Discuss the monotonicity of the following function without using differentiation.

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 $[3,+\infty)$.

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.

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

$$f(y)-f(x)=(y-x)[1-9/xy]$$

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

This inequality can be solved without (infinitesimal) calculus

$$1-9/xy\geq0$$

$$\frac{xy-9}{xy}\geq0$$

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