گام به گام تمرین صفحه 62 درس 2 حسابان (1) (تابع)

1) ارتباط میان معلم ریاضی و تمام شاگردان یک کلاس.

2) خیر؛ زیرا \(\left( {0\;,\;\frac{2}{5}} \right) \in f\) ولی \(\left( {\frac{2}{5}\;,\;0} \right) \notin g\) توابع f و g به خاطر یک به یک نبودن وارون پذیر نیستند.

3)

(الف \(f\left( x \right) = {\left( {x + 5} \right)^2}\;\;\;,\;\;\;\;x \ge  - 5\)

\(\begin{array}{l}y = {\left( {x + 5} \right)^2} \Rightarrow  \pm \sqrt y  = x + 5\\\;\mathop  \Rightarrow \limits^{\left( i \right)} \;\sqrt y  = x + 5\\ \Rightarrow x = \sqrt y  - 5 \Rightarrow {f^{ - 1}}\left( y \right) = \sqrt y  - 5\\ \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt x  - 5\\\left\{ \begin{array}{l}{D_{{f^{ - 1}}}} = \left[ {0\;,\;\infty } \right)\\{R_{{f^{ - 1}}}} = \left[ { - 5\;,\;\infty } \right)\end{array} \right.\end{array}\)

تابع f وارون پذیر است

(ب \(f\left( x \right) =  - \left| {x - 1} \right| + 1\;\;\;,\;\;\;\;x \ge 2\)

\(\begin{array}{l}\left\{ \begin{array}{l}{D_f} = \left[ {2\;,\;\infty } \right)\\{R_f} = \left( { - \infty \;,\;0} \right]\quad \left( i \right)\end{array} \right.\\y =  - \left| {x - 1} \right| + 1 \Rightarrow y - 1 =  - \left| {x - 1} \right| \Rightarrow \;1 - y\; = \left| {x - 1} \right|\\ \Rightarrow  \pm \left( {\;1 - y} \right) = x - 1\;\mathop  \Rightarrow \limits^{\left( i \right)} \;\left( {\;1 - y} \right) = x - 1 \Rightarrow x = 2 - y\\{f^{ - 1}}\left( y \right) = 2 - y \Rightarrow {f^{ - 1}}\left( x \right) = 2 - x\\\left\{ \begin{array}{l}{D_{{f^{ - 1}}}} = \left( { - \infty \;,\;0} \right]\\{R_{{f^{ - 1}}}} = \left[ {2\;,\;\infty } \right)\end{array} \right.\end{array}\)

تابع f وارون پذیر است

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

تابع f وارون پذیر نیست

(ت \(f\left( x \right) = \sqrt {x + 2}  - 3\)

\(\begin{array}{l}\left\{ \begin{array}{l}{D_f} = \left[ { - 2\;,\;\infty } \right)\\{R_f} = \left[ { - 3\;,\;\infty } \right)\quad \left( i \right)\end{array} \right.\\y = \sqrt {x + 2}  - 3 \Rightarrow y + 3 = \sqrt {x + 2}  \Rightarrow {\left( {y + 3} \right)^2} = x + 2\\ \Rightarrow x = {\left( {y + 3} \right)^2} - 2\\\left\{ \begin{array}{l}{D_{{f^{ - 1}}}} = \left[ { - 3\;,\;\infty } \right)\\{R_{{f^{ - 1}}}} = \left[ { - 2\;,\;\infty } \right)\end{array} \right.\end{array}\)

تابع f وارون پذیر است

4)

الف)

\(\begin{array}{l}h\left( t \right) = 100 - 5{t^2} \Rightarrow 0 = 100 - 5{t^2}\\ \Rightarrow {t^2} = 20 \Rightarrow \left\{ \begin{array}{l}t =  \pm \sqrt {20} \\t \ge 0\end{array} \right. \Rightarrow t = \sqrt {20} \\\left\{ \begin{array}{l}{D_h} = \left[ {0\;,\;\sqrt {20} } \right]\\{R_h} = \left[ {0\;,\;100} \right]\end{array} \right.\end{array}\)

ب) زیرا به هر ارتفاعی از برد، تنها و تنها یک زمان از دامنه نظیر می شود.

پ)

\(\begin{array}{l}\left\{ \begin{array}{l}{D_h} = \left[ {0\;,\;\sqrt {20} } \right]\\{R_h} = \left[ {0\;,\;100} \right]\end{array} \right.\\h\left( t \right) = y = 100 - 5{t^2} \Rightarrow {t^2} = \frac{{100 - y}}{5} \Rightarrow t = \sqrt {\frac{{100 - y}}{5}} \\ \Rightarrow {h^{ - 1}}\left( t \right) = \sqrt {\frac{{100 - t}}{5}} \\\left\{ \begin{array}{l}{D_{{h^{ - 1}}}} = \left[ {0\;,\;100} \right]\\{R_{{h^{ - 1}}}} = \left[ {0\;,\;\sqrt {20} } \right]\end{array} \right.\end{array}\)

5)

6)

\(\begin{array}{l}f\left( x \right) =  - \frac{1}{2}x + 3\\y =  - \frac{1}{2}x + 3 \Rightarrow y - 3 =  - \frac{1}{2}x \Rightarrow x =  - 2\left( {y - 3} \right)\\ \Rightarrow {f^{ - 1}}\left( y \right) =  - 2y + 6 \Rightarrow {f^{ - 1}}\left( x \right) =  - 2x + 6\end{array}\)