Here is the question (written)
The functions \(g\) and \(h\) are such that
\(g(x)=3x+2\) and \(h(x)=ax+b\)
Where \(a\) and \(b\) are constants.
\(h(4)=22\)
\(g^{-1}(14)=h(1)\)
Answer:
First, let's find \(g^{-1}(x)\) since we don't know much about \(h(x)\)
let \(g(x)=y\)
\(y=3x+2\) (Switch variables, finding inverse method)
\(x=3y+2\) Solve for y and that is \(g^{-1}(x)\)
\(y=\frac{x-2}{3}\)
\(g^{-1}(x)\)\(=\frac{x-2}{3}\)
Ok given \(g^{-1}(14)=h(1)\) then
\(g^{-1}(14)=\frac{14-2}{3}=4\)
Thus \(h(1)=4\)
Now we know that \(h(x)=ax+b\)
\(h(1)=a+b\) then \(a+b=4\) (Since \(h(1)=4\))
Also given \(h(4)=22\) then
\(4a+b=22\)
solve the system of two equations
\(a+b=4\) (1)
\(4a+b=22\) (2)
Multiply (1) by -1 and add it to (2)
\(-a-b=-4\)
\(4a+b=22\)
\(3a=18\)
\(a=6\)
\(a+b=4\) then \(6+b=4\) then \(b=-2\)
Thus,
\(a=6\)
\(b=-2\)
.