math

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f(x) = 2x+1
g(x)= 3x-2
find:
a) f -1(x)
b) f { g(x) }
c) g{ f(x) }
d) 2f { g(x) } = 3g { f(x) }

 

[Deleted member 4993]

f(x) = 2x+1
g(x)= 3x-2
find:
a) f -1(x)
b) f { g(x) }
c) g{ f(x) }
d) 2f { g(x) } = 3g { f(x) }

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[Dr.Peterson]

f(x) = 2x+1
g(x)= 3x-2
find:
a) f -1(x)
b) f { g(x) }
c) g{ f(x) }
d) 2f { g(x) } = 3g { f(x) }

First, you need to clarify what these mean. We don't normally use braces {} in writing function notation, and "f-1" is not meaningful. And the last is an equation, not an expression; I suspect a typo.

I would guess that these mean

a) f^-1(x), the inverse function
b) f(g(x)), a composition
c) g(f(x)), another composition
d) 2f(g(x)) + 3g(f(x)), a sum of multiples of the compositions

Now, do you know what composition means? That's a place to start. To find f(g(x)), you substitute the value of g(x) for x in the definition of f(x).

 

[HallsofIvy]

To find \(\displaystyle f^{-1}\) when f(x)= 2x+ 1 write y= 2x+ 1 and swap x and y (because if \(\displaystyle y=f(x)\) then \(\displaystyle x= f^{-1}(y)\)) so x= 2y+ 1 and solve for y.

For the "compositions" f(g(x)) or g(f(x)) you do basically what they say: first find g(x) then find f of that value. With f(x)+ 2x+ 1 and g(x)= 3x- 2, f(g(x))= 2(3x- 2)+1 and g(f(x))= 3(2x+ 1)- 2.[/tex][/tex]