گام به گام تمرین صفحه 29 درس 1 ریاضی (3) (تابع)

1)

الف)

\(\begin{array}{l}y = f(x) = \frac{{ - 8x + 3}}{2}\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x = \frac{{ - 8y + 3}}{2} \Rightarrow  - 8y + 3 = 2x\\ \Rightarrow  - 8y = 2x - 3\\ \Rightarrow y = \frac{{2x - 3}}{{ - 8}} = \frac{{ - 2x + 3}}{8}\\ \Rightarrow {f^{ - 1}}(x) = \frac{{ - 2x + 3}}{8}\end{array}\)

ب)

\(\begin{array}{l}y = g(x) =  - 5 - \sqrt {3x + 1} \;\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x =  - 5 - \sqrt {3y + 1}  \Rightarrow \sqrt {3y + 1}  =  - x - 5\\ \Rightarrow 3y + 1 = {( - x - 5)^2} = {(x + 5)^2}\\ \Rightarrow 3y = {(x + 5)^2} - 1 \Rightarrow y = \frac{{{{(x + 5)}^2} - 1}}{3}\\ \Rightarrow {g^{ - 1}}(x) = \frac{{{{(x + 5)}^2} - 1}}{3}\end{array}\)

2)

الف)

\(\begin{array}{l}fog(x) = f(g(x)) =  - \frac{{{7}}}{2}( - \frac{{2x + 6}}{{{7}}}) - 3 = x + 3 - 3 = x\\gof(x) = g(f(x)) =  - \frac{{2(\frac{{ - 7x}}{2} - 3) + 6}}{7} = \frac{{7x - 6 + 6}}{7} = x\end{array}\)

ب)

\(\begin{array}{l}fog(x) = f(g(x)) =  - \sqrt {8 + {x^2} - 8}  =  - \sqrt {{x^2}} \quad \mathop  = \limits^{x \le  \circ } \quad  - ( - x) = x\\gof(x) = g(f(x)) = {( - \sqrt {x - 8} )^2} + 8 = x - 8 + 8 = x\end{array}\)

3)

\(\begin{array}{l}y = f(x) = \frac{9}{5}x + 32\;\;,\;\;x \leftrightarrow y\\x = \frac{9}{5}y + 32 \Rightarrow 5x = 9y + 160\\ \Rightarrow 9y = 5x - 160\\ \Rightarrow y = \frac{{5x - 160}}{9}\\ \Rightarrow {f^{ - 1}}(x) = \frac{5}{9}x - \frac{{160}}{9}\end{array}\)

میزان تغییرات درجه نسبت به فارنهایت را نشان می دهد.

4)

الف)

\(\begin{array}{l}f(x) = \left| x \right|\quad  \to \quad x \ge 0\\y = f(x) = x\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x = y \Rightarrow y = x\\ \Rightarrow {f^{ - 1}}(x) = x\end{array}\)

ب)

\(\begin{array}{l}g(x) =  - {x^2}\quad  \to \quad x \le 0\\y = g(x) =  - {x^2}\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x =  - {y^2} \Rightarrow y = \sqrt { - x} \\ \Rightarrow {g^{ - 1}}(x) = \sqrt { - x} \end{array}\)

پ)

\(\begin{array}{l}h(x) = {(x + 2)^2} - 1\quad  \to \quad x \ge 0\\y = h(x) = {(x + 2)^2} - 1\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x = {(y + 2)^2} - 1 \Rightarrow {(y + 2)^2} = x + 1\\ \Rightarrow y = \sqrt {x + 1}  - 2\\ \Rightarrow {h^{ - 1}}(x) = \sqrt {x + 1}  - 2\end{array}\)

5)

6)

\(\begin{array}{l}f(x) = {x^2} - 4x + 5 = {(x - 2)^2} + 1\quad \\ \to \quad \left\{ \begin{array}{l}{D_f} = \left[ {2\;,\; + \infty } \right)\\{R_f} = \left[ {1\;,\; + \infty } \right)\end{array} \right.\\y = {(x - 2)^2} + 1\;\;,\;\;x \leftrightarrow y\\ \Rightarrow x = {(y - 2)^2} + 1 \Rightarrow {(y - 2)^2} = x - 1\\ \Rightarrow y = \sqrt {x - 1}  + 2\\ \Rightarrow {f^{ - 1}}(x) = \sqrt {x - 1}  + 2\end{array}\)

7)

الف)

\(\begin{array}{l}\left\{ \begin{array}{l}f(x) = \frac{1}{8}x - 3 \Rightarrow {f^{ - 1}}(x) = 8x + 24\\g(x) = {x^3} \Rightarrow {g^{ - 1}}(x) = \sqrt[3]{x}\end{array} \right.\\y = fog(x) = \frac{1}{8}{x^3} - 3\mathop  \Rightarrow \limits^{x \leftrightarrow y} x = \frac{1}{8}{y^3} - 3\\ \Rightarrow y = \sqrt[3]{{8x + 24}} \Rightarrow {(fog)^{ - 1}}(x) = \sqrt[3]{{8x + 24}}\\{(fog)^{ - 1}}(5) = \sqrt[3]{{8 \times 5 + 24}} = \sqrt[3]{{64}} = 4\end{array}\)

ب)

\(\begin{array}{l}{f^{ - 1}}o{f^{ - 1}}(6) = {f^{ - 1}}({f^{ - 1}}(6)) = {f^{ - 1}}(8 \times 6 + 24)\\ = {f^{ - 1}}(72) = 8 \times 72 + 24 = 600\end{array}\)

پ)

\(\begin{array}{l}{g^{ - 1}}o{f^{ - 1}}(5) = {g^{ - 1}}({f^{ - 1}}(5)) = {g^{ - 1}}(8 \times 5 + 24)\\ = {g^{ - 1}}(64) = \sqrt[3]{{64}} = 4\end{array}\)