جواب کاردرکلاس صفحه 27 درس 1 ریاضی دوازدهم تجربی (تابع)

الف

\(\begin{array}{l}y = f(x) = - \frac{1}{2}x + 3x = - \frac{1}{2}y + 3\\\\ \Rightarrow - \frac{1}{2}y = x - 3\\\\ \Rightarrow y = - 2x + 6 \Rightarrow {f^{ - 1}}(x) = - 2x + 6\\\\\left\{ \begin{array}{l}{D_f} = \mathbb{R}\\{R_f} = \mathbb{R}\end{array} \right.\begin{array}{*{20}{c}}{}&{}&{}\end{array},\begin{array}{*{20}{c}}{}&{}&{}\end{array}\left\{ \begin{array}{l}{D_{{f^{ - 1}}}} = \mathbb{R}\\{R_{{f^{ - 1}}}} = \mathbb{R}\end{array} \right.\end{array}\)

ب

\(\begin{array}{l}y = g(x) = 1 + \sqrt {x - 2} x = 1 + \sqrt {y - 2} \Rightarrow \sqrt {y - 2} = x - 1\\\\ \Rightarrow y - 2 = {(x - 1)^2} \Rightarrow y = {(x - 1)^2} + 2\\\\ \Rightarrow {g^{ - 1}}(x) = {(x - 1)^2} + 2\\\\\left\{ \begin{array}{l}{D_g} = \left[ {2\;,\; + \infty } \right)\\{R_g} = \left[ {1\;,\; + \infty } \right)\end{array} \right.\begin{array}{*{20}{c}}{}&{}&{}\end{array},\begin{array}{*{20}{c}}{}&{}&{}\end{array}\left\{ \begin{array}{l}{D_{{g^{ - 1}}}} = \left[ {1\;,\; + \infty } \right)\\{R_{{g^{ - 1}}}} = \left[ {2\;,\; + \infty } \right)\end{array} \right.\end{array}\)

پ

تابع \(h(x) = {x^2} + 1\) فقط به ازای یک از بازه های \(\left[ {0\;,\; + \infty } \right)\) و \(\left( { - \;\infty \;,\;0} \right]\) می تواند یک به یک باشد؛ تابع وارون\(h(x) = {x^2} + 1\)  را به ازای بازه \(\left[ {0\;,\; + \infty } \right)\) بدست می آوریم :

\(\begin{array}{l}y = h(x) = {x^2} + 1x = {y^2} + 1 \Rightarrow {y^2} = x - 1\\\\ \Rightarrow y = \sqrt {x - 1} \Rightarrow {h^{ - 1}}(x) = \sqrt {x - 1} \\\\\left\{ \begin{array}{l}{D_h} = \left[ {0\;,\; + \infty } \right)\\{R_h} = \left[ {1\;,\; + \infty } \right)\end{array} \right.\begin{array}{*{20}{c}}{}&{}&{}\end{array},\begin{array}{*{20}{c}}{}&{}&{}\end{array}\left\{ \begin{array}{l}{D_{{h^{ - 1}}}} = \left[ {1\;,\; + \infty } \right)\\{R_{{h^{ - 1}}}} = \left[ {0\;,\; + \infty } \right)\end{array} \right.\end{array}\)