Question #151211

Find the inverse, g−1(t), when g(t) is given by 1/5t+10.

g−1(t)=

Expert's answer

Solution:

g(t)=15t+10g(t)= \dfrac{1}{5t+10}


let`s call g(t) as y:


y=15t+10y= \dfrac{1}{5t+10} then (1y)−10=5t( \dfrac{1}{y}) -10=5t


t=15y−2t = \dfrac{1}{5y}-2


Answer:

g(t)−1=15t−2g(t)^{-1} = \dfrac{1}{5t}-2 is inverse.



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