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# i found it and have no idea how to answer it

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https://vle.mathswatch.co.uk/images/questions/question16870.png

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.

$$h(4)=22$$

$$g^{-1}(14)=h(1)$$

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,

$$a=6$$

$$b=-2$$

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Mar 16, 2020
#2
+349
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thank you

TacoBell  Mar 18, 2020