Let f(x)=yf(x) = yf(x)=y, then we have
y=x+12 ⟹ 2y=x+1;x=2y−1y = \frac{x+1}{2} \implies 2y = x+1; x = 2y-1y=2x+1⟹2y=x+1;x=2y−1
Let f1(x)=x;x=2y−1f^{1}(x) = x; x = 2y-1f1(x)=x;x=2y−1, replacing y with x,
f1(x)=2x−1 ⟹ f1(−2)=2(−2)−1=−5f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5f1(x)=2x−1⟹f1(−2)=2(−2)−1=−5
