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To find the inverse, set y=3x+2x−1y = \dfrac{3x+2}{x-1} and solve for xx in terms of yy. Cross-multiply: y(x−1)=3x+2⇒yx−y=3x+2y(x-1) = 3x + 2 \Rightarrow yx - y = 3x + 2. Collect xx terms: yx−3x=y+2⇒x(y−3)=y+2yx - 3x = y + 2 \Rightarrow x(y - 3) = y + 2. Hence x=y+2y−3x = \dfrac{y + 2}{y - 3}. Interchange variables to express f−1f^{-1}: f−1(x)=x+2x−3f^{-1}(x) = \dfrac{x+2}{x-3}. However, check carefully: our algebra gave x=(y+2)/(y−3)x = (y+2)/(y-3), so replacing yy with xx produces f−1(x)=x+2x−3f^{-1}(x)=\dfrac{x+2}{x-3}. Compare to options: that corresponds to choice C if denominator is 3−x3 - x or x−3x - 3? We found x−3x-3 in the denominator; option C is (x+2)/(3−x)(x+2)/(3-x), which equals −x+2x−3-\dfrac{x+2}{x-3} and would introduce a sign error. Re-check manipulation: yx−y=3x+2⇒yx−3x=y+2⇒x(y−3)=y+2⇒x=y+2y−3yx - y = 3x + 2 \Rightarrow yx - 3x = y + 2 \Rightarrow x(y-3)=y+2 \Rightarrow x=\dfrac{y+2}{y-3}. Thus the correct inverse is x+2x−3\boxed{\dfrac{x+2}{x-3}}, which is not option C as printed—it matches option A if A were (x+2)/(x−3)(x+2)/(x-3). Since option A is (x+2)/(x−3)(x+2)/(x-3) (verify): Option A given is x+2x−3\dfrac{x+2}{x-3}. Therefore the correct choice is A. Domain note: x≠3x\ne 3 for the inverse, corresponding to f(1)f(1) undefined; ranges and domains swap under inversion.

(Note: I selected the option matching (x+2)/(x−3)(x+2)/(x-3); if your choices display differently, pick the one with denominator x−3x-3.)