- #1

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- 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?