De um fun\u00e7\u00e3o $f$ de dom\u00ednio $\\left[ { – \\pi ,\\pi } \\right]$, sabe-se que a sua derivada \u00e9:<\/p>\n

$$f'(x) = 2x – 2\\cos \\left( {2x} \\right)$$<\/p><\/blockquote>\n

\u00ad<\/p>\n

\n
1. O valor pedido \u00e9: $$\\mathop {\\lim }\\limits_{x \\to 0} \\frac{{f(x + \\pi ) – f(\\pi )}}{x} = f'(\\pi ) = 2\\pi\u00a0 – 2\\cos \\left( {4\\pi } \\right) = 2\\pi\u00a0 – 2$$
   \n\u00ad<\/li>\n
2. Como $$\\begin{array}{*{20}{l}}
   \n{f”(x)}& = &{\\left( {2x – 2\\cos \\left( {2x} \\right)} \\right)’} \\\\
   \n{}& = &{2 + 2 \\times 2\\operatorname{sen} \\left( {2x} \\right)} \\\\
   \n{}& = &{2 + 4\\operatorname{sen} \\left( {2x} \\right),\\forall x \\in \\left[ { – \\pi ,\\pi } \\right]}
   \n\\end{array}$$
   \nent\u00e3o $$\\begin{array}{*{20}{l}}
   \n{f”(x) = 0}& \\Leftrightarrow &{2 + 4\\operatorname{sen} \\left( {2x} \\right) = 0 \\wedge x \\in \\left[ { – \\pi ,\\pi } \\right]} \\\\
   \n{}& \\Leftrightarrow &{\\operatorname{sen} \\left( {2x} \\right) =\u00a0 – \\frac{1}{2} \\wedge x \\in \\left[ { – \\pi ,\\pi } \\right]} \\\\
   \n{}& \\Leftrightarrow &{\\left( {2x =\u00a0 – \\frac{\\pi }{6} + 2k\\pi\u00a0 \\vee 2x = \\frac{{7\\pi }}{6} + 2k\\pi ,k \\in \\mathbb{Z}} \\right) \\wedge x \\in \\left[ { – \\pi ,\\pi } \\right]} \\\\
   \n{}& \\Leftrightarrow &{\\left( {x =\u00a0 – \\frac{\\pi }{{12}} + k\\pi\u00a0 \\vee x = \\frac{{7\\pi }}{{12}} + k\\pi ,k \\in \\mathbb{Z}} \\right) \\wedge x \\in \\left[ { – \\pi ,\\pi } \\right]} \\\\
   \n{}& \\Leftrightarrow &{x \\in \\left\\{ { – \\frac{{5\\pi }}{{12}}, – \\frac{\\pi }{{12}},\\frac{{7\\pi }}{{12}},\\frac{{11\\pi }}{{12}}} \\right\\}}
   \n\\end{array}$$
   \nAssim, temos:
   \n\u00ad<\/p>\n\n\n\n\n\n

   $x$<\/td>\n	${ – \\pi }$<\/td>\n	<\/td>\n	${ – \\frac{{5\\pi }}{{12}}}$<\/td>\n	<\/td>\n	${ – \\frac{\\pi }{{12}}}$<\/td>\n	<\/td>\n	${\\frac{{7\\pi }}{{12}}}$<\/td>\n	<\/td>\n	${\\frac{{11\\pi }}{{12}}}$<\/td>\n	<\/td>\n	$\\pi $<\/td>\n<\/tr>\n
   Sinal de\u00a0$f”(x) = 2 + 4\\operatorname{sen} \\left( {2x} \\right)$<\/td>\n	$+$<\/td>\n	$+$<\/td>\n	$0$<\/td>\n	$-$<\/td>\n	$0$<\/td>\n	$+$<\/td>\n	$0$<\/td>\n	$-$<\/td>\n	$0$<\/td>\n	$+$<\/td>\n	$+$<\/td>\n<\/tr>\n
   Gr\u00e1fico de $f$<\/td>\n	$ \\cup $<\/td>\n	P.I.<\/td>\n	$ \\cap $<\/td>\n	P.I.<\/td>\n	$ \\cup $<\/td>\n	P.I.<\/td>\n	$ \\cap $<\/td>\n	P.I.<\/td>\n	$ \\cup $<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n \u00ad<\/li>\n A reta tangente referida tem declive ${m_t} = 3$.Logo, a abcissa do ponto de tang\u00eancia \u00e9 ${x_0}$, tal que $f'({x_0}) = {m_t} = 3$.\n Trata-se, portanto, de determinar graficamente a solu\u00e7\u00e3o da equa\u00e7\u00e3o $2x – 2\\cos \\left( {2x} \\right) = 3$ no intervalo $\\left[ { – \\pi ,\\pi } \\right]$.<\/p>\n O valor procurado \u00e9 $x = 1,03$ (com aproxima\u00e7\u00e3o \u00e0s cent\u00e9simas), abcissa do ponto A: \n\u00ad<\/li>\n<\/ol>\n