Lista de exercícios do ensino médio para impressão
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) $\,(f \circ g)(x)\,=\,f \left[g(x)\right]\,=\,g(x)\,+\,1\,=\,(x^2\,+x\,+\,1)\,+\,1\,=\,x^2\,+\,x\,+\,2$
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) $\,(h \circ f)(x)\,=\,h \left[f(x)\right]\,=\,1 \,-f(x)\,=\,1\,-\,(x\,+\,1)\,=\,1\,-\,x\,-1\,=\,-x$
Logo $ \left\{\begin{array}{rcr} &h \circ f : \, \mathbb{R} \rightarrow \mathbb{R}\; \phantom{XXXXX} \\ &(h \circ f)(x)\,=\,-x \\ \end{array} \right.$
d) $\,(f \circ h)(x)\,=\,f \left[h(x)\right]\,=\,h(x)\,+\,1\,=\,(1\,-\,x)\,+\,1\,=\,2\,-\,x$
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:

a) $\,(f \circ g)(x)\,=\,f \left[g(x)\right]\,=\,\left\{\begin{array}{rcr} g(x)\,+\,3 & \mbox{, se}\;\; g(x) \leqslant 3 \\ g(x)\,-\,4 & \mbox{, se}\;\; g(x) > 3 \\ \end{array} \right. \; \Longrightarrow$
$\,\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 $
$\,\Longrightarrow (f \circ g)(x)\,=\, \left\{\begin{array}{rcr} 2x\,-\,4 \phantom{X} & \mbox{, se}\;\; x \leqslant 5 \\ 2x\,-\,11 \phantom{X} & \mbox{, se}\;\; x > 5 \\ \end{array} \right. \;$
b) $\,(g \circ f)(x)\,=\,g \left[f(x)\right]\,=\, \,2 \centerdot f(x) \,-\,7\,= \left\{\begin{array}{rcr} 2 \centerdot (x\,+\,3)\, - \,7 & \mbox{, se}\;\; x \leqslant 3 \\ 2 \centerdot (x\,-\,4) \,-\, 7 & \mbox{, se}\;\; x > 3 \\ \end{array} \right. \; \Longrightarrow$
$\,\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:
$\,\left\{\begin{array}{rcr} \, & g\,:\, \mathbb{R}_+ \rightarrow \mathbb{R}_+\; \\ \, & g(x)\,=\,+\,\sqrt{x} \\ \end{array} \right.\,$


×
(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:
a)
$\,1 \mbox{ e } \frac{1}{2}\phantom{XX}$
b)
$\,-1 \mbox{ e } \frac{1}{2}\,$
c)
1 e 2 
d)
1 e -2
e)
-1 e qualquer
 
 

 



resposta: (B)
×
(ITA - 1992) Considere as funções $\phantom{X} f\;:\;\mathbb{R^*}\,\rightarrow \,\mathbb{R}\;$,$\;\;g\;:\mathbb{R}\,\rightarrow\; \mathbb{R}\;\;$ e $\;\;h\,:\,\mathbb{R^*}\,\rightarrow\,\mathbb{R}\phantom{X}$ definidas por:
$\phantom{X}f(x)\,=\,{\large 3}^{\,{\huge x\,+\,\frac{1}{x}}}\,$,
$\phantom{X}g(x)\,=\,x^2\,$;
$\phantom{X}h(x)\,=\,{\large \frac{81}{x}}\,$

O conjunto dos valores de $\phantom{X} x \phantom{X}$ em $\phantom{X}\mathbb{R^*} \phantom{X}$ tais que $\phantom{X} (f\circ g)(x)\,=\,(h\circ f)(x) \phantom{X}$, é subconjunto de:
a)
$\,[0\,,\,3]\,$
d)
$\,[-2\,,\,2]\,$
b)
$\,[3\,,\,7]\,$
e)
nenhuma das anteriores
c)
$\,[-6\,,\,1]\,$

 



resposta: alternativa C
×
Veja exercÍcio sobre:
composição de funções
equação de funções
função composta