\u00a0Ora, ${D_h} = \\left\\{ {x \\in \\mathbb{R}:\\left( {x + 1 \\ne 0 \\wedge x <\u00a0 – 2} \\right) \\vee \\left( {x + 3 \\ne 0 \\wedge x \\geqslant\u00a0 – 2} \\right)} \\right\\} = \\mathbb{R}$.\nA fun\u00e7\u00e3o \u00e9 cont\u00ednua nos intervalos $\\left] { – \\infty , – 2} \\right[$ e $\\left] { – 2, + \\infty } \\right[$, pois \u00e9 o quociente de fun\u00e7\u00f5es polinomiais (cont\u00ednuas em $\\mathbb{R}$), n\u00e3o se anulando a fun\u00e7\u00e3o divisor nos intervalos considerados.<\/p>\n
Investiguemos a continuidade da fun\u00e7\u00e3o em $x =\u00a0 – 2$:
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ – }} h(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ – }} \\frac{{x – 2}}{{x + 1}} = 4$$
\n$$\\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ + }} h(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ – }} \\frac{{ – 2x}}{{x + 3}} = 4$$
\nA fun\u00e7\u00e3o \u00e9 cont\u00ednua em $x =\u00a0 – 2$, pois $$\\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ – }} h(x) = \\mathop {\\lim }\\limits_{x \\to\u00a0 – {2^ + }} h(x) = f( – 2) = \\frac{{ – 2 \\times ( – 2)}}{{ – 2 + 3}} = 4$$
\nPortanto, a fun\u00e7\u00e3o $h$ \u00e9 cont\u00ednua em $\\mathbb{R}$.
\n\u00ad<\/p>\n<\/li>\n
A fun\u00e7\u00e3o $h$ \u00e9 cont\u00ednua em $\\left[ { – \\frac{5}{2},\\frac{1}{2}} \\right]$, pois \u00e9 cont\u00ednua em $\\mathbb{R}$.\nComo $$f( – \\frac{5}{2}) = \\frac{{ – 2,5 – 2}}{{ – 2,5 + 1}} = 3$$ e $$f(\\frac{1}{2}) = \\frac{{ – 2 \\times 0,5}}{{0,5 + 3}} =\u00a0 – \\frac{2}{7}$$ ent\u00e3o $$f(\\frac{1}{2}) < 0 < f( – \\frac{5}{2})$$<\/p>\n
Logo, de acordo com o teorema de Bolzano-Cauchy, $\\exists x \\in \\left] { – \\frac{5}{2},\\frac{1}{2}} \\right[:f(x) = 0$.<\/p>\n
Consequentemente, a fun\u00e7\u00e3o tem um zero (1)<\/span><\/strong> (pelo menos) no intervalo $\\left] { – \\frac{5}{2},\\frac{1}{2}} \\right[$.<\/p>\n<\/li>\n<\/ol>\n\u00ad
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