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 Mar 16, 2020
 #1
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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,

\(a=6\)

\(b=-2\)

.
 Mar 16, 2020
 #2
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+1

thank you

TacoBell  Mar 18, 2020

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