- #1
aisha
- 584
- 0
the inverse of 1/(2x+6) x cannot=-3
is -3 + 1/2x ? Is this correct?
is -3 + 1/2x ? Is this correct?
Click For Summary
Homework Help Overview
The discussion revolves around finding the inverse of the function \( f(x) = \frac{1}{2x+6} \) and whether the proposed inverse \( -3 + \frac{1}{2}x \) is correct. Participants are exploring the concept of function inverses and the process of verifying them through composite functions.
Discussion Character
- Conceptual clarification, Mathematical reasoning, Problem interpretation
Approaches and Questions Raised
- Participants are questioning the correctness of the proposed inverse and discussing the conditions under which the original function is defined. There are requests for clarification on how to verify inverses using composite functions, with some participants expressing confusion about handling fractions in this context.
Discussion Status
Some participants have provided hints and partial guidance on how to check the inverses using composite functions. However, there is no explicit consensus on the correctness of the proposed inverse, and multiple interpretations of the problem are being explored.
Contextual Notes
There are constraints mentioned regarding the values of \( x \) that can be used, specifically that \( x \) cannot equal -3 or 0, which may affect the discussion on the function's domain and its inverse.
- #2
Xasuke
- 13
- 0
yep.
oh, x =\= 0 ..
oh, x =\= 0 ..
- #3
aisha
- 584
- 0
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.
g(f(x))=x and f(g(x))=x then both are inverse of each other.
- #4
daster
Hint:
f(a)=b
f-1(b)=a
f(a)=b
f-1(b)=a
- #5
aisha
- 584
- 0
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!
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!
Last edited:
- #6
dextercioby
Science Advisor
- 13,410
- 4,207
Okay,i'll be a nice guy... 
[tex]y=\frac{1}{2x+6}\Rightarrow 2x+6=\frac{1}{y}\Rightarrow x=\frac{1}{2y}-3[/tex]
So the function and the inverses are:
[tex]f(x)=\frac{1}{2x+6};f^{-1}(x)=\frac{1}{2x}-3[/tex]
U wan to compute 2 functions:
[tex]f(f^{-1}(x))=...??;f^{-1}(f(x))=...??[/tex]
I'll take the first and leave you with the second.
[tex]f(f^{-1}(x))=\frac{1}{2f^{-1}(x)+6}=\frac{1}{2(\frac{1}{2x}-3)+6}=<br /> \frac{1}{\frac{1}{x}-6+6}=x[/tex]
I hope u saw the pattern and you won't have any trouble with the second.
Daniel.
[tex]y=\frac{1}{2x+6}\Rightarrow 2x+6=\frac{1}{y}\Rightarrow x=\frac{1}{2y}-3[/tex]
So the function and the inverses are:
[tex]f(x)=\frac{1}{2x+6};f^{-1}(x)=\frac{1}{2x}-3[/tex]
U wan to compute 2 functions:
[tex]f(f^{-1}(x))=...??;f^{-1}(f(x))=...??[/tex]
I'll take the first and leave you with the second.
[tex]f(f^{-1}(x))=\frac{1}{2f^{-1}(x)+6}=\frac{1}{2(\frac{1}{2x}-3)+6}=<br /> \frac{1}{\frac{1}{x}-6+6}=x[/tex]
I hope u saw the pattern and you won't have any trouble with the second.
Daniel.
- #7
check
- 146
- 0
Yeesh. I saw the title of this thread and was very confused for a second. lol