گام به گام تمرین صفحه 52 درس 2 حسابان (1) (تابع)

1)

(الف  \(f\left( x \right) = \frac{{x - 1}}{{2 - x}}\)

\({D_f} = \mathbb{R} - \left\{ 2 \right\}\)

(ب  \(f\left( x \right) = \frac{{ - 3x}}{{{x^2} + 1}}\)

\({D_f} = \mathbb{R}\)

(پ  \(f\left( x \right) = \frac{{2x + 3}}{{{x^2} + x - 12}}\)

\({D_f} = \mathbb{R} - \left\{ { - 4\;,\;3} \right\}\)

(ت  \(f\left( x \right) = \sqrt {3x + 1}\)

\({D_f} = \left[ { - \frac{1}{3}\;,\;\infty } \right)\)

(ث  \(f\left( x \right) = 2\sqrt x  - 3\)

\({D_f} = \left[ {0\;,\;\infty } \right)\)

(ج  \(f\left( x \right) = \sqrt {8 - x}\)

\({D_f} = \left( { - \infty \;,\;8} \right]\)

2)

با قرینه کردن تابع f نسبت به محور x ، تابع g بدست می آید:

3)

4)

(الف  \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}\begin{array}{l}\frac{1}{x}\;\;\;\;\;\;\;\;\;\;x \ge 0\\\end{array}\\{x - 2\;\;\;\;\;\;x \le 0}\end{array}} \right.\)

\(\left\{ \begin{array}{l}{D_f} = \mathbb{R}\\{R_f} = \left( { - \;\infty \;,\; - 2} \right]\; \cup \left( {0\;,\; + \infty } \right)\end{array} \right.\)

(ب  \(f\left( x \right) = \sqrt {x - 2}  + 5\)

\(\left\{ \begin{array}{l}{D_f} = \left[ {2\;,\; + \infty } \right)\\{R_f} = \left[ {5\;,\; + \infty } \right)\end{array} \right.\)

(پ  \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{\sqrt x  + 2\:\,\,\,\,\:\:\:\:\:\:\:\:\:x > 0}\end{array}}\\{\sqrt {x + 2} \:\:\:\:\:\: - 2 \le x \le 0}\end{array}} \right.\)

\(\left\{ \begin{array}{l}{D_f} = \left[ { - 2\;,\; + \infty } \right)\\{R_f} = \left[ {0\;,\;\sqrt 2 } \right]\; \cup \left( {0\;,\;2} \right)\end{array} \right.\)

(ت \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\begin{array}{*{20}{l}}{ - \frac{1}{x}\,\,\,\,\:\:\:\:\:\:\:\:\:x > 0}\end{array}}\\{ - \sqrt x \:\:\:\:\,\,\,\,\,\,\,\,\:x \ge 0}\end{array}} \right.\)

\(\left\{ \begin{array}{l}{D_f} = \mathbb{R}\\{R_f} = \mathbb{R}\end{array} \right.\)

5) معادلاتی که تابع هستند:

الف) \(3x + 2y = 12\)

پ) \(y =  - 2\)

ج) \(y = \left| x \right|\)

6)

الف)

\(\begin{array}{l}f\left( x \right) = \frac{{255x}}{{100 - x}} \Rightarrow x = 50\\ \Rightarrow f\left( {50} \right) = \frac{{255\left( {50} \right)}}{{100 - 50}} = 255\end{array}\)

255 میلیون تومان

ب) \({D_f} = \left[ {0\;,\;100} \right)\)

7)

(الف \( f\left( x \right) = \left[ x \right] + 1\;\;\;,\;\;\; - 2 \le x \le 3\)

\(\begin{array}{l} - 2 \le x <  - 1 \Rightarrow \left[ x \right] =  - 2 \Rightarrow f\left( x \right) =  - 2 + 1 =  - 1\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 1 + 1 = 0\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) = 0 + 1 = 1\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) = 1 + 1 = 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) = 2 + 1 = 3\end{array}\)

(ب \(f\left( x \right) = \left[ {\frac{1}{2}x} \right]\;\;\;,\;\;\; - 4 \le x \le 4\)

\(\begin{array}{l} - 2 \le x <  - 1 \Rightarrow \left[ x \right] =  - 2 \Rightarrow f\left( x \right) =  - 2 + 1 =  - 1\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 1 + 1 = 0\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) = 0 + 1 = 1\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) = 1 + 1 = 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) = 2 + 1 = 3\end{array}\)

8)

\(\begin{array}{l}y = \left[ x \right] - 3\\ - 1 \le x < 0 \Rightarrow \left[ x \right] =  - 1 \Rightarrow f\left( x \right) =  - 4\\0 \le x < 1 \Rightarrow \left[ x \right] = 0 \Rightarrow f\left( x \right) =  - 3\\1 \le x < 2 \Rightarrow \left[ x \right] = 1 \Rightarrow f\left( x \right) =  - 2\\2 \le x < 3 \Rightarrow \left[ x \right] = 2 \Rightarrow f\left( x \right) =  - 1\\3 \le x < 4 \Rightarrow \left[ x \right] = 3 \Rightarrow f\left( x \right) = 0\\\\y = \left[ {x - 3} \right]\\ - 1 \le x < 0 \Rightarrow  - 4 \le x - 3 <  - 3 \Rightarrow \left[ {x - 3} \right] =  - 4 \Rightarrow f\left( x \right) =  - 4\\0 \le x < 1 \Rightarrow  - 3 \le x - 3 <  - 2 \Rightarrow \left[ {x - 3} \right] =  - 3 \Rightarrow f\left( x \right) =  - 3\\1 \le x < 2 \Rightarrow  - 2 \le x - 3 <  - 1 \Rightarrow \left[ {x - 3} \right] =  - 2 \Rightarrow f\left( x \right) =  - 2\\2 \le x < 3 \Rightarrow  - 1 \le x - 3 < 0 \Rightarrow \left[ {x - 3} \right] =  - 1 \Rightarrow f\left( x \right) =  - 1\\3 \le x < 4 \Rightarrow 0 \le x - 3 < 1 \Rightarrow \left[ {x - 3} \right] = 0 \Rightarrow f\left( x \right) = 0\end{array}\)

این دو تابع با هم برابرند.

9)

الف)

\(\begin{array}{l}n\left( t \right) = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow \\n\left( 5 \right) = \frac{{9500\left( 5 \right) - 2000}}{{4 + 5}} = \left[ {5055/\bar 5} \right] = 5055\end{array}\)

ب)

\(\begin{array}{l}n\left( t \right) = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow 5500 = \frac{{9500t - 2000}}{{4 + t}} \Rightarrow \\22000 + 5500t = 9500t - 2000\\ \Rightarrow 4000t = 24000 \Rightarrow t = 6\end{array}\)