گام به گام تمرین صفحه 69 درس 3 ریاضی یازدهم (تابع)

1)

الف) \(g\left( x \right) =  - \left| x \right|\)

ب) \(h\left( x \right) =  - \left| {x - 3} \right|\)

پ) \(l(x) = 2\left| {x - 2} \right|\)

2)

الف)

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) = \left| x \right|\quad ,\quad {D_f} = \mathbb{R}\\g\left( x \right) = \frac{1}{x}\quad ,\quad {D_g} = \mathbb{R} - \left\{ 0 \right\}\end{array} \right.\\\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = \left| x \right| + \frac{1}{x}\\ \Rightarrow {D_{f + g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ 0 \right\}\\\\\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right) = \left| x \right| - \frac{1}{x}\\ \Rightarrow {D_{f - g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ 0 \right\}\\\\\left( {f \cdot g} \right)\left( x \right) = f\left( x \right) \times g\left( x \right) = \left| x \right| \times \frac{1}{x} = \frac{{\left| x \right|}}{x}\\ \Rightarrow {D_{f \cdot g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ 0 \right\}\\\\\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\left| x \right|}}{{\frac{1}{x}}} = x\left| x \right|\\ \Rightarrow {D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {\left. x \right|g\left( x \right) = 0} \right\} = \\\mathbb{R} - \left\{ 0 \right\} - \left\{ {} \right\} = \mathbb{R} - \left\{ 0 \right\}\end{array}\)

ب)

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) = {x^2} - 4\quad ,\quad {D_f} = \mathbb{R}\\g\left( x \right) = x + 2\quad ,\quad {D_g} = \mathbb{R}\end{array} \right.\\\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = {x^2} + x - 2\\ \Rightarrow {D_{f + g}} = {D_f} \cap {D_g} = \mathbb{R}\\\\\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right) = {x^2} - x - 6\\ \Rightarrow {D_{f - g}} = {D_f} \cap {D_g} = \mathbb{R}\\\\\left( {f \cdot g} \right)\left( x \right) = f\left( x \right) \times g\left( x \right) = {x^3} + 2{x^2} - 4x - 8\\ \Rightarrow {D_{f \cdot g}} = {D_f} \cap {D_g} = \mathbb{R}\\\\\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{{x^2} - 4}}{{x + 2}} = x - 2\\ \Rightarrow {D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {\left. x \right|g\left( x \right) = 0} \right\} = \mathbb{R} - \left\{ { - 2} \right\}\end{array}\)

پ)

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) = \sqrt x \quad ,\quad {D_f} = \left[ {0\;,\;\infty } \right)\\g\left( x \right) =  - \sqrt x \quad ,\quad {D_g} = \left[ {0\;,\;\infty } \right)\end{array} \right.\\\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = \sqrt x  - \sqrt x  = 0\\ \Rightarrow {D_{f + g}} = {D_f} \cap {D_g} = \left[ {0\;,\;\infty } \right)\\\\\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right) = \sqrt x  - \left( { - \sqrt x } \right) = 2\sqrt x \\ \Rightarrow {D_{f - g}} = {D_f} \cap {D_g} = \left[ {0\;,\;\infty } \right)\\\\\left( {f \cdot g} \right)\left( x \right) = f\left( x \right) \times g\left( x \right) = \left( {\sqrt x } \right) \times \left( { - \sqrt x } \right) =  - x\\ \Rightarrow {D_{f \cdot g}} = {D_f} \cap {D_g} = \left[ {0\;,\;\infty } \right)\\\\\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\sqrt x }}{{ - \sqrt x }} =  - 1\\ \Rightarrow {D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {\left. x \right|g\left( x \right) = 0} \right\}\\ = \left[ {0\;,\;\infty } \right) - \left\{ 0 \right\} = \left( {0\;,\;\infty } \right)\end{array}\)

ت)

\(\begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) = \frac{{x - 2}}{{x + 5}}\quad ,\quad {D_f} = \mathbb{R} - \left\{ { - 5} \right\}\\g\left( x \right) = {x^2} + 3x - 10\quad ,\quad {D_g} = \mathbb{R}\end{array} \right.\\\left( {f + g} \right)\left( x \right) = f\left( x \right) + g\left( x \right) = \frac{{x - 2}}{{x + 5}} + {x^2} + 3x - 10\\ \Rightarrow {D_{f + g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ { - 5} \right\}\\\\\left( {f - g} \right)\left( x \right) = f\left( x \right) - g\left( x \right) = \frac{{x - 2}}{{x + 5}} - {x^2} - 3x + 10\\ \Rightarrow {D_{f - g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ { - 5} \right\}\\\\\left( {f \cdot g} \right)\left( x \right) = f\left( x \right) \times g\left( x \right) = \left( {\frac{{x - 2}}{{x + 5}}} \right) \times \left( {{x^2} + 3x - 10} \right)\\ = \left( {\frac{{x - 2}}{{x + 5}}} \right) \times \left( {x - 2} \right)\left( {x + 5} \right) = {\left( {x - 2} \right)^2}\\ \Rightarrow {D_{f \cdot g}} = {D_f} \cap {D_g} = \mathbb{R} - \left\{ { - 5} \right\}\\\\\left( {\frac{f}{g}} \right)\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\frac{{x - 2}}{{x + 5}}}}{{{x^2} + 3x - 10}}\\ = \frac{{x - 2}}{{\left( {x + 5} \right)\left( {x - 2} \right)\left( {x + 5} \right)}} = \frac{1}{{{{\left( {x + 5} \right)}^2}}}\\ \Rightarrow {D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {\left. x \right|g\left( x \right) = 0} \right\}\\ = \mathbb{R} - \left\{ { - 5} \right\} - \left\{ { - 5\;,\;2} \right\} = \mathbb{R} - \left\{ { - 5\;,\;2} \right\}\end{array}\)

ث)

\(\begin{array}{l}\left\{ \begin{array}{l}f = \left\{ {\left( {2\;,\;5} \right)\;,\;\left( {3\;,\;4} \right)\;,\;\left( {0\;,\; - 2} \right)} \right\}\quad ,\quad {D_f} = \left\{ {0\;,\;2\;,\;3} \right\}\\g = \left\{ {\left( { - 1\;,\;2} \right)\;,\;\left( {0\;,\;3} \right)\;,\;\left( {2\;,\;4} \right)\;,\;\left( {3\;,\;0} \right)} \right\}\quad ,\quad {D_g} = \left\{ { - 1\;,\;0\;,\;2\;,\;3} \right\}\end{array} \right.\\{D_{f + g}} = {D_f} \cap {D_g} = \left\{ {0\;,\;2\;,\;3} \right\}\\ \Rightarrow f + g = \left\{ {\left( {0\;,\; - 2 + 3} \right)\;,\;\left( {2\;,\;5 + 4} \right)\;,\;\left( {3\;,\;4 + 0} \right)} \right\}\\ = \left\{ {\left( {0\;,\;1} \right)\;,\;\left( {2\;,\;9} \right)\;,\;\left( {3\;,\;4} \right)} \right\}\\\\{D_{f - g}} = {D_f} \cap {D_g} = \left\{ {0\;,\;2\;,\;3} \right\}\\ \Rightarrow f - g = \left\{ {\left( {0\;,\; - 2 - 3} \right)\;,\;\left( {2\;,\;5 - 4} \right)\;,\;\left( {3\;,\;4 - 0} \right)} \right\}\\ = \left\{ {\left( {0\;,\; - 5} \right)\;,\;\left( {2\;,\;1} \right)\;,\;\left( {3\;,\;4} \right)} \right\}\\\\{D_{f \cdot g}} = {D_f} \cap {D_g} = \left\{ {0\;,\;2\;,\;3} \right\}\\ \Rightarrow f \cdot g = \left\{ {\left( {0\;,\; - 2 \times 3} \right)\;,\;\left( {2\;,\;5 \times 4} \right)\;,\;\left( {3\;,\;4 \times 0} \right)} \right\}\\ = \left\{ {\left( {0\;,\; - 6} \right)\;,\;\left( {2\;,\;20} \right)\;,\;\left( {3\;,\;0} \right)} \right\}\\\\{D_{\frac{f}{g}}} = {D_f} \cap {D_g} - \left\{ {\left. x \right|g\left( x \right) = 0} \right\}\\ = \left\{ {0\;,\;2\;,\;3} \right\} - \left\{ 0 \right\} = \left\{ {0\;,\;2} \right\}\\ \Rightarrow \frac{f}{g} = \left\{ {\left( {0\;,\;\frac{{ - 2}}{3}} \right)\;,\;\left( {2\;,\;\frac{5}{4}} \right)} \right\}\end{array}\)

3)

الف) \(r\left( x \right) = 2\sqrt x\)

ب) \(s\left( x \right) =  - \sqrt {x - 2}\)

پ) \(l\left( x \right) =  - 3\sqrt x\)

ت) \(u\left( x \right) = 1 - \sqrt x\)

ث) \(v\left( x \right) = 1 - \sqrt {x - 3}\)

4)

5)

\(\left. \begin{array}{l}f\left( x \right) = x\\g\left( x \right) =  - x\\h\left( x \right) =  - 2x\end{array} \right\} \Rightarrow g\left( x \right) = f\left( x \right) + h\left( x \right) = x + \left( { - 2x} \right) =  - x\)