- #1
AN630078
- 242
- 25
- 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?
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?