Question

Помогите Пожалуйста!!! Пж пж пж​

Answer 1

$f(x) = {x}^{2}$

$g(x) = 2x + 1$

$h(x) = \sqrt{x - 1}$

$a)$

$y = f(h(x)) = f( \sqrt{x - 1} ) = {( \sqrt{x - 1}) }^{2} = x - 1$

$y = f(g(x)) = f(2x + 1) = {(2x + 1)}^{2} = {4x}^{2} + 4x + 1$

$b)$

$f(h(1)) = f( \sqrt{1 - 1} ) = f(0) = {0}^{2} = 0$

$f(g(1)) = f(2 \times 1 + 1) = f(2 + 1) = f(3) = {3}^{2} = 9$

$f(g(1)) > f(h(1))$

$c) \: \: \: \: \: g(f(h(x))) = g(f( \sqrt{x - 1} )) = g( ({ \sqrt{x - 1}) }^{2} )= g(x - 1) = 2x + 1 - 1 = 2x$

$g(x) = 2x + 1$

Обратная будет:

$\frac{1}{2x + 1}$