Question

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

Answer 1

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