Dadas as funções a seguir encontre as fórmulas para as funções (f+g)(x),(f-g)(x),(f g)(x), (f / g)(x),(f \circ g)(x) e (g \circ f)(x).
a) f(x)=x^{2} e g(x)=\sqrt{x+1}
b) f(x)=(x-1)^{2} e g(x)=7 x-4
c) f(x)=\sqrt{x+5} e g(x)=\frac{1}{x}
d) f(x)=\sqrt{x+1} e g(x)=|x-1|
e) f(x)=x^{3} e g(x)=\sqrt[3]{1-x^{3}}
f) f(x)=10^{x} e g(x)=3 x^{2}

Passo 1

O enunciado pede para, dadas duas funções f e g, obter as fórmulas de \,(f+g)(x),\;(f-g)(x),\;(f\,g)(x),\;(f/g)(x),\;(f\circ g)(x) e (g\circ f)(x).
Para qualquer par (f,g) vale
(f+g)(x)=f(x)+g(x),\quad (f-g)(x)=f(x)-g(x),\quad (f\,g)(x)=f(x)\,g(x),
\bigl(f/g\bigr)(x)=\frac{f(x)}{g(x)},\quad (f\circ g)(x)=f\bigl(g(x)\bigr),\quad (g\circ f)(x)=g\bigl(f(x)\bigr).
Nos Passos seguintes aplicamos essas definições a cada par dado.

Passo 2

Para f(x)=x^{2} e g(x)=\sqrt{x+1}:
(f+g)(x)=x^{2}+\sqrt{x+1},\quad (f-g)(x)=x^{2}-\sqrt{x+1},
(f\,g)(x)=x^{2}\,\sqrt{x+1},\quad \bigl(f/g\bigr)(x)=\frac{x^{2}}{\sqrt{x+1}},
(f\circ g)(x)=\bigl(\sqrt{x+1}\bigr)^{2}=x+1,\quad (g\circ f)(x)=\sqrt{x^{2}+1}.

Passo 3

Para f(x)=(x-1)^{2} e g(x)=7x-4:
(f+g)(x)=(x-1)^{2}+7x-4,\quad (f-g)(x)=(x-1)^{2}-\bigl(7x-4\bigr),
(f\,g)(x)=(x-1)^{2}\,(7x-4),\quad \bigl(f/g\bigr)(x)=\frac{(x-1)^{2}}{7x-4},
(f\circ g)(x)=\bigl(7x-4-1\bigr)^{2}=(7x-5)^{2},\quad (g\circ f)(x)=7\,(x-1)^{2}-4.

Passo 4

Para f(x)=\sqrt{x+5} e g(x)=\frac{1}{x}:
(f+g)(x)=\sqrt{x+5}+\frac{1}{x},\quad (f-g)(x)=\sqrt{x+5}-\frac{1}{x},
(f\,g)(x)=\sqrt{x+5}\,\frac{1}{x}=\frac{\sqrt{x+5}}{x},\quad \bigl(f/g\bigr)(x)=\frac{\sqrt{x+5}}{1/x}=x\,\sqrt{x+5},
(f\circ g)(x)=\sqrt{\frac{1}{x}+5}=\sqrt{\frac{1+5x}{x}},\quad (g\circ f)(x)=\frac{1}{\sqrt{x+5}}.

Passo 5

Para f(x)=\sqrt{x+1} e g(x)=|x-1|:
(f+g)(x)=\sqrt{x+1}+|x-1|,\quad (f-g)(x)=\sqrt{x+1}-|x-1|,
(f\,g)(x)=\sqrt{x+1}\,|x-1|,\quad \bigl(f/g\bigr)(x)=\frac{\sqrt{x+1}}{|x-1|},
(f\circ g)(x)=\sqrt{\;|x-1|+1\;}=\sqrt{|x-1|+1},\quad (g\circ f)(x)=\bigl|\sqrt{x+1}-1\bigr|.

Passo 6

Para f(x)=x^{3} e g(x)=\sqrt[3]{\,1 - x^{3}\,}:
(f+g)(x)=x^{3}+\sqrt[3]{1-x^{3}},\quad (f-g)(x)=x^{3}-\sqrt[3]{1-x^{3}},
(f\,g)(x)=x^{3}\,\sqrt[3]{1-x^{3}},\quad \bigl(f/g\bigr)(x)=\frac{x^{3}}{\sqrt[3]{1-x^{3}}},
(f\circ g)(x)=\bigl(\sqrt[3]{1-x^{3}}\bigr)^{3}=1-x^{3},\quad (g\circ f)(x)=\sqrt[3]{\,1 - (x^{3})^{3}\,}=\sqrt[3]{\,1 - x^{9}\,}.

Passo 7

Para f(x)=10^{x} e g(x)=3x^{2}:
(f+g)(x)=10^{x}+3x^{2},\quad (f-g)(x)=10^{x}-3x^{2},
(f\,g)(x)=10^{x}\,3x^{2}=3x^{2}10^{x},\quad \bigl(f/g\bigr)(x)=\frac{10^{x}}{3x^{2}},
(f\circ g)(x)=10^{3x^{2}},\quad (g\circ f)(x)=3\bigl(10^{x}\bigr)^{2}=3\cdot10^{2x}.

Resposta

a) (f+g)(x)=x^{2}+\sqrt{x+1}, (f-g)(x)=x^{2}-\sqrt{x+1}, (f\,g)(x)=x^{2}\sqrt{x+1}, \bigl(f/g\bigr)(x)=\frac{x^{2}}{\sqrt{x+1}}, (f\circ g)(x)=x+1, (g\circ f)(x)=\sqrt{x^{2}+1} b) (f+g)(x)=(x-1)^{2}+7x-4, (f-g)(x)=(x-1)^{2}-(7x-4), (f\,g)(x)=(x-1)^{2}(7x-4), \bigl(f/g\bigr)(x)=\frac{(x-1)^{2}}{7x-4}, (f\circ g)(x)=(7x-5)^{2}, (g\circ f)(x)=7(x-1)^{2}-4 c) (f+g)(x)=\sqrt{x+5}+\frac{1}{x}, (f-g)(x)=\sqrt{x+5}-\frac{1}{x}, (f\,g)(x)=\frac{\sqrt{x+5}}{x}, \bigl(f/g\bigr)(x)=x\sqrt{x+5}, (f\circ g)(x)=\sqrt{\frac{1+5x}{x}}, (g\circ f)(x)=\frac{1}{\sqrt{x+5}} d) (f+g)(x)=\sqrt{x+1}+|x-1|, (f-g)(x)=\sqrt{x+1}-|x-1|, (f\,g)(x)=\sqrt{x+1}\,|x-1|, \bigl(f/g\bigr)(x)=\frac{\sqrt{x+1}}{|x-1|}, (f\circ g)(x)=\sqrt{|x-1|+1}, (g\circ f)(x)=\bigl|\sqrt{x+1}-1\bigr| e) (f+g)(x)=x^{3}+\sqrt[3]{1-x^{3}}, (f-g)(x)=x^{3}-\sqrt[3]{1-x^{3}}, (f\,g)(x)=x^{3}\sqrt[3]{1-x^{3}}, \bigl(f/g\bigr)(x)=\frac{x^{3}}{\sqrt[3]{1-x^{3}}}, (f\circ g)(x)=1-x^{3}, (g\circ f)(x)=\sqrt[3]{1-x^{9}} f) (f+g)(x)=10^{x}+3x^{2}, (f-g)(x)=10^{x}-3x^{2}, (f\,g)(x)=3x^{2}10^{x}, \bigl(f/g\bigr)(x)=\frac{10^{x}}{3x^{2}}, (f\circ g)(x)=10^{3x^{2}}, (g\circ f)(x)=3\cdot10^{2x}

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