\n$C$ \u00e9 a curva representativa da fun\u00e7\u00e3o $$f:x \\to \\frac{1}{{1 + x}}$$<\/p>\n<\/blockquote>\n
\u00ad<\/p>\n
\n- A derivada de uma fun\u00e7\u00e3o num ponto, caso exista, \u00e9 igual ao declive da reta tangente ao gr\u00e1fico da fun\u00e7\u00e3o nesse ponto. Assim, procuramos ${x_0}$ tal que $f'({x_0}) =\u00a0 – 1$: $$\\begin{array}{*{20}{l}}
\n{f'({x_0}) =\u00a0 – 1}& \\Leftrightarrow &{ – \\frac{1}{{{{\\left( {1 + {x_0}} \\right)}^2}}} =\u00a0 – 1} \\\\
\n{}& \\Leftrightarrow &{{{\\left( {1 + {x_0}} \\right)}^2} = 1} \\\\
\n{}& \\Leftrightarrow &{{x_0} =\u00a0 – 2 \\vee {x_0} = 0}
\n\\end{array}$$
\nPortanto, os pontos pedidos s\u00e3o: $A\\left( { – 2, – 1} \\right)$ e $B(0,1)$.
\n\u00ad<\/li>\n - Como $$\\begin{array}{*{20}{l}}
\n{f'({x_0}) = 1}& \\Leftrightarrow &{ – \\frac{1}{{{{\\left( {1 + {x_0}} \\right)}^2}}} = 1} \\\\
\n{}& \\Leftrightarrow &{{{\\left( {1 + {x_0}} \\right)}^2} =\u00a0 – 1} \\\\
\n{}& \\Leftrightarrow &{{x_0} \\in \\left\\{ {} \\right\\}}
\n\\end{array}$$
\nn\u00e3o existem tangentes \u00e0 curva $C$ paralelas \u00e0 reta de equa\u00e7\u00e3o $y = x$, pois $f'({x_0}) \\ne 1,\\forall {x_0} \\in \\mathbb{R}\\backslash \\left\\{ { – 1} \\right\\}$.
\n\u00ad<\/li>\n - <\/li>\n<\/ol>\n