جواب تمرین صفحه 61 درس 2 ریاضی و آمار دهم انسانی (تابع)

\(\begin{array}{l}f(x) = mx + n\\\\f(1) = 1 \Rightarrow m(1) + n = 1 \Rightarrow m + n = 1\\\\f(2) = 4 \Rightarrow m(2) + n = 4 \Rightarrow 2m + n = 4\\\\ \Rightarrow \left\{ \begin{array}{l}m + n = 1\\\\2m + n = 4\end{array} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\begin{array}{*{20}{c}}{( - 1) \times }\\{}\\{}\end{array}\left\{ \begin{array}{l}m + n = 1\\\\2m + n = 4\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l} - m - n = - 1\\\\2m + n = 4\end{array} \right.\\\\ \Rightarrow - m - n + 2m + n = - 1 + 4\\\\m = 3\\\\m + n = 1 \Rightarrow 3 + n = 1\\\\ \Rightarrow n = 1 - 3 = - 2\\\\\left\{ \begin{array}{l}m = 3\\\\n = - 2\end{array} \right.\\\\ \Rightarrow f(x) = 3x - 2\end{array}\)

\(m = \frac{{3 - 1}}{{2 - 4}} = \frac{2}{{ - 2}} = - 1\)

\(:\,y - {y_0} = m(x - {x_0})\) معادله خط

\(\begin{array}{l} \Rightarrow y - 3 = - 1(x - 2) \Rightarrow y = - x + 5\\\\ \Rightarrow f(x) = - x + 5\end{array}\)

\(m = \frac{{8 - 5}}{{2 - 1}} = 3\)

\(:\,y - {y_0} = m(x - {x_0})\) معادله خط

\(\begin{array}{l} \Rightarrow y - 5 = 3(x - 1) \Rightarrow y = 3x + 2\\\\ \Rightarrow f(x) = 3x + 2\\\\f( - 3) = 3( - 3) + 2 = - 9 + 2 = - 7\\\\f(5) = 3(5) + 2 = 15 + 2 = 17\end{array}\)

توابع خطی بی شماری را می توان رسم کرد. به عنوان نمونه، 3 تابع در زیر رسم شده است:

\(m = \frac{{7 - 0}}{{2 - 0}} = \frac{7}{2}\)

\(:\,y - {y_0} = m(x - {x_0})\) معادله خط

\(\begin{array}{l} \Rightarrow y - 0 = \frac{7}{2}(x - 0) \Rightarrow y = \frac{7}{2}x\\\\ \Rightarrow f(x) = \frac{7}{2}x\\\\f(0/1) = \frac{7}{2}(0/1) = \frac{7}{2} \times \frac{1}{{10}} = \frac{7}{{20}}\\\\f( - 0/1) = \frac{7}{2}( - 0/1) = \frac{7}{2} \times \frac{{ - 1}}{{10}} = - \frac{7}{{20}}\\\\f(0/1) - f( - 0/1) = \frac{7}{{20}} - ( - \frac{7}{{20}}) = \\\\\frac{7}{{20}} + \frac{7}{{20}} = \frac{{14}}{{20}} = \frac{7}{{10}} = 0/7\end{array}\)

\(\begin{array}{l}F = \frac{9}{5}C + 32 = \frac{9}{5}(20) + 32 = \\\\\frac{{180}}{5} + 32 = 36 + 32 = 68\end{array}\)

الف

C(x)=3000+50x

درآمد : R(x)=70x

سود : P(x)=70x-(3000+50x)=20x-3000

ب

\(\begin{array}{l}P(x) = 0 \Rightarrow 20x - 3000 = 0\\\\ \Rightarrow 20x = 3000 \Rightarrow x = \frac{{3000}}{{20}} = 150\end{array}\)

پس باید یکی بیشتر، یعنی 151 کالا بفروشد تا سوددهی آغاز شود.