Question

Fie f,g:|R->|R. f(x)=3x+5, g(x)=mx+m-2. Sa se determine m∈|R, pentru care g = f^{-1}

Answer 1

Cautam functia:
$f^{-1}(x)$
astfel incat:
[tex]f^{-1}(f(x))=x \rightarrow f^{-1}(3x+5)=x \rightarrow f^{-1}(y)=\frac{y-5}{3} [/tex]
Verificare:
[tex]f^{-1}(f(x))=x \rightarrow f^{-1}(3*x+5)=x \rightarrow \frac{3*x+5-5}{3}=x\\ \rightarrow x=x \ (A)[/tex]
Acum g:
[tex]g(x)=f^{-1}(x) \Leftrightarrow m*x+m-2=\frac{x-5}{3}\\ \Leftrightarrow m*(x+1)=\frac{x-5}{3}+2 \Leftrightarrow m*(x+1)=\frac{x+1}{3}\\ \Leftrightarrow m=\frac{x+1}{3*(x+1)} \Leftrightarrow m=\frac{1}{3}[/tex]

Answer 2

[tex]f,g:\mathbb_{R} \rightarrow \mathbb_{R}, $ $ $ $f(x) = 3x+5, $ $ $ $ g(x) = mx+m-2 \\ \\ 3x+5 = y \Rightarrow 3x = y-5 \Rightarrow x = \dfrac{y-5}{3} \Rightarrow x = \dfrac{y}{3}-\dfrac{5}{3}\Rightarrow \\ \\ \Rightarrow x = \dfrac{1}{3}\cdot y-\dfrac{5}{3} \Rightarrow f^{-1}(x) = \dfrac{1}{3}\cdot x-\dfrac{5}{3} \\ \\ g(x) = f^{-1}(x) \Rightarrow mx+(m-2) = \dfrac{}{}\dfrac{1}{3}\cdot x -\dfrac{5}{3} [/tex]

$\left\{ \begin{array}{ll} m = \dfrac{1}{3} \\ m-2 = -\dfrac{5}{3} \end{array} \right \Rightarrow \left\{ \begin{array}{ll} m = \dfrac{1}{3} \\ m= -\dfrac{5}{3}+2 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} m = \dfrac{1}{3} \\ m = -\dfrac{5}{3}+\dfrac{6}{3}\end{array} \right \Rightarrow \\ \\ \Rightarrow \left\{ \begin{array}{ll} m = \dfrac{1}{3} \\\\ m = \dfrac{1}{3} \end{array} \right| \Rightarrow \boxed{\boxed{m = \dfrac{1}{3}}}$