Is -3 + 1/2x the Correct Inverse of 1/(2x+6)?

[aisha]

the inverse of 1/(2x+6) x cannot=-3

is -3 + 1/2x ? Is this correct?

 

[Xasuke]

yep.
oh, x =\= 0 ..

 

[aisha]

can someone show me how to check my self my using composites

g(f(x))=x and f(g(x))=x then both are inverse of each other.

 

[daster]

Hint:
f(a)=b
f^-1(b)=a

 

[aisha]

I need more than that I don't need a hint I need to see how its done because if my answer is right then how come I don't know how to write out the composite function so that f(g(x)) and g(f(x)) both equal x?

Can someone please show me how its done?

I know how to do it I am able to do it for f(x)=x^2 and g(x)=x+1 but in my question the fractions are throwing me off I don't know how to write it out, someone please help me! PLEASE! :uhh:

 

[dextercioby]

Okay,i'll be a nice guy... :tongue2:
$$y=\frac{1}{2x+6}\Rightarrow 2x+6=\frac{1}{y}\Rightarrow x=\frac{1}{2y}-3$$
So the function and the inverses are:
$$f(x)=\frac{1}{2x+6};f^{-1}(x)=\frac{1}{2x}-3$$
U wan to compute 2 functions:
$$f(f^{-1}(x))=...??;f^{-1}(f(x))=...??$$
I'll take the first and leave you with the second.
$$f(f^{-1}(x))=\frac{1}{2f^{-1}(x)+6}=\frac{1}{2(\frac{1}{2x}-3)+6}= \frac{1}{\frac{1}{x}-6+6}=x$$

I hope u saw the pattern and you won't have any trouble with the second.

Daniel.

 

[check]

Yeesh. I saw the title of this thread and was very confused for a second. lol