- #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?
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SUMMARY
The correct inverse of the function \( f(x) = \frac{1}{2x+6} \) is \( f^{-1}(x) = \frac{1}{2x} - 3 \). To verify that these functions are indeed inverses, one must compute the composite functions \( f(f^{-1}(x)) \) and \( f^{-1}(f(x)) \). The calculation shows that \( f(f^{-1}(x)) = x \), confirming the relationship. The discussion highlights the importance of understanding composite functions in verifying inverse functions.
PREREQUISITES- Understanding of inverse functions and their properties
- Knowledge of composite functions
- Familiarity with algebraic manipulation involving fractions
- Basic skills in function notation
- Practice finding inverses of various functions, including rational functions
- Learn how to compute composite functions in detail
- Explore the concept of function transformations
- Study the properties of functions and their graphs
Students, educators, and anyone studying algebra, particularly those focused on functions and their inverses.
- #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
- 13,404
- 4,183
Okay,i'll be a nice guy... 
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}=<br /> \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.
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}=<br /> \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.
- #7
check
- 144
- 0
Yeesh. I saw the title of this thread and was very confused for a second. lol