g(t)=15t+10g(t)= \dfrac{1}{5t+10}g(t)=5t+101
let`s call g(t) as y:
y=15t+10y= \dfrac{1}{5t+10}y=5t+101 then (1y)−10=5t( \dfrac{1}{y}) -10=5t(y1)−10=5t
t=15y−2t = \dfrac{1}{5y}-2t=5y1−2
g(t)−1=15t−2g(t)^{-1} = \dfrac{1}{5t}-2g(t)−1=5t1−2 is inverse.
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