Verified answer

[tex]f(x) = {x}^{2} [/tex]

[tex]g(x) = 2x + 1[/tex]

[tex]h(x) = \sqrt{x - 1} [/tex]

[tex]a)[/tex]

[tex]y = f(h(x)) = f( \sqrt{x - 1} ) = {( \sqrt{x - 1}) }^{2} = x - 1[/tex]

[tex]y = f(g(x)) = f(2x + 1) = {(2x + 1)}^{2} = {4x}^{2} + 4x + 1[/tex]

[tex] \\ \\ [/tex]

[tex]b)[/tex]

[tex]f(h(1)) = f( \sqrt{1 - 1} ) = f(0) = {0}^{2} = 0[/tex]

[tex]f(g(1)) = f(2 \times 1 + 1) = f(2 + 1) = f(3) = {3}^{2} = 9[/tex]

[tex]f(g(1)) > f(h(1))[/tex]

[tex] \\ \\ [/tex]

[tex]c) \: \: \: \: \: g(f(h(x))) = g(f( \sqrt{x - 1} )) = g( ({ \sqrt{x - 1}) }^{2} )= g(x - 1) = 2x + 1 - 1 = 2x[/tex]

[tex] \\ \\ [/tex]

[tex]g(x) = 2x + 1[/tex]

Обратная будет:

[tex] \frac{1}{2x + 1} \\ [/tex]