(i)$(f \circ g)(x) = f(g(x)) =f( \log_b ( \frac{x+3} 2) + 1) = 2*b^{(\log_b ( \frac
{x+3} 2) + 1) -1} - 3 = (x+3) -3= x$.

(ii)$(g \circ f)(x) = g(f(x)) =g(2b^{x-1} -3) = \log_b ( \frac {(2b^{x-1} -3)+3} 2) + 1 = (x-1)+1 = x $.

We check both with Mapping Diagrams.

Which do you find more convincing visually?

(ii)$(g \circ f)(x) = g(f(x)) =g(2b^{x-1} -3) = \log_b ( \frac {(2b^{x-1} -3)+3} 2) + 1 = (x-1)+1 = x $.

We check both with Mapping Diagrams.

Which do you find more convincing visually?

Click on play arrow to animate or stop value of $x$ in mapping diagram frame.

Select the point $x$ in the mapping diagram to move the point by mouse or use the scroll up or down key to move by increments.