# Composite and Inverse Functions Problem

• AN630078
A function is specified by its domain, its range, and the rule that takes each value of the domain to the corresponding value in the range.In this case, the domain of f(x) is ##x > 0##, the range of f(x) is ##(-\infty, \infty)##, and the rule is ##f(x) = \log_{10}(x)##.In this case, the domain of f^-1(x) is ##(-\infty, \infty)##, the range of f^-1(x) is ##x > 0##, and the rule is ##f^{-1}(x) = 10^x##.

#### AN630078

Homework Statement
Hello, I have a question on composite and inverse functions which I have attempted to answer fully but I am little unsure of my solutions, especially my comments in part a on the functions of f and g, I do not know whether I am required to add anything further. I would be very grateful of any advice or evaluation of my methodology. 1. a If f:x→2x+1 and g:x→1/2(x-1)
Find fg(x) and gf(x).
What can you say about the functions f and g

b.If f:x→log 10 (x)
How can the graph of f^-1 be found from f? What are the ranges and domains of f and f^-1?
Relevant Equations
f(x)
1. a.
fg(x)=2(1/2(x-1))+1
fg(x)=2(x/2-1/2)+1
fg(x)=x-1+1
fg(x)=x

gf(x)=1/2((2x+1)-1)
gf(x)=1/2(2x+1-1)
gf(x)=x+1/2-1/2
gf(x)=x

The functions functions f(x) and g(x) are inverses of each other. This can be demonstarted by
f(x)=2x+1
y=2x+1
x=2y+1
x-1=2y
(x-1)/2=y
Thus, y=1/2(x-1) = g(x)
And, g(x)=1/2(x-1)
y=1/2(x-1)
x=1/2(y-1)
2x=y-1
2x-1=y=f(x)
Besides from stating that the functions are inverses of each other, is there anything else that I could comment about the two functions which would be relevant here?

b. f^-1 is the inverse function of f(x), therefore, the graph of f^-1 can be found by reflecting the graph of f(x) in the line y=x.
The inverse function of f(x) can be found by rewriting f in terms of y, swapping x and y and then making y the subject.
i.e. y=log 10 (x)
x=log10(y)
If log a(b)=c then b=a^c
Thus, y=10^x
f^-1=10^x

The domain of f(x) would be x>0, since the graph of f(x) reaches the y-axis where x=0 but it is never less than zero, x must be positive.
The range of f(x) would be f(x)ϵ ℝ, because the values of y increase to infinity and decrease to negative infinity, so the range is any value of y.

The domain of f^-1(x) would be xϵ ℝ because the function has no undefined points or domain constraints.
The range of f^-1(x) would be f^-1(x)>0 since the function never falls below the x axis, therefore y must be positive.

I am not sure if the domains and ranges would be correct, however, the reversal of domains and ranges for the function is true for all inverses so perhaps I have not made an error. I wondered if the range of f(x) would be f(x)ϵ ℝ because the graph is a vertical asymptote to the y-axis, I did not know if this imposed a restriction on the range?

AN630078 said:
Homework Statement:: Hello, I have a question on composite and inverse functions which I have attempted to answer fully but I am little unsure of my solutions, especially my comments in part a on the functions of f and g, I do not know whether I am required to add anything further. I would be very grateful of any advice or evaluation of my methodology. 1. a If f:x→2x+1 and g:x→1/2(x-1)
Find fg(x) and gf(x).
What can you say about the functions f and g

b.If f:x→log 10 (x)
How can the graph of f^-1 be found from f? What are the ranges and domains of f and f^-1?
Relevant Equations:: f(x)

1. a.
fg(x)=2(1/2(x-1))+1
fg(x)=2(x/2-1/2)+1
fg(x)=x-1+1
fg(x)=x

gf(x)=1/2((2x+1)-1)
gf(x)=1/2(2x+1-1)
gf(x)=x+1/2-1/2
gf(x)=x

The functions functions f(x) and g(x) are inverses of each other. This can be demonstarted by
f(x)=2x+1
y=2x+1
x=2y+1
x-1=2y
(x-1)/2=y
Thus, y=1/2(x-1) = g(x)
And, g(x)=1/2(x-1)
y=1/2(x-1)
x=1/2(y-1)
2x=y-1
2x-1=y=f(x)
Besides from stating that the functions are inverses of each other, is there anything else that I could comment about the two functions which would be relevant here?
I can't think of anything else you could add, other than to note the domain and range of each function.
I would add, though, that your notation could be confused by some. When you write fg(x), that usually means the product of the two functions, not the composition of f and g.
Instead of fg(x), you should write f(g(x)).
AN630078 said:
b. f^-1 is the inverse function of f(x), therefore, the graph of f^-1 can be found by reflecting the graph of f(x) in the line y=x.
The inverse function of f(x) can be found by rewriting f in terms of y, swapping x and y and then making y the subject.
i.e. y=log 10 (x)
x=log10(y)
If log a(b)=c then b=a^c
Thus, y=10^x
f^-1=10^x
This work would look a lot nicer if you used LaTeX.
The last line above could be written as ##f^{-1}(x) = 10^x##.
See our tutorial in the link at the lower left corner of the page.
AN630078 said:
The domain of f(x) would be x>0, since the graph of f(x) reaches the y-axis where x=0 but it is never less than zero, x must be positive.
The graph of ##f(x) = \log_{10}(x)## never reaches the y-axis. The graph is asymptotic to the y-axis.
AN630078 said:
The range of f(x) would be f(x)ϵ ℝ, because the values of y increase to infinity and decrease to negative infinity, so the range is any value of y.

The domain of f^-1(x) would be xϵ ℝ because the function has no undefined points or domain constraints.
The range of f^-1(x) would be f^-1(x)>0 since the function never falls below the x axis, therefore y must be positive.

I am not sure if the domains and ranges would be correct, however, the reversal of domains and ranges for the function is true for all inverses so perhaps I have not made an error. I wondered if the range of f(x) would be f(x)ϵ ℝ because the graph is a vertical asymptote to the y-axis, I did not know if this imposed a restriction on the range?
Your domains and ranges are correct here.

Mark44 said:
I can't think of anything else you could add, other than to note the domain and range of each function.
I would add, though, that your notation could be confused by some. When you write fg(x), that usually means the product of the two functions, not the composition of f and g.
Instead of fg(x), you should write f(g(x)).
This work would look a lot nicer if you used LaTeX.
The last line above could be written as ##f^{-1}(x) = 10^x##.
See our tutorial in the link at the lower left corner of the page.
The graph of ##f(x) = \log_{10}(x)## never reaches the y-axis. The graph is asymptotic to the y-axis.
Your domains and ranges are correct here.
Thank you very much for your informative reply. Right, I may add the domain and range of each function for f(x) and g(x) in part a. Really, I used the notation fg(x) and gf(x) because this was what the question specified and additionally what I had been taught in my textbook, which explained that one should simplify f(g(x)) to fg(x) etc. However, I will use the notation you have suggested if this is correct when defining the product of two functions.
Yes, I agree regarding LaTeX and I have been trying to teach myself, however, sometimes I have been unsuccessful when trying to implement it in my postings so I have abandoned my attempts. Although, I am determined to learn properly and will make use of the LaTeX guide thank you for the suggestion.
Right, sorry my mistake. Even though the graph of f(x) is asymptotic to the y-axis, would the domain still be x>0, since x must be positive as the function is never reaches the y-axis, so it is never less than 0.