Text Given f(x)=2x−1x+3f(x)=\dfrac{2x-1}{x+3}f(x)=x+32x−1, find the inverse function and compute f−1(0)f^{-1}(0)f−1(0). Solution Set y=2x−1x+3y=\dfrac{2x-1}{x+3}y=x+32x−1 and solve for xxx: y(x+3)=2x−1 ⇒ yx+3y=2x−1 ⇒ x(y−2)=−1−3y,y(x+3)=2x-1\;\Rightarrow\; yx+3y=2x-1\;\Rightarrow\; x(y-2)=-1-3y,y(x+3)=2x−1⇒yx+3y=2x−1⇒x(y−2)=−1−3y, x=−1−3yy−2=3y+12−y.x=\frac{-1-3y}{y-2}=\frac{3y+1}{2-y}.x=y−2−1−3y=2−y3y+1. Hence f−1(x)=3x+12−xf^{-1}(x)=\dfrac{3x+1}{2-x}f−1(x)=2−x3x+1, and f−1(0)=0+12−0=12.f^{-1}(0)=\frac{0+1}{2-0}=\frac12.f−1(0)=2−00+1=21. f−1(x)=3x+12−x,f−1(0)=12 \boxed{\,f^{-1}(x)=\dfrac{3x+1}{2-x},\quad f^{-1}(0)=\tfrac12\,}f−1(x)=2−x3x+1,f−1(0)=21