I need to prove that the set $M=\left\{\cfrac{6}{{\left(x-2\right)}^{3}}:\quad x≤0\right\}$ is bounded and find inf M and sup M.
The derivative $\cfrac{d}{dx}\cfrac{6}{{\left(x-2\right)}^{3}}=\cfrac{-18\quad}{{\left(x-2\right)}^{4}}≤0$ for all $x≤0$. Therefore $\cfrac{6}{{\left(x-2\right)}^{3}}$ is decreasing for all $x≤0$.
From this, I understand that M will have its highest value when $x=-\infty$ (which is $0$) and its lowest value when $x=0$ (which is $\cfrac{-3}{4}$). Therefore, sup M$=0$ and inf M$\cfrac{-3}{4}$.
Now, how do I prove this?
You have that $$\lim_{x\to-\infty }\frac{6}{(x-2)^3}=0\quad \text{and}\quad \lim_{x\to 0}\frac{6}{(x-2)^3}=\frac{-3}{4}.$$ and thus the function is bounded, and thus admit an infimum and a supremum.
Now, the derivative is strictly negative for all $x$, therefore the function has no local extremum on $(-\infty ,0)$. Therefore, $$\frac{-3}{4}\leq \frac{6}{(x-2)^3}\leq 0,$$ and thus $\frac{-3}{4}$ it the infimum and $0$ the supremum.