1. If g(x) is the inverse of f(x) then which of the following statements is true?

[peterjackson5123]

I f(g(x)) = x


II The domain of f(x) is the same as the range of g(x)


III The graph of f(x) and the graph of g(x) are symmetric about the line y = x

1. I and II only b. I and III only c. II and III only d. All of them are true

 

[Deleted member 4993]

I f(g(x)) = x


II The domain of f(x) is the same as the range of g(x)


III The graph of f(x) and the graph of g(x) are symmetric about the line y = x

1. I and II only b. I and III only c. II and III only d. All of them are true

Please show us what you have tried and exactly where you are stuck.

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[Dr.Peterson]

1. If g(x) is the inverse of f(x) then which of the following statements is true?
I f(g(x)) = x
II The domain of f(x) is the same as the range of g(x)
III The graph of f(x) and the graph of g(x) are symmetric about the line y = x

1. I and II only b. I and III only c. II and III only d. All of them are true

Give it a try:

If g is the inverse of f, is f(g(x)) = x for all x?​

​

If g is the inverse of f, do f and g necessarily have the same domain?​

​

If g is the inverse of f, is the graph of one necessarily the reflection of the other in the line y=x?​

By the way, it's not a good idea to put part of the problem only in the title, because once a thread gets more than a page long, people will easily fail to read the title as part of your question. I've done that too many times.

 

[HallsofIvy]

Give it a try:

If g is the inverse of f, is f(g(x)) = x for all x?​

​

If g is the inverse of f, do f and g necessarily have the same domain?​

​

The question does not ask if they have the same domain. It is asking if the domain of one is the range of the other.​

​

If g is the inverse of f, is the graph of one necessarily the reflection of the other in the line y=x?

By the way, it's not a good idea to put part of the problem only in the title, because once a thread gets more than a page long, people will easily fail to read the title as part of your question. I've done that too many times.