# For the functions: f(x) =1/x^2 g(x)= 7x-6 h(x)= 1-2x, Find. a) f o g(x) b) g o f o (x)c) Given f(x) = 3x + 5 h(x)=3x^(2) + 3x + 2 Find a function g(x) so that f o g(x) = h(x)

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`f@g(x)=f(g(x))`

`f(g(x))=1-(g(x))^2`

`=1-(7x-6)^2`

`=1-(49x^2-84x+36)`

`=-49x^2+84x-35`

`g@f(x)=g(f(x))`

`g(f(x))=7f(x)-6`

`=7(1-x^2)-6`

`=7-7x^2-6`

`=1-7x^2`

`f(x)=3x+5`

Think of this as:

f is a rule, or sort of a procedure.

You start with x, multiply by 3, then add 5

but you don't have to start with x

you can start with a number, for example, 2

then f, our rule, does the following to 2:

start with 2, multiply by 3, then add 5

so f(2)=3(2)+5=11

or you could start with something more complicated-looking, like g(x)

but still you are just doing f to the thing "g(x)":

start with g(x), multiply by 3, then add 5

so f(g(x))=3g(x)+5

so if we want `f@g(x)=h(x)`

really we want:

`3g(x)+5=3x^2+3x+2`

subtract 5 from both sides:

`3g(x)=3x^2+3x-3`

divide both sides by 3:

`g(x)=x^2+x-1`

So that is the function g(x) such that f(g(x))=h(x)