If f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3}f(x)=x−32x+1, what is f−1(x)f^{-1}(x)f−1(x)?Ax−32x+1\frac{x - 3}{2x + 1}2x+1x−3Bx−12x+3\frac{x - 1}{2x + 3}2x+3x−1C2x−1x+3\frac{2x - 1}{x + 3}x+32x−1D3x+1x−2\frac{3x + 1}{x - 2}x−23x+1check_circleExplanationSwap and solve: x=2y+1y−3⇒x(y−3)=2y+1⇒xy−2y=3x+1⇒y=3x+1x−2x = \frac{2y + 1}{y - 3} \Rightarrow x(y - 3) = 2y + 1 \Rightarrow xy - 2y = 3x + 1 \Rightarrow y = \frac{3x + 1}{x - 2}x=y−32y+1⇒x(y−3)=2y+1⇒xy−2y=3x+1⇒y=x−23x+1.