Resoluções de Análise Matemática, Demidovitch Cap. I Ex.1-10

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1.** Demonstrar que se \(a\) e \(b\) são números reais, então \[ || a | - |b|| \le | a - b | \le | a |+| b |. \]

Resolução: Como \(a=(a-b)+b\), então \(|a|\le|a-b|+|b|.\) Donde \(|a-b|\ge|a|-|b|\) e \(|a-b|=|b-a| \ge |b|-|a|. \) Portanto, \(|a-b| \ge |a|-|b|.\) Além disso, \(|a-b|=|a+(-b)| \le |a|+|-b| = |a|+|b|.\)

2. Demonstrar as seguintes igualdades:
a) \(|ab| = |a|\cdot|b|;\)

Resolução: Para começar vamos analisar o primeiro membro, \( |ab|= \left\{ \begin{array}{rll}
a\cdot b, & \hbox{se} & a \ge 0,b \ge 0 \\
-a\cdot b, & \hbox{se} & a \lt 0,b \ge 0 \\
a\cdot (-b), & \hbox{se} & a \ge 0,b \lt 0 \\
-a\cdot (-b), & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
\(=\left\{ \begin{array}{rll}
ab, & \hbox{se} & a \ge 0,b \ge 0 \\
-ab, & \hbox{se} & a \lt 0,b \ge 0 \\
-ab, & \hbox{se} & a \ge 0,b \lt 0 \\
ab, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
Agora, vamos analisar o segundo membro:
\( |a|= \left\{ \begin{array}{rll}
a, & \hbox{se} & a \ge 0\\
-a, & \hbox{se} & a \lt 0
\end{array}\right.\) e
\(|b|=\left\{ \begin{array}{rll}
b, & \hbox{se} & b \ge 0 \\
-b, & \hbox{se} & b \ge 0
\end{array}\right.\) Assim, \( |a|\cdot|b|=\left\{ \begin{array}{rll}
a\cdot b, & \hbox{se} & a \ge 0,b \ge 0 \\
-a\cdot b, & \hbox{se} & a \lt 0,b \ge 0 \\
a\cdot (-b), & \hbox{se} & a \ge 0,b \lt 0 \\
-a\cdot (-b), & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
\(=\left\{ \begin{array}{rll}
ab, & \hbox{se} & a \ge 0,b \ge 0 \\
-ab, & \hbox{se} & a \lt 0,b \ge 0 \\
-ab, & \hbox{se} & a \ge 0,b \lt 0 \\
ab, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\) Entretanto, \(|ab| = |a|\cdot|b|.\)

b) \(|a|^2=a^2;\)
Resolução: Visto que \[|a|=\left\{ \begin{array}{rll}
a, & \hbox{se} & a \ge 0\\
-a, & \hbox{se} & a \lt 0
\end{array}\right.\] Podemos afirmar seguramente que \(|a|^2=\left\{ \begin{array}{rll}
a^2, & \hbox{se} & a \ge 0\\
(-a)^2, & \hbox{se} & a \lt 0
\end{array}\right.\) \(=\left\{ \begin{array}{rll}
a^2, & \hbox{se} & a \ge 0\\
a^2, & \hbox{se} & a \lt 0
\end{array}\right.\) \(=a^2.\)

c) \(|\frac{a}{b}| = \frac{|a|}{|b|} \quad (b \ne 0);\)
Resolução: Para começar vamos analisar o primeiro membro, \( |\frac{a}{b}|= \left\{ \begin{array}{rll}
\frac{a}{b}, & \hbox{se} & a \ge 0,b \ge 0 \\
\frac{-a}{b}, & \hbox{se} & a \lt 0,b \ge 0 \\
\frac{a}{-b}, & \hbox{se} & a \ge 0,b \lt 0 \\
\frac{-a}{-b}, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
\(=\left\{ \begin{array}{rll}
\frac{a}{b}, & \hbox{se} & a \ge 0,b \ge 0 \\
-\frac{a}{b}, & \hbox{se} & a \lt 0,b \ge 0 \\
-\frac{a}{b}, & \hbox{se} & a \ge 0,b \lt 0 \\
\frac{a}{b}, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
Agora, vamos analisar o segundo membro:
\( |a|= \left\{ \begin{array}{rll}
a, & \hbox{se} & a \ge 0\\
-a, & \hbox{se} & a \lt 0
\end{array}\right.\) e
\(|b|=\left\{ \begin{array}{rll}
b, & \hbox{se} & b \gt 0 \\
-b, & \hbox{se} & b \gt 0
\end{array}\right.\) Assim, \( \frac{|a|}{|b|}=\left\{ \begin{array}{rll}
\frac{a}{b}, & \hbox{se} & a \ge 0,b \gt 0 \\
\frac{-a}{b}, & \hbox{se} & a \lt 0,b \gt 0 \\
\frac{a}{-b}, & \hbox{se} & a \ge 0,b \lt 0 \\
\frac{-a}{-b}, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\)
\(=\left\{ \begin{array}{rll}
\frac{a}{b}, & \hbox{se} & a \ge 0,b \gt 0 \\
-\frac{a}{b}, & \hbox{se} & a \lt 0,b \gt 0 \\
-\frac{a}{b}, & \hbox{se} & a \ge 0,b \lt 0 \\
\frac{a}{b}, & \hbox{se} & a \lt 0,b \lt 0
\end{array}\right.\) Entretanto, \(|\frac{a}{b}| = \frac{|a|}{|b|}.\)

d) \(\sqrt{a^2}=|a|;\)
Resolução: Primeiro vamos elevar ao quadrado ambos membros:
\(\sqrt{a^2}=|a|\) \(\Longrightarrow \) \((\sqrt{a^2})^2=|a|^2,\) Agora ao simplificarmos a raiz quadrada teremos: \(a^2=|a|^2\), que ja foi demonstrado na alinea b).


3. Resolva as desigualidades:
a) \(|x-1| \lt 3 \);
Resolução: \(|x-1| \lt 3 \) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x-1 \lt 3, & \hbox{se} & x-1 \ge 0\\
-(x-1) \lt 3, & \hbox{se} & x-1 \lt 0
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x \lt 3+1, & \hbox{se} & x\ge 1\\
-x+1 \lt 3, & \hbox{se} & x\lt 1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x\lt 4, & \hbox{se} & x \ge 1\\
-x\lt 3-1, & \hbox{se} & x \lt 1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x \lt 4, & \hbox{se} & x \ge 1\\
x \gt -2, & \hbox{se} & x \lt 1
\end{array}\right. ;\) Entretanto, \(-2\lt x\lt 4.\)


b) \(|x+1| \gt 2 \);
Resolução: \(|x+1| \gt 2 \) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x+1 \gt 2, & \hbox{se} & x+1 \ge 0\\
-(x+1) \gt 2, & \hbox{se} & x+1 \lt 0
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x \gt 2-1, & \hbox{se} & x\ge -1\\
-x-1 \gt 2, & \hbox{se} & x\lt -1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x\gt 1, & \hbox{se} & x \ge -1\\
-x\gt 2+1, & \hbox{se} & x \lt -1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x \gt 1, & \hbox{se} & x \ge -1\\
x \lt -3, & \hbox{se} & x \lt -1
\end{array}\right. ;\) Entretanto, \(x\lt -3, \ x\gt 1.\)



c) \(|2x+1| \lt 1 \);
Resolução: \(|2x+1| \lt 1 \) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
2x+1 \lt 1, & \hbox{se} & 2x+1 \ge 0\\
-(2x+1) \lt 1, & \hbox{se} & 2x+1 \lt 0
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
2x \lt 1-1, & \hbox{se} & 2x\ge -1\\
-2x-1 \lt 1, & \hbox{se} & 2x\lt -1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
2x\lt 0, & \hbox{se} & x \ge \frac{-1}{2}\\
-2x\lt 1+1, & \hbox{se} & x \lt \frac{-1}{2}
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x \lt 0, & \hbox{se} & x \ge -\frac{1}{2}\\
x \gt -1, & \hbox{se} & x \lt -\frac{-1}{2}
\end{array}\right. ;\) Entretanto, \(-1\lt x\lt 0.\)


d) \(|x-1| \lt |x+1| \);
Resolução: \(|x-1| \lt |x+1| \) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
x-1 \lt x+1, & \hbox{se} & x-1 \ge 0, x+1 \ge 0 \\
-(x-1) \lt x+1, & \hbox{se} & x-1 \lt 0, x+1 \ge 0
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
-1 \lt 1, & \hbox{se} & x\ge 1, x \ge -1\\
-x+1 \lt x+1, & \hbox{se} & x\lt 1, x \ge -1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
-1 \lt 1, & \hbox{se} & x\ge 1, x \ge -1\\
-x-x\lt 1-1, & \hbox{se} & x\lt 1, x \ge -1
\end{array}\right.\) \(\Longrightarrow \) \(\left\{ \begin{array}{rll}
-1 \lt 1, & \hbox{se} & x\ge 1, x \ge -1\\
x \gt 0, & \hbox{se} & x\lt 1, x \ge -1
\end{array}\right. ;\) Entretanto, \(x\gt 0.\)

4. Achar \(f(-1),\) \(f(0),\) \(f(1),\) \(f(2),\) \(f(3),\) \(f(4),\) se \(f(x)=x^3-6x^2+11x-6.\)

Resolução:
\(f(-1)=(-1)^3-6\cdot (-1)^2+11\cdot (-1)-6\) \(=-1-6\cdot 1-11-6\) \(=-1-6-11-6\) \(=-24.\)

\(f(0)=0^3-6\cdot 0^2+11\cdot 0-6\) \(=0-6\cdot 0+0-6\) \(=-6.\)

\(f(1)=1^3-6\cdot 1^2+11\cdot 1-6\) \(=1-6\cdot 1+11-6\) \(=0.\)

\(f(2)=2^3-6\cdot 2^2+11\cdot 2-6\) \(=8-6\cdot 4+22-6\) \(=8-24+22-6\) \(=0.\)

\(f(3)=3^3-6\cdot 3^2+11\cdot 3-6\) \(=27-6\cdot 9+33-6\) \(=27-54+33-6\)\(=0.\)

\(f(4)=4^3-6\cdot 4^2+11\cdot 4-6\) \(=64-6\cdot 16+44-6\) \(=64-96+44-6\)\(=6.\)

5. Achar \(f(0),\) \(f(-\frac{3}{4}),\) \(f(-x),\) \(f(\frac{1}{x}),\) \(\frac{1}{f(x)},\) se \(f(x)=\sqrt{1+x^2}.\)
Resolução:

\(f(0)=\sqrt{1+0^2}\) \(=\sqrt{1+0}\) \(=\sqrt{1}\) \(=1.\)
\(f(-\frac{3}{4})=\sqrt{1+(-\frac{3}{4})^2}\) \(=\sqrt{1+\frac{9}{16}}\) \(=\sqrt{\frac{16+9}{16}}\) \(=\sqrt{\frac{25}{16}}\) \(=\frac{5}{4}\)
\(f(-x)=\sqrt{1+(-x)^2}\) \(=\sqrt{1+x^2.}\)
\(f(\frac{1}{x})=\sqrt{1+(\frac{1}{x})^2}\) \(=\sqrt{1+\frac{1}{x^2}}\) \(=\sqrt{\frac{x^2+1}{x^2}}\) \(=\frac{\sqrt{x^2+1}}{x^2}\) \(=\frac{\sqrt{x^2+1}}{|x|}\) \(=|x|^{-1}\cdot \sqrt{x^2+1}.\)
\(\frac{1}{f(x)}\) \( =\frac{1}{\sqrt{1+x^2}.} \)

6. Seja \(f(x)=arccos(lg x).\) Achar \(f\left(\frac{1}{x}\right),\) \(f(1),\) \(f(10).\)
Resolução:

\(f\left(\frac{1}{10}\right)=arccos\left[lg\left(\frac{1}{10}\right)\right]\) \(=arccos(log10^{-1})\)
\(=arccos(-1)\) \(=\pi+2k\pi, k\in Z\)
\(f(1)=arccos(lg1)\) \(=arccos(log10^0)\) \(=arccos(0)\) \(=\frac{\pi}{2}+k\pi, k\in Z\)
\(f(1)=arccos(lg10)\) \(=arccos(log10^1)\) \(=arccos(1)\) \(=2\pi+2k\pi, k\in Z\)

7. A função \(f(x)\) é linear. Achar esta função, se \(f(-1)=2\) e \(f(2)=-3.\)
Resolução:

\(f(x)=ax+b\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a\cdot(-1)+b=2 \\
a\cdot(2)+b=-3
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
-a+b=2 \\
2a+b=-3
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a=b-2 \\
2\cdot (b-2)+b=-3
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a=b-2 \\
2b-4+b=-3
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a=b-2 \\
3b=1
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a=\frac{1}{3}-2 \\
b=\frac{1}{3}
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a=\frac{-5}{3} \\
b=\frac{1}{3}
\end{array}\right.\); Entretanto, \(f(x)=\frac{-5}{3}x+\frac{1}{3}.\)

8. Achar uma função racional inteira \(f(x)\) de segundo grau, se \(f(0)=1,\) \(f(1)=0,\) e \(f(3)=5.\)
Resolução:

\(f(x)=ax^2+bx^2+c\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a\cdot 0^2+b\cdot 0+c=1 \\
a\cdot 1^2+b\cdot 1+c=0 \\
a\cdot 3^2+b\cdot 3+c=5
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a\cdot 0+0+c=1 \\
a\cdot 1+b+c=0 \\
a\cdot 9+3b+c=5
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
0+0+c=1 \\
a+b+c=0 \\
9a+3b+c=5
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a+b+1=0 \\
9a+3b+1=5
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=-1-b \\
9\cdot(-1-b)+3b=5-1
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=-1-b \\
-9-9b+3b=4
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=-1-b \\
-6b=4+9
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=-1-(-\frac{13}{6}) \\
b=-\frac{13}{6}
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=-1+\frac{13}{6} \\
b=-\frac{13}{6}
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
c=1 \\
a=\frac{7}{6} \\
b=-\frac{13}{6}
\end{array}\right.\); Entretanto, \(f(x)=\frac{-7}{6}x^2-\frac{13}{6}x+1.\)

9. Sabe-se que \(f(4)=-2,\) \(f(5)=6.\) Achar o valor aproximado de \(f(4,3),\) considerando que a função \(f(x)\)
no segmento \(4\le x \le 5\) é linear (interpolação linear da função).
Resolução:

\(f(x)=ax+b\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
a\cdot(4)+b=-2 \\
a\cdot(5)+b=6
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
4a+b=-2 \\
5a+b=6
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
b=-2-4a \\
5a+(-2-4a)=6
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
b=-2-4a \\
5a-2-4a=6
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
b=-2-4a \\
a=6+2
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
b=-2-4\cdot 8 \\
a=8
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
b=-34 \\
a=8
\end{array}\right.\); Dai que, \(f(x)=8x-34\); Entretanto: \(f(4,3)=8\cdot 4,3-34\) \(=34,4-34\) \(=0.4.\)

10. Escrever a função \[f(x)=\left\{ \begin{array}{rll}
0, & \hbox{se} & x \ge 0,\\
x, & \hbox{se} & x \lt 0,
\end{array}\right.\] atreves de uma formula, utilizando o sinal de grandeza absoluta.
Resolução:

\(f(x)=\left\{ \begin{array}{rll}
0, & \hbox{se} & x \le 0 \\
x, & \hbox{se} & x \gt 0
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
x-x, & \hbox{se} & x \le 0 \\
\frac{x+x}{2}, & \hbox{se} & x \gt 0
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
x+(-x), & \hbox{se} & x \le 0 \\
\frac{1}{2}(x+x), & \hbox{se} & x \gt 0
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
x+|x|, & \hbox{se} & x \le 0 \\
\frac{1}{2}(x+|x|), & \hbox{se} & x \gt 0
\end{array}\right.\) \(\Longrightarrow\) \(\left\{ \begin{array}{rll}
\frac{1}{2}(x+|x|), & \hbox{se} & x \le 0 \\
\frac{1}{2}(x+|x|), & \hbox{se} & x \gt 0
\end{array}\right.\) \(\Longrightarrow\) \(f(x)=\frac{1}{2}(x+|x|).\)

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