\(\frac{x + 3}{x - 2},x ≠ 2\)
\(\frac{x - 3}{x + 2},x ≠ -2\)
\(\frac{3x - 2}{x+1},x ≠ -1\)
\(\frac{3x + 2}{x - 1},x ≠ 1\)
Correct answer is D
\(f :x→\frac{x + 2}{x - 3},x ≠ 3, f = ?\)
Let \(f :x=y\)
\(y=\frac{x + 2}{x - 3}\)
\(=x+2=y(x-3)\)
\(=x-xy=-3y-2\)
\(=x(1-y)=-3y-2\)
\(=x=\frac{-3y - 2}{1 - y}=\frac{-(3y + 2)}{- (y - 1)}\)
\(=x=\frac{3y + 2}{y - 1}\)
\(∴f ^{-1} : x=\frac{3x + 2}{x - 1},x ≠ 1\)