How to show that a function is strictly decreasing. The function $f \colon \mathbb{R} \to \mathbb{R}$ is defined by $f(x)=\frac{x^2+2x}{x^2-1}$.
How would you show that $f(x)$ is a strictly decreasing function.
Actually your function $f(x) = \frac{x^2+2x}{x^2-1}$ is NOT strictly decreasing in its whole domain $\mathbb{R} \smallsetminus \{-1,1\}$. For instance, $f(\frac{1}{2}) = \frac{5}{4}(-\frac{4}{3}) = -\frac{5}{3} &lt;  \frac{8}{3} = f(2)$ but $\frac{1}{2} &lt; 2$.

It is true that $f(x)$ is strictly decreasing in $(-\infty, -1)$, and in $(-1,1)$, and in $(1, +\infty)$. You can prove that because the derivative $f'(x) = -2\frac{x^2+x+1}{(x^2-1)^2} &lt; 0$ for all $x \in \mathbb{R} \smallsetminus \{-1,1\}$ and $f(x)$ is continuous in $(-\infty, -1)$, and in $(-1,1)$, and in $(1, +\infty)$. But this does not imply that $f(x)$ is strictly decreasing in $\mathbb{R} \smallsetminus \{-1, 1\}$ since $x = -1$ and $x = 1$ are infinite discontinuities. More precisely, $\lim_{x\to \pm 1^-} f(x) = -\infty$ and  $\lim_{x\to \pm 1^+} f(x) = +\infty$.
Hint: Prove that $$f'(x)=-2\,{\frac {x+{x}^{2}+1}{ \left( {x}^{2}-1 \right) ^{2}}}&lt;0,$$
for suitable $x.$