گام به گام تمرین صفحه 69 درس 2 حسابان (1) (تابع)

1)

\(\begin{array}{l}{D_f} = \mathbb{R}\\{D_g} = \mathbb{R}\\\frac{f}{g}\left( x \right) = \frac{{4x}}{{x - 2}} \Rightarrow {D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {x\left| {g\left( x \right) = 0} \right.} \right\} = \mathbb{R} - \left\{ 2 \right\}\\\left( {f - g} \right)\left( x \right) = 5x + 2 \Rightarrow {D_{f - g}} = {D_f} \cap {D_g} = \mathbb{R}\\fog\left( x \right) = f\left( {g\left( x \right)} \right) = 4\left( {2 - x} \right) =  - 4x + 8\\ \Rightarrow {D_{fog}} = \left\{ {x \in {D_g}\left| {g\left( x \right) \in {D_f}} \right.} \right\} = \left\{ {x \in \mathbb{R}\left| {\left( {2 - x} \right) \in \mathbb{R}} \right.} \right\} = \mathbb{R}\end{array}\)

2)

\(\begin{array}{l}{D_f} = \mathbb{R} - \left\{ 3 \right\}\quad :x \ne 3\\{D_g} = \mathbb{R} - \left\{ 0 \right\}\;\quad :x \ne 0\\fog\left( x \right) = f\left( {g\left( x \right)} \right) = f\left( {\frac{4}{x}} \right) = \frac{1}{{\frac{4}{x} - 3}} = \frac{x}{{4x - 3}}\\{D_{fog}} = \left\{ {x \in {D_g}\left| {g\left( x \right) \in {D_f}} \right.} \right\} = \\\left\{ {x \ne 0\left| {\frac{4}{x} \ne 3} \right.} \right\} = \left\{ {x \ne 0\left| {x \ne \frac{3}{4}} \right.} \right\} = \mathbb{R} - \left\{ {0\;,\;\frac{3}{4}} \right\}\end{array}\)

3) گزاره های درست موارد ب – پ و ج می باشند.

4)

\(\begin{array}{l}{D_f} = A\\{D_g} = \mathbb{N}\\{D_{f + g}} = {D_f} \cap {D_g} = A\\ \Rightarrow f + g = \left\{ {\left( {1\;,\;2 + 2} \right)\;,\;\left( {2\;,\;3 + 4} \right)\;,\;\left( {3\;,\;5 + 6} \right)\;,\;\left( {4\;,\;7 + 8} \right)} \right\}\\ = \left\{ {\left( {1\;,\;4} \right)\;,\;\left( {2\;,\;7} \right)\;,\;\left( {3\;,\;11} \right)\;,\;\left( {4\;,\;15} \right)} \right\}\\{D_{gof}} = \left\{ {x \in {D_f}\left| {f\left( x \right) \in {D_g}} \right.} \right\} = \left\{ {x \in A\left| {f\left( x \right) \in \mathbb{N}} \right.} \right\} = A\\ \Rightarrow gof = \left\{ {\left( {1\;,\;4} \right)\;,\;\left( {2\;,\;6} \right)\;,\;\left( {3\;,\;10} \right)\;,\;\left( {4\;,\;14} \right)} \right\}\end{array}\)

5)

\(\begin{array}{l}f + g = \left\{ {\left( { - 4\;,\;6} \right)\;,\;\left( {0\;,\;2} \right)\;,\;\left( {3\;,\; - 5} \right)} \right\}\\f - g = \left\{ {\left( { - 4\;,\;20} \right)\;,\;\left( {0\;,\;8} \right)\;,\;\left( {3\;,\; - 5} \right)} \right\}\\\frac{f}{g} = \left\{ {\left( { - 4\;,\; - \frac{{13}}{7}} \right)\;,\;\left( {0\;,\; - \frac{5}{3}} \right)} \right\}\end{array}\)

6)

\(\begin{array}{l}{D_f} = \mathbb{R}\\{D_g} = \left[ { - 2\;,\;2} \right]\\\\{D_{fog}} = \left\{ {x \in {D_g}\left| {g\left( x \right) \in {D_f}} \right.} \right\} = \left\{ {x \in \left[ { - 2\;,\;2} \right]\left| {\sqrt {4 - {x^2}}  \in \mathbb{R}} \right.} \right\} = \left[ { - 2\;,\;2} \right]\\fog\left( x \right) = f\left( {g\left( x \right)} \right) = f\left( {\sqrt {4 - {x^2}} } \right) = \sqrt {{{\left( {\sqrt {4 - {x^2}} } \right)}^2} + 5}  = \sqrt {9 - {x^2}} \\\\{D_{gof}} = \left\{ {x \in {D_f}\left| {f\left( x \right) \in {D_g}} \right.} \right\} = \left\{ {x \in \mathbb{R}\left| {\sqrt {{x^2} + 5}  \in \left[ { - 2\;,\;2} \right]} \right.} \right\}\\ = \left\{ {x \in \mathbb{R}\left| {x \notin \mathbb{R}} \right.} \right\} = \emptyset \\\sqrt {{x^2} + 5}  \in \left[ { - 2\;,\;2} \right]\;:\; - 2 \le \sqrt {{x^2} + 5}  \le 2 \Rightarrow 0 \le \sqrt {{x^2} + 5}  \le 2\\ \Rightarrow 0 \le {x^2} + 5 \le 4 \Rightarrow  - 5 \le {x^2} \le  - 1\;\; \otimes \\gof\left( x \right) = g\left( {f\left( x \right)} \right) = g\left( {\sqrt {{x^2} + 5} } \right)\\ = \sqrt {4 - {{\left( {\sqrt {{x^2} + 5} } \right)}^2}}  = \sqrt { - 1 - {x^2}} \;\; \otimes \end{array}\)

7) دامنه تقسیم قبل از ساده شدن محاسبه می شود؛ در نتیجه:

\(\begin{array}{l}{D_f} = \mathbb{R}\\{D_g} = \mathbb{R}\\{D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {x\left| {g\left( x \right) = 0} \right.} \right\} = \mathbb{R} - \left\{ { - 3} \right\}\end{array}\)

8)

\(\begin{array}{l}f\left( x \right) = 2x + 5\\y = 2x + 5 \Rightarrow 2x = y - 5 \Rightarrow x = \frac{{y - 5}}{2}\\ \Rightarrow {f^{ - 1}}\left( y \right) = \frac{{y - 5}}{2}\\ \Rightarrow {f^{ - 1}}\left( x \right) = \frac{{x - 5}}{2}\\fo{f^{ - 1}}\left( x \right) = f\left( {{f^{ - 1}}\left( x \right)} \right) = \\f\left( {\frac{{x - 5}}{2}} \right) = 2\left( {\frac{{x - 5}}{2}} \right) + 5 = x\\{f^{ - 1}}of\left( x \right) = {f^{ - 1}}\left( {f\left( x \right)} \right) = \\{f^{ - 1}}\left( {2x + 5} \right) = \frac{{\left( {2x + 5} \right) - 5}}{2} = x\end{array}\)

9)

\(\begin{array}{l}f\left( x \right) =  - \frac{2}{5}x + 2\\g\left( x \right) = \frac{3}{2}x - 3\\\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = \frac{{11}}{{10}}x - 1\\\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right) =  - \frac{{19}}{{10}}x + 5\\\left( {f.g} \right)\left( x \right) = f\left( x \right).g\left( x \right) =  - \frac{3}{5}{x^2} + \frac{{21}}{5}x - 6\end{array}\)

10)

الف)

\(\left( {f + g} \right)\left( 2 \right) = f\left( 2 \right) + g\left( 2 \right) = 2 + 1 = 3\)

ب)

\(\left( {f + g} \right)\left( { - 3} \right) = f\left( { - 3} \right) + g\left( { - 3} \right) =  \otimes\)

پ)

\(\left( {fg} \right)\left( {\frac{1}{2}} \right) = f\left( {\frac{1}{2}} \right)g\left( {\frac{1}{2}} \right) = \sqrt {\frac{5}{2}}  \times \frac{1}{2} = \frac{{\sqrt {10} }}{4}\)

ت)

\(\left( {fog} \right)\left( { - 4} \right) = f\left( {g\left( { - 4} \right)} \right) = f\left( 5 \right) = \sqrt 7\)

ث)

\(\left( {\frac{f}{g}} \right)\left( 0 \right) = \frac{{f\left( 0 \right)}}{{g\left( 0 \right)}} = \frac{{\sqrt 2 }}{1} = \sqrt 2\)

ج)

\(\left( {gof} \right)\left( { - 1} \right) = g\left( {f\left( { - 1} \right)} \right) = g\left( 1 \right) = 0\)

11)

\(\begin{array}{l}f\left( x \right) = ax + b \Rightarrow y = ax + b \Rightarrow y - b = ax\\ \Rightarrow x = \frac{{y - b}}{a} = \frac{1}{a}y - \frac{b}{a} = a'y + b'\\ \Rightarrow {f^{ - 1}}\left( y \right) = a'y + b' \Rightarrow {f^{ - 1}}\left( x \right) = a'x + b'\\\\f\left( x \right) = ax + b \Rightarrow {f^{ - 1}}\left( x \right) = a'x + b'\end{array}\)

12)

\(\begin{array}{l}f\left( x \right) = \frac{5}{9}\left( {x - 32} \right) \Rightarrow y = \frac{5}{9}\left( {x - 32} \right)\\ \Rightarrow \frac{9}{5}y = x - 32 \Rightarrow x = \frac{9}{5}y + 32\\ \Rightarrow {f^{ - 1}}\left( y \right) = \frac{9}{5}y + 32 \Rightarrow {f^{ - 1}}\left( x \right) = \frac{9}{5}x + 32\end{array}\)

13)

الف)

\(\begin{array}{l}f\left( n \right) = \left\{ \begin{array}{l}10,000\quad ,n = 1\\7,000\quad \;,n > 1\end{array} \right.\\g\left( n \right) = \left\{ \begin{array}{l}20,000\quad ,n = 1\\9,000\quad \;,n > 1\end{array} \right.\end{array}\)

ب)

\(h\left( n \right) = f\left( n \right) + g\left( n \right) = \left\{ \begin{array}{l}30,000\quad ,n = 1\\16,000\quad \;,n > 1\end{array} \right.\)

ت)