اگر $f(x)=\frac{۲x-۱}{x+۲}$ و $g(x)=x+۴$ باشند، جواب معادلهٔ $(gof)(x)=(fog)(x)$ کدام است؟
توابع $fog$ و $gof$ را تشکیل میدهیم: $\begin{align} & f(x)=\frac{2x-1}{x+2}\,\,,\,\,g(x)=x+4 \\ & (fog)(x)=f(g(x))=f(x+4)=\frac{2(x+4)-1}{x+4+2}=\frac{2x+7}{x+6} \\ & (gof)(x)=g(f(x))=g(\frac{2x-1}{x+2})=\frac{2x-1}{x+2}+4=\frac{2x-1+4x+8}{x+2}=\frac{6x+7}{x+2} \\ \end{align}$ بنابراین: $\begin{align} & (fog)(x)=(gof)(x)\Rightarrow \frac{2x+7}{x+6}=\frac{6x+7}{x+2} \\ & \Rightarrow (2x+7)(x+2)=(6x+7)(x+6) \\ & \Rightarrow 2{{x}^{2}}+4x+7x+14=6{{x}^{2}}+36x+7x+42 \\ & \Rightarrow 4{{x}^{2}}+32x+28=0\xrightarrow{\div 4}{{x}^{2}}+8x+7=0 \\ & \Rightarrow (x+7)(x+1)=0\Rightarrow \left\{ \begin{matrix} x=-7 \\ x=-1 \\\end{matrix} \right. \\ \end{align}$