Найдите границы выражения 4-6/(n^2+1), если -1<=n<=3

$-1 \leq n \leq 3$
$0 \leq n^2 \leq max(1^2, 3^2)$
$0 \leq n^2 \leq 9$
$0+1 \leq n^2+1 \leq 9+1$
$1 \leq n^2+1 \leq 10$

0" alt="\frac{1}{1} \geq \frac{1}{n^2+1} \geq \frac{1}{10}>0" align="absmiddle" class="latex-formula">
$\frac{1}{10} \leq \frac{1}{n^2+1} \leq 1$
$\frac{6}{10} \leq \frac{6}{n^2+1} \leq 6$
$-\frac{3}{5} \geq -\frac{6}{n^2+1} \geq -6$
$4-\frac{3}{5} \geq 4-\frac{6}{n^2+1} \geq 4-6$
$3.4 \geq 4-\frac{6}{n^2+1} \geq -2$
$-2 \leq 4-\frac{6}{n^2+1} \leq 3.4$