Exercícios sobre equação de funções

Sejam $\,f\,:\, \mathbb{R} \rightarrow \mathbb{R}\;$, $\,g\,:\, \mathbb{R} \rightarrow \mathbb{R}\;$ e $\,h\,:\, \mathbb{R} \rightarrow \mathbb{R}\;$ três funções definidas por $\,f(x)\,=\,x\,+\,1\;$,$\,g(x)\,=\,x^2\,\,+x\,+\,1\;$ e $\,h(x)\,=\,1\,-\,x\;$. Determine $\,g \circ f\;$, $\;f \circ g\;$, $\;h \circ f\;$, $\;f \circ h\;$,$\;h \circ g\;\,$ e $\;\,g \circ h\,$.

resposta: Resolução:

a)	$\,(g \circ f)(x)\,=\,g \left[f(x)\right]\,=\,\left(f(x)\right)^2\,+\,f(x)\,+1\,=$
	$\,=\,(x + 1)^2\,+\,(x+1)\,+\,1\,=\,(x^2\,+\,2x\,+\,1)\,+\,(x+1)\,+\,1\,=$
	$\,=\,x^2\,+\,3x\,+\,3\,$
	Logo $ \left\{\begin{array}{rcr} &g \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(g \circ f)(x)\,=\,x^2\,+\,3x\,+\,3 \\ \end{array} \right.$
b)	$\,(f \circ g)(x)\,=\,f \left[g(x)\right]\,=\,g(x)\,+\,1\,=\,(x^2\,+x\,+\,1)\,+\,1\,=\,x^2\,+\,x\,+\,2$
b)	Logo $ \left\{\begin{array}{rcr} &f \circ g : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(f \circ g)(x)\,=\,x^2\,+\,2x\,+\,2 \\ \end{array} \right.$
c)	$\,(h \circ f)(x)\,=\,h \left[f(x)\right]\,=\,1 \,-f(x)\,=\,1\,-\,(x\,+\,1)\,=\,1\,-\,x\,-1\,=\,-x$
c)	Logo $ \left\{\begin{array}{rcr} &h \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(h \circ f)(x)\,=\,-x \\ \end{array} \right.$
d)	$\,(f \circ h)(x)\,=\,f \left[h(x)\right]\,=\,h(x)\,+\,1\,=\,(1\,-\,x)\,+\,1\,=\,2\,-\,x$
d)	Logo $ \left\{\begin{array}{rcr} &f \circ h : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(f \circ h)(x)\,=\,2\,-x \\ \end{array} \right.$
e)	$\,(h \circ g)(x)\,=\,h \left[g(x)\right]\,=\,1 \,-g(x)\,=\,1\,-\,(x^2\,+\,x\,+1)\,=$
	$\,=\,1\,-\,x^2\,-x\,-1\,=\,-x^2 - x$
	Logo $ \left\{\begin{array}{rcr} &h \circ g : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(h \circ g)(x)\,=\,-x^2\,-\,x \\ \end{array} \right.$
f)	$\,(g \circ h)(x)\,=\,g \left[h(x)\right]\,=\,\left(h(x)\right)^2\,+\,h(x)\,+\,1\,=$
	$\,=\,(1\,-\,x)^2 \,+\,(1\,-\,x)\,+\,1\,=\,(1\,-\,2x\,+\,x^2)\,+\,(1\,-\,x)\,+\,1\,=$
	$\,=\,x^2\,-\,3x\,+\,3$
	Logo $ \left\{\begin{array}{rcr} &g \circ h : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(g \circ h)(x)\,=\,x^2\,-\,3x\,+\,3 \\ \end{array} \right.$
	É muito importante notar que $\; \left\{\begin{array}{rcr} g \circ f & \neq & f \circ g \\ h \circ f & \neq & f \circ h \\ h \circ g & \neq &g \circ h \\ \end{array} \right.$

Seja $\,f\,:\, \mathbb{R} \rightarrow \mathbb{R}\;$ a função definida por $\,f(x)\,=\,x\,+\,1\;$. Determine $\;\,f \circ f\;\,$, $\;\;f \circ f\, \circ f\;\,$ e $\;\;f \circ f\, \circ f\, \circ f\;$.

resposta: Resolução:

a)	$\,(f \circ f)(x)\,=\,f \left[f(x)\right]\,=\, f(x)\,+\,1\,=\,(x\,+\,1)\,+\,1=\,x\,+\,2$
	Portanto: $ \left\{\begin{array}{rcr} & f \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XX} \\ &(f \circ f)(x)\,=\,x\,+\,2 \\ \end{array} \right.$
b)	$\,(f \circ f \circ f)(x)\,=\,(f \circ f) \left[f(x)\right]\,=\,f(x)\,+\,2\,=\,(x\,+\,1)\,+\,2\,=\,x\,+\,3$
	Portanto: $ \left\{\begin{array}{rcr} &f \circ f \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XX} \\ &(f \circ f \circ f)(x)\,=\,x\,+\,3 \\ \end{array} \right.$
c)	$\,(f \circ f \circ f \circ f)(x)\,=\,(f \circ f \circ f) \left[f(x)\right]\,=\,f(x)\,+\,3\,=\,(x\,+\,1)\,+\,3\,=$
	$\,=\,x\,+\,4\,$
	Portanto $ \left\{\begin{array}{rcr} &f \circ f \circ f \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XX} \\ &(f \circ f \circ f \circ f)(x)\,=\,x\,+\,4 \\ \end{array} \right.$

Sejam $\,f\;$ e $\,g\;$ duas funções de $\,\mathbb{R} \rightarrow \mathbb{R}\,$ definidas por:

$\;f(x)\,=\; \left\{\begin{array}{rcr} x\,+\,3\; \mbox{, se}& x \leqslant 3 \\ x\,-\,4\; \mbox{, se} & x \geqslant 3 \\ \end{array} \right.$

$\,g(x)\,=\,2x\,-\,7\,$,$\;\;\vee \negthickspace \negthickspace \negthickspace \negthinspace - x \,\in\, \, \mathbb{R}$

Determine $\;f \circ g\;$ e $\,g \circ f\;$.

resposta: Resolução:

$\,(f \circ g)(x)\,=\,f \left[g(x)\right]\,=\,\left\{\begin{array}{rcr} g(x)\,+\,3 & \mbox{, se}\;\; g(x) \leqslant 3 \\ g(x)\,-\,4 & \mbox{, se}\;\; g(x) > 3 \\ \end{array} \right. \; \Longrightarrow$

$\,\Longrightarrow (f \circ g)(x)\,=\, \left\{\begin{array}{rcr} (2x\,-\,7)\,+\,3 & \mbox{, se}\;\; g(x) \leqslant 3 \phantom{XX} \\ (2x\,-\,7)\,-\,4 & \mbox{, se}\;\; (2x\,-\,7) > 3 \\ \end{array} \right. \; \Longrightarrow $

$\,\Longrightarrow (f \circ g)(x)\,=\, \left\{\begin{array}{rcr} 2x\,-\,4 \phantom{X} & \mbox{, se}\;\; x \leqslant 5 \\ 2x\,-\,11 \phantom{X} & \mbox{, se}\;\; x > 5 \\ \end{array} \right. \;$

$\,(g \circ f)(x)\,=\,g \left[f(x)\right]\,=\, \,2 \centerdot f(x) \,-\,7\,= \left\{\begin{array}{rcr} 2 \centerdot (x\,+\,3)\, - \,7 & \mbox{, se}\;\; x \leqslant 3 \\ 2 \centerdot (x\,-\,4) \,-\, 7 & \mbox{, se}\;\; x > 3 \\ \end{array} \right. \; \Longrightarrow$

$\,\Longrightarrow (g \circ f)(x)\,=\; \left\{\begin{array}{rcr} 2x\,-\,1 \; & \mbox{, se}\;\; x \leqslant 3 \\ 2x\,-\,15 & \mbox{, se}\;\; x > 3 \\ \end{array} \right. \; $

Portanto:

$\,f \circ g \; : \,\mathbb{R} \rightarrow \mathbb{R}\,$
$(f \circ g)(x)\,=\, \left\{\begin{array}{rcr} 2x\,-\,4 \; & \mbox{, se}\;\; x \leqslant 5 \\ 2x\,-\,11 & \mbox{, se}\;\; x > 5 \\ \end{array} \right. \;$

$\,g \circ f \; : \,\mathbb{R} \rightarrow \mathbb{R}\,$
$(g \circ f)(x)\,=\; \left\{\begin{array}{rcr} 2x\,-\,1 \; & \mbox{, se}\;\; x \leqslant 3 \\ 2x\,-\,15 & \mbox{, se}\;\; x > 3 \\ \end{array} \right. \; $

Seja $\,f\,:\, \mathbb{R}_+ \rightarrow \mathbb{R}_+\;$ a função definida por $\,f(x)\,=\,x^2\;$.
Determine uma função $\,g\,:\, \mathbb{R}_+ \rightarrow \mathbb{R}_+\;$ tal que a função composta $\;(f \circ g)\;$ seja uma função identidade.

resposta: Resolução:
De $\;(f \circ g)\,=\,id\;$ decorre que:
$\,(f \circ g)(x)\,=\,id(x) \, \mbox{, } \; \vee \negthickspace \negthickspace \negthickspace \negthinspace - x \,\in\, \, \mathbb{R}_+ \, \Rightarrow $
$\Rightarrow\,f \left[g(x)\right]\,=\,x\, \mbox{, }\vee \negthickspace \negthickspace \negthickspace \negthinspace - x \,\in\, \, \mathbb{R}\,\Rightarrow $
$\Rightarrow \left[g(x)\right]^2 \,=\,x\,\mbox{, }\;\vee \negthickspace \negthickspace \negthickspace \negthinspace - x \,\in\, \, \mathbb{R}_+ \,$(pois $\,f(x)\,=\,x^2\,\mbox{, } \vee \negthickspace \negthickspace \negthickspace \negthinspace - x \,\in\, \, \mathbb{R}_+$)$\;\Rightarrow$
$\,\Rightarrow\,g(x)\,=\,+ \sqrt{x}\,$, (pois $\,g(x) \geqslant 0$).
Portanto:
$\,\left\{\begin{array}{rcr} \, & g\,:\, \mathbb{R}_+ \rightarrow \mathbb{R}_+\; \\ \, & g(x)\,=\,+\,\sqrt{x} \\ \end{array} \right.\,$

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(FUVEST) Se $\;f\,:\, {\rm I\!R}\; \rightarrow \; {\rm I\!R} \;$ é da forma $\,f(x)\,=\,ax\,+\,b\;$ e verifica $\,(f \circ f)(x)\,=\,x\,+\,1\;$, para todo $\,x\,$ real, então $\,a\,$ e $\,b\,$ valem, respectivamente:

$\,1 \mbox{ e } \frac{1}{2}\phantom{XX}$

$\,-1 \mbox{ e } \frac{1}{2}\,$

1 e 2

1 e -2

-1 e qualquer

resposta: (B)
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