Question

Sejam as funções reais $f$ e $g$ definidas por:
$f(x)=\begin{cases}x^2+2 \qquad \text{Se} \qquad x \leq -1 \\\\\frac{1}{x-2} \qquad \text{Se} \qquad -1< x < 1\\\\ 4-x^2 \qquad\text{Se} \qquad x \geq 1 \end{cases}$
$\text{e} \qquad g(x) = 2-3x$

Obtenha as leis que definem $f\circ~g \qquad \text{e} \qquad g\circ f$

Obs.: preciso dos cálculos. O gabarito encontra-se em anexo.
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Answer 1

Explicação passo a passo:

Calculando f o g(x):

Para que a composta f o g esteja definida, devemos garantir que Im(g) ⊂ D(f).

• Todo y ∈ ℝ pode ser escrito como y = 2 − 3x para algum x ∈ ℝ. A saber, x = (2 − y)/3. Logo, Im(g) = ℝ.
• f está definida em todo o ℝ, dadas as leis que a definem. Logo D(f) = ℝ.

Portanto, Im(g) ⊂ D(f).

Segue que

    $f\circ g(x)=f[g(x)]=\left\{\begin{array}{ll}[g(x)]^2+2,&\quad\mathrm{se~}g(x)\le -1\\\\ \dfrac{1}{g(x)-2},&\quad\mathrm{se~}-1 < g(x) < 1\\\\ 4-[g(x)]^2,&\quad \mathrm{se~}g(x)\ge 1\end{array}\right.$

    $=\left\{\begin{array}{ll}(2-3x)^2+2,&\quad\mathrm{se~}2-3x\le -1\\\\ \dfrac{1}{(2-3x)-2},&\quad\mathrm{se~}-1 < 2-3x < 1\\\\ 4-(2-3x)^2,&\quad \mathrm{se~}2-3x\ge 1\end{array}\right.$

    $=\left\{\begin{array}{ll}(4-12x+9x^2)+2,&\quad\mathrm{se~}2+1\le 3x\\\\ \dfrac{1}{(2-3x)-2},&\quad\mathrm{se~}-1-2 < -3x < 1-2\\\\ 4-(4-12x+9x^2),&\quad \mathrm{se~}2-1\ge 3x\end{array}\right.$

    $=\left\{\begin{array}{ll}6-12x+9x^2,&\quad\mathrm{se~}3\le 3x\\\\ \dfrac{1}{-3x},&\quad\mathrm{se~}-3 < -3x < -1\\\\ 12x-9x^2,&\quad \mathrm{se~}1\ge 3x\end{array}\right.$

    $=\left\{\begin{array}{ll}6-12x+9x^2,&\quad\mathrm{se~}3x\ge 3\\\\ \dfrac{1}{-3x},&\quad\mathrm{se~}1 < 3x < 3\\\\ 12x-9x^2,&\quad \mathrm{se~}3x\le 1\end{array}\right.$

    $\therefore~~f\circ g(x)=\left\{\begin{array}{ll}6-12x+9x^2,&\quad\mathrm{se~}x \ge 1\\\\ -\,\dfrac{1}{3x},&\quad\mathrm{se~}\frac{1}{3} < x < 1\\\\ 12x-9x^2,&\quad \mathrm{se~}x\le \frac{1}{3}\end{array}\right.\qquad\quad \checkmark$

Calculando g o f(x):

Agora, devemos garantir que Im(f) ⊂ D(g).

Pelas leis que definem f, Im(f) ⊂ ℝ = D(g). Logo a composta está bem definida.

Segue que

    $g\circ f(x)=g[f(x)]=\left\{\begin{array}{ll}g(x^2+2),&\quad \mathrm{se~}x\le -1\\\\ g\!\left(\dfrac{1}{x-2}\right),&\quad \mathrm{se~}-1 < x < 1\\\\ g(4-x^2),&\quad \mathrm{se~}x\ge 1 \end{array}\right.$

    $=\left\{\begin{array}{ll}2-3(x^2+2),&\quad \mathrm{se~}x\le -1\\\\ 2-3\!\left(\dfrac{1}{x-2}\right),&\quad \mathrm{se~}-1 < x < 1\\\\ 2-3(4-x^2),&\quad \mathrm{se~}x\ge 1 \end{array}\right.$

    $=\left\{\begin{array}{ll}2-3x^2-6,&\quad \mathrm{se~}x\le -1\\\\ 2-\dfrac{3}{x-2},&\quad \mathrm{se~}-1 < x < 1\\\\ 2-12+3x^2,&\quad \mathrm{se~}x\ge 1 \end{array}\right.$

    $=\left\{\begin{array}{ll}-3x^2-4,&\quad \mathrm{se~}x\le -1\\\\ \dfrac{2(x-2)-3}{x-2},&\quad \mathrm{se~}-1 < x < 1\\\\ 3x^2-10,&\quad \mathrm{se~}x\ge 1 \end{array}\right.$

    $=\left\{\begin{array}{ll}-3x^2-4,&\quad \mathrm{se~}x\le -1\\\\ \dfrac{2x-4-3}{x-2},&\quad \mathrm{se~}-1 < x < 1\\\\ 3x^2-10,&\quad \mathrm{se~}x\ge 1 \end{array}\right.$

    $\therefore~~g\circ f(x)=\left\{\begin{array}{ll}-3x^2-4,&\quad \mathrm{se~}x\le -1\\\\ \dfrac{2x-7}{x-2},&\quad \mathrm{se~}-1 < x < 1\\\\ 3x^2-10,&\quad \mathrm{se~}x\ge 1 \end{array}\right.\qquad\quad \checkmark$

Dúvidas? Comente.

Bons estudos!