Respuesta :

caylus
Hello,

1)
I suppose we must write correcty and corrects the mistakes!

f(x)=x+3

y=x+3
->x=y+3 exchange x and y
->y=x-3 solve for y

f^(-1)(x)=x-3=g(x)

Here are the mistakes:

(fog)(x)=f(g(x))=f(x-3)=(x-3)+3=x and not -x !

(gof)(x)=g(f(x))=g(x+3)=(x+3)-3=x and not -x!!

f(x) and g(x) ARE inverses each another.


2)
f(x)=1/(4x) x≠0

y=1/(4x)
->x=1/(4y)
->y=1/(4x)
g(x)=f^(-1)(x)=1/(4x)=f(x), x≠0

[tex](gof)(x)=(fog)(x)=f(g(x))=f(\dfrac{1}{4x})= \dfrac{1}{4* \dfrac{1}{4x}} = \dfrac{1}{ \frac{1}{x}} =1*x=x [/tex]







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