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

[latex]-1 \leq n \leq 3[/latex] [latex]0 \leq n^2 \leq max(1^2, 3^2)[/latex] [latex]0 \leq n^2 \leq 9[/latex] [latex]0+1 \leq n^2+1 \leq 9+1[/latex] [latex]1 \leq n^2+1 \leq 10[/latex] [latex]\frac{1}{1} \geq \frac{1}{n^2+1} \geq \frac{1}{10}>0[/latex] [latex]\frac{1}{10} \leq \frac{1}{n^2+1} \leq 1[/latex] [latex]\frac{6}{10} \leq \frac{6}{n^2+1} \leq 6[/latex] [latex]-\frac{3}{5} \geq -\frac{6}{n^2+1} \geq -6[/latex] [latex]4-\frac{3}{5} \geq 4-\frac{6}{n^2+1} \geq 4-6[/latex] [latex] 3.4 \geq 4-\frac{6}{n^2+1} \geq -2[/latex] [latex]-2 \leq 4-\frac{6}{n^2+1} \leq 3.4[/latex]