<s>[INST] 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? 
 [/INST] 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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