\begingroup$ What I wrote gives complete information about intervals of increase and decrease. Of course, one can't answer the question if $x < 3 < y$, because it depends on the values of $x$ and $y$. Plot the graph and you will see this. $\endgroup$ – user49640 Jun 16 '17 at 18:28
  • $\begingroup$ Alright, fair enough. $\endgroup$ – Chris Jun 16 '17 at 18:39

As $f$ is odd, it suffices to consider the case $x>0$.

One can verify in a straightforward manner that

$$f(x):=x+\dfrac{9}{x}=6 \cosh \left(\ln\left(\dfrac{x}{3}\right)\right).$$

Let us write this equality as:

$$f(x)=g(\cosh(\ln(h(x)))) \ \ \ \text{where} \ g, \ h \ \ \text{resp. denote multiplication by} \ \ 6 \ \text{and} \ 1/3.$$

Two cases:

  • for $x>3$ (where $\ln(x/3)> 0$), $f$ is a composition of 4 increasing functions (because the hyperbolic cosine is computed on $> 0$ values, thus is an increasing function.

  • for $0<x<3$ (where $\ln(x/3)<0$), $f$ is a composition of three increasing functions ($g,h$ and $\ln$) and a decreasing function (cosh is decreasing on $(-\infty,0)$), thus, $f$ a decreasing function.

  • $\begingroup$ for 0 <x < 3 , why are you sure that $ln (z/3)$ will be a negative function? $\endgroup$ – user426277 Jun 16 '17 at 16:21
  • $\begingroup$ It is $ln (x/3)$ (maybe in a earlier version, I had used a $z$ ?) $\endgroup$ – Jean Marie Jun 16 '17 at 16:23

Assume that $y\geq x$. Then study the sign of


The $(y-x)\geq0$. So, all it matters is the sign of $1-9/xy$.

This inequality can be solved without (infinitesimal) calculus



Now divide the plane in regions using the two axes $x=0$, $y=0$ and the hyperbola $xy=9$.

  • $\begingroup$ the last line is not clear for me. $\endgroup$ – user426277 Jun 16 '17 at 16:15
  • $\begingroup$ @Idonotknow The left-hand side has three factors $(xy-9)\frac{1}{x}\frac{1}{y}$. Its sign is determined by the signs of the factors. The sign of each factor changes at the places where they vanish. The places where they vanish are a parabola and two lines. The regions in between will have constant sign. $\endgroup$ – OR. Jun 16 '17 at 17:33

It's sufficient to discuss the behavior for $x>0,$ because the rest follows from $f(-x)=-f(x).$ Observe $f(x)=6+(\sqrt{x}-3/\sqrt{x})^2.$ Now $\sqrt{x}-3/\sqrt{x}$ is negative and growing for $x<3,$ so the square (and $f(x)$) is decreasing. For $x\ge3,$ $\sqrt{x}-3/\sqrt{x}$ is non-negative and growing, so the square is growing.


If $f$ is strictly increasing on an interval $I$, then $f(x)> f(y)$ for $x,y\in I$ with $x>y$.

Note that $f$ is undefined when $x=0$

\begin{align} f(x)-f(y)&=x-y+\frac{9}{x}-\frac{9}{y}\\ &=(x-y)\left(1-\frac{9}{xy}\right)\\ &=\frac{(x-y)\left(x-\frac{9}{y}\right)}{x}\\ \end{align}

For $x>0$, $\displaystyle f(x)-f(3)=\frac{(x-3)^2}{x}\ge 0$ and so $(3,f(3))$ is a local minimum.

If $x>y\ge3$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)>0$.

If $3\ge x>y>0$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)<0$.

For $x<0$, $\displaystyle f(x)-f(-3)=\frac{(x+3)^2}{x}\le 0$ and so $(-3,f(-3))$ is a local maximum.

If $0>x>y\ge-3$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)<0$.

If $-3\ge x>y$, $\displaystyle f(x)-f(y)=(x-y)\left(1-\frac{9}{xy}\right)>0$.

$f$ is strictly increasing on $(-\infty,-3]\cup[3,\infty)$.

$f$ is strictly decreasing on $[-3,0)\cup(0,3]$.


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