جواب فعالیت صفحه 65 درس 3 ریاضی یازدهم تجربی (تابع)

1 اگر f(x) = 2x – 1 و g(x) = x – 2، آن گاه مجموع، تفاضل، حاصل ضرب و حاصل تقسیم آنها \((\frac{f}{g})\) را به دست آورید و دامنهٔ هر یک را مشخص کنید.

حل:

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

2 توابع f و g به صورت زیر تعریف شده اند. 

\(\begin{array}{l}f = \left\{ {\left( { - 2,3} \right)\;,\;\left( { - 1,5} \right)\;,\;\left( {0, - 2} \right)\;,\;\left( {1,2} \right)} \right\}\\g = \left\{ {\left( { - 2,\frac{1}{3}} \right)\;,\;\left( { - 1,0} \right)\;,\;\left( {1,\frac{2}{3}} \right)\;,\;\left( {4, - 6} \right)} \right\}\end{array}\)

با توجه به توابع f و g جاهای خالی را پر کنید.

\(\begin{array}{l}{D_f} = \left\{ { - 2, - 1,0,1} \right\}\\{D_g} = \left\{ {.............} \right\}\\{D_f} \cap {D_g} = \left\{ { - 2, - 1,1} \right\}\end{array}\)

\(\begin{array}{l}f + g = \left\{ {\left( { - 2,3 + \frac{1}{3}} \right)\;,\;\left( { - 1,5 + ...} \right)\;,\;\left( {...,2 + \frac{2}{3}} \right)} \right\} = \left\{ {\left( { - 2,...} \right)\;,\;\left( { - 1,5} \right)\;,\;\left( {1,\frac{8}{3}} \right)} \right\}\\g - f = \left\{ {\left( { - 2,\frac{1}{3} - 3} \right)\;,\;\left( { - 1,0 - ...} \right)\;,\;\left( {1,... - 2} \right)} \right\} = \left\{ {\left( { - 2, - \frac{8}{3}} \right)\;,\;\left( { - 1,...} \right)\;,\;\left( {1,...} \right)} \right\}\end{array}\)

\(f\,.\,g = \left\{ {\left( { - 2,3 \times \frac{1}{3}} \right)\;,\;\left( {...,5 \times 0} \right)\;,\;\left( {...,2 \times \frac{2}{3}} \right)} \right\} = \left\{ {\left( { - 2,1} \right)\;,\;\left( {...,...} \right)\;,\;\left( {1,\frac{4}{3}} \right)} \right\}\)

\({D_{f/g}} = \left\{ { - 2, - 1,1} \right\} - \left\{ { - 1} \right\} = \left\{ { - 2,1} \right\}\)

\(\begin{array}{l}\frac{f}{g} = \left\{ {\left( { - 2,\frac{3}{{\frac{1}{3}}}} \right)\;,\;\left( {1,\frac{2}{{...}}} \right)} \right\} = \left\{ {\left( { - 2,...} \right)\;,\;\left( {1,...} \right)} \right\}\\{D_{g/f}} = \left\{ { - 2, - 1,1} \right\} - ... = ....\\\frac{g}{f} = ...\end{array}\)