Example  TRIG.INV.1:    Suppose $f(x)=2tan(x−1)−3$ for $x\in(−\fracπ2+1,\fracπ2+1)$. Verify that $g(x)=arctan(\frac{x+3}2)+1$ is the inverse function for $f$ .

(i)$(f \circ g)(x) = f(g(x)) =f( arctan(\frac{x+3}2)+1) = 2tan((arctan(\frac{x+3}2)+1) -1)- 3 = 2 \frac{x+3}2 -3= x$.
(ii)$(g \circ f)(x) = g(f(x)) =g(2tan(x−1)−3) = arctan (\frac {(2tan(x−1)−3)+3}2) + 1 = (x-1)+1 = x$.

We check both with Mapping Diagrams and the cartesian graphs..
Which do you find more convincing visually?

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