Answer:
0
Step-by-step explanation:
We are given that
\(G(x)=3x+1\)
We have to find the value of \(G^{-1}(1)\)
To find the value of \(G^{-1}(1)\) we will find first \(G^{-1}(x)\)
Let \(y=G(x)=3x+1\)
\(3x=y-1\)
\(x=\frac{1}{3}(y-1)\)
Now, replace x by y and y replace by x
Then, we get
\(y=\frac{1}{3}(x-1)\)
Now, substitute \(G^{-1}(x)=y\)
\(G^{-1}(x)=\frac{1}{3}(x-1)\)
Now, substitute x=1
Then, we get
\(G^{-1}(1)=\frac{1}{3}(1-1)=0\)
Hence, the value of \(G^{-1}(1)=0\)
