Considere as funções

Considere as funções
$$\begin{array}{*{35}{l}}
f:x\to \frac{4-\ln (2-x)}{3}  \\
g:x\to 2+3{{e}^{2x-1}}  \\
h:x\to {{\log }_{2}}(2x-2)-{{\log }_{2}}(x+2)-2  \\
\end{array}$$

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1. Os domínios das funções são:
   \[\begin{array}{*{35}{l}}
   {{D}_{f}} & = & \left\{ x\in \mathbb{R}:2-x>0 \right\}  \\
   {} & = & \left\{ x\in \mathbb{R}:x<2 \right\}  \\
   {} & = & \left] -\infty ,2 \right[  \\
   \end{array}\]
   \[\begin{array}{*{35}{l}}
   {{D}_{g}} & = & \left\{ x\in \mathbb{R}:(2x-1)\in \mathbb{R} \right\}  \\
   {} & = & \mathbb{R}  \\
   \end{array}\]
   \[\begin{array}{*{35}{l}}
   {{D}_{h}} & = & \left\{ x\in \mathbb{R}:2x-2>0\wedge x+2>0 \right\}  \\
   {} & = & \left\{ x\in \mathbb{R}:x>1\wedge x>-2 \right\}  \\
   {} & = & \left] 1,+\infty  \right[  \\
   \end{array}\]
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2. Ora, \[\begin{array}{*{35}{l}}
   y=\frac{4-\ln (2-x)}{3} & \Leftrightarrow  & \ln (2-x)=4-3y  \\
   {} & \Leftrightarrow  & 2-x={{e}^{4-3y}}  \\
   {} & \Leftrightarrow  & x=2-{{e}^{-3y+4}}  \\
   \end{array}\]
   Assim, \[\begin{array}{*{20}{l}}{{{D’}_f}}& = &{\left\{ {x \in \mathbb{R}:( – 3x + 4) \in \mathbb{R}} \right\}{\rm{\;}}}\\{}& = &{\mathbb{R}{\rm{\;}}}\end{array}\]
   Logo, a função inversa de $f$ pode ser assim caraterizada: \[\begin{array}{*{35}{l}}
   {{f}^{-1}}: & \mathbb{R}\to \left] -\infty ,2 \right[  \\
   {} & x\to 2-{{e}^{-3x+4}}  \\
   \end{array}\]

   Ora, \[\begin{array}{*{35}{l}}
   y=2+3{{e}^{2x-1}} & \Leftrightarrow  & {{e}^{2x-1}}=\frac{y-2}{3}  \\
   {} & \Leftrightarrow  & 2x-1=\ln \left( \frac{y-2}{3} \right)  \\
   {} & \Leftrightarrow  & x=\frac{1}{2}+\frac{1}{2}\ln \left( \frac{y-2}{3} \right)  \\
   \end{array}\]
   Assim, \[\begin{array}{*{20}{l}}{{{D’}_g}}& = &{\left\{ {x \in \mathbb{R}:\frac{{x – 2}}{3} > 0} \right\}{\rm{\;}}}\\{}& = &{\left] {2, + \infty {\rm{\;}}} \right[{\rm{\;}}}\end{array}\]
   Logo, a função inversa de $g$ pode ser assim caraterizada: \[\begin{array}{*{35}{l}}
   {{g}^{-1}}: & \left] 2,+\infty  \right[\to \mathbb{R}  \\
   {} & x\to \frac{1}{2}+\frac{1}{2}\ln \left( \frac{x-2}{3} \right)  \\
   \end{array}\]
   ­

3. Como \[\begin{array}{*{35}{l}}
   f(x)=0 & \Leftrightarrow  & \frac{4-\ln (2-x)}{3}=0  \\
   {} & \Leftrightarrow  & \ln (2-x)=4  \\
   {} & \Leftrightarrow  & 2-x={{e}^{4}}  \\
   {} & \Leftrightarrow  & x=2-{{e}^{4}}  \\
   \end{array}\]
   a função $f$ apenas possui um zero: $x=2-{{e}^{4}}$.

   Como \[\begin{array}{*{35}{l}}
   g(x)=0 & \Leftrightarrow  & 2+3{{e}^{2x-1}}=0  \\
   {} & \Leftrightarrow  & x\in \left\{ {} \right\},\text{ pois }{{e}^{2x-1}}>0,\forall x\in \mathbb{R}  \\
   \end{array}\]
   a função $g$ não tem zeros.

   Como \[\begin{array}{*{35}{l}}
   h(x)=0 & \Leftrightarrow  & {{\log }_{2}}(2x-2)-{{\log }_{2}}(x+2)-2=0  \\
   {} & \Leftrightarrow  & {{\log }_{2}}(2x-2)-{{\log }_{2}}4={{\log }_{2}}(x+2)  \\
   {} & \Leftrightarrow  & {{\log }_{2}}\left( \frac{2x-2}{4} \right)={{\log }_{2}}(x+2)  \\
   {} & \Leftrightarrow  & \begin{matrix}
   \frac{2x-2}{4}=x+2 & \wedge  & x\in {{D}_{h}}  \\
   \end{matrix}  \\
   {} & \Leftrightarrow  & \begin{matrix}
   2x-2=4x+8 & \wedge  & x\in \left] 1,+\infty  \right[  \\
   \end{matrix}  \\
   {} & \Leftrightarrow  & \begin{matrix}
   x=-5 & \wedge  & x\in \left] 1,+\infty  \right[  \\
   \end{matrix}  \\
   {} & \Leftrightarrow  & x\in \left\{ {} \right\}  \\
   \end{array}\]
   a função $h$ não tem zeros.
   ­

4. Ora,
   \[\begin{array}{*{20}{l}}
   {h(x) \leqslant  – 2}& \Leftrightarrow &{{{\log }_2}(2x – 2) – {{\log }_2}(x + 2) – 2 \leqslant  – 2} \\
   {}& \Leftrightarrow &{{{\log }_2}(2x – 2) \leqslant {{\log }_2}(x + 2)} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {2x – 2 \leqslant x + 2}& \wedge &{x \in {D_h},{\text{ pois }}x \to {{\log }_2}x{\text{ }}{\text{ é estritamente crescente}}}
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x \leqslant 4}& \wedge &{x \in \left] {1, + \infty } \right[}
   \end{array}} \\
   {}& \Leftrightarrow &{x \in \left] {1,4} \right]}
   \end{array}\]