Hi, I need help with Inverse Functions and thank you

Question

State if given functions are inverse

1. g(x)=4-3/2x

f(x)=1/2x+3/2

2.g(n)=-12-2n/3

f(n)=-5+6n/5

Find the inverse of each function

1. h(x)=³√x-3
2. g(x)=1/x-2
3. h(x)=2x³+3

Find the inverse of each function and graph the function

f(x)=-1-1/5x

Answer 1

text has been corrected to clarify my assumptions where parentheses go in terms of algebra operations

for the first two, the answer can be found in two ways. 1. take f(g(x)), or f(g(n)), if f and g are inverse

f(g(x))=x, if not then they are not inverse, or 2. pick f(x) and find inverse f^-1(x) and see if it is g(x)

let's try 2., I do it this way

f(x)=(1/2)*x+3/2

x=(1/2)*f^-1(x)+3/2, range is now domain, domain is now range

f^-1(x)=2*(x-3/2) which is not equal to g(x)

let's check my work, does f(f^-1(x))=x

1/2*(2*(x-3/2)+3/2=x ,,,check

h(x)=(³√x)-3,,,solve x=(³√h^-1(x))-3 for h^-1(x), notice domain and range have no limitations

g(x)=(1/x)-2 here there is a limitation, x=0 is not allowed as a g(x) domain value and g(x)=-2 is a horizontal asymptote.. For g^-1(x), the value 0 will be horizonal asymptote and x= -2 will be a disallowed value in the domain of g^-1(x),,,,, solve x=[1/g^-1(x)]-2,,,,,g^-1(x)=1/(x+2)

h(x)=2x³+3,,,,,,(x-3)/2=[h^-1(x)]³ take cubed root for h^-1(x)

f(x)=-1-1/5x,,,,here again x=0 not valid for f(x) and horizonatal asymptote is -1

x=-1-1/[5*f^-1(x)]

f^-1(x)= -1/[5*(x+1)]

see that graphs are symmetric with respect to line y=x (it isn't very apparent in this case but they are)

symmetric around y=x means pair (a,b) from f(x) is matched by pair (b,a) for f^-1(x)

Answer 2

Inverse of h(x)=³√x-3 is h^-1(x)= x^3 + 3

Inverse of g(x)= 1/x - 2 is g^-1(x)= 1/(x + 2)

Inverse of h(x)= 2x³+3 is h^-1(x)= [³√4(x - 3)]/2