show that f and g are inverse function of each other?

Question

1.) f(x)=3x : g(x) = x/3
2.)f(x) = 2-3x ; g(x) = 2 - x/3
3.)f(x) = the square root of x^2 + 1 ; g(x) = the square root x^2 - 1

.

Answer 1

f and g are inverses if f(g(x)) =x

1. f(g(x)) = 3g(x) = 3. x/3 =x

2. f(g(x)) = 2 - 3.g(x) = 2 - 3. ((2-x)/3)= 2 - (2-x) = x
Make sure you have brackets round the 2-x

3. f(g(x)) = sqrt( g(x)^2 +1)
= sqrt( (sqrt(x^2 - 1))^2+1)
=sqrt( x^2 -1 +1)
sqrt(x^2)
=x

so in each case f and g are inverses

Answer 2

Suppose f(a)=b . We call g(x) the inverse of f(x) , if for every b given by a "f(a)=b" statement , we get g(b)=a . So , in other words , g( f(a) )=g(b)=a . Or , g(f(x))=x . Simply replace the x in g(x) , with f(x) , and if g(f(x))= x , then f(x) is the inverse of g(x) (and vice-versa) . For example : g(x)=sqrt(x*x-1) , f(x)=sqrt(x*x+1) . Lets see what happens when we replace the x in g(x) with f(x) .
g(f(x))=sqrt(f(x)^2-1)=sqrt( (x*x+1) -1)=x .The others are solved this way to .

Answer 3

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Answer 4

Easy to prove.

IF
f( g(x) ) = x

Then they are inverses