Discuss the monotonicity of the following function without using differentiation. Can I discuss the monotonicity of the following function without using differentiation?
$$f(x) = x + \frac{9}{x}$$
Could anyone help me? 
 A: 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 $[3,+\infty)$.
A: 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.  
A: 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 $xy=9$.
A: 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.
A: 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 $[-3,0)\cup(0,3]$.
