Question

Дано функции f(x)= √ x+1 и g (x) = x^ 2 - 2x. задайте формулой функцию: 1)y = f (3x) 2)y = g (-x) 3)y = f (g (x)) 4)y = g (f (x)) 5)y = f (f (x)) 6)y = g (g (x))

Answer 1

$f(x) = \sqrt {x+1}, \ g(x) = x^2 -2x. \\ \\ 1) \ y = f(3x)=\sqrt{3x+1}; \\ 2) \ y = g(-x) = x^2 + 2x; \\ 3) \ y = f(g(x))=\sqrt{x^2-2x+1} = \sqrt{(x-1)^2}=|x-1|; \\ 4) \ y = g(f(x)) = (\sqrt{x+1})^2 - 2\sqrt{x+1} = x + 1 - 2\sqrt{x+1}; \\ 5) \ y = f(f(x))= \sqrt{\sqrt{x+1}+1}; \\ 6) \ y = g(g(x)) = (x^2-2x)^2 - 2(x^2-2x) = x^4 - 4x^3 + 4 x^2 - 2x^2 +\\ + 4x = x^4 - 4x^3 + 2 x^2 +4x.$