Inverse Functions: Determine f(f-1(x)) and f-1(f(x))

[aisha]

Determine f(f-1(x)) and f-1(f(x)) of the following functions. What do you notice?

a.) f(x) = 7x-3
b.)f(x) = 2/3 x+2 (there were no brackets the x+2 is just beside the fraction)

Rule: If f(g(x)) = x and f(f(x)) = x, then f(x) and g(x) are inverse of each other.

This is my exercise question I don't understand what to do I get the rule but how am getting an answer of x? I have the solutions also, but I don't know what to do with the numbers am I solving for something?

 

[learningphysics]

Determine f(f-1(x)) and f-1(f(x)) of the following functions. What do you notice?

a.) f(x) = 7x-3
b.)f(x) = 2/3 x+2 (there were no brackets the x+2 is just beside the fraction)

Rule: If f(g(x)) = x and f(f(x)) = x, then f(x) and g(x) are inverse of each other.

This is my exercise question I don't understand what to do I get the rule but how am getting an answer of x? I have the solutions also, but I don't know what to do with the numbers am I solving for something?

Did you work out f(f-1(x)) and f-1(f(x))? Which numbers are you referring too?

 

[Gokul43201]

Determine f(f-1(x)) and f-1(f(x)) of the following functions. What do you notice?

a.) f(x) = 7x-3
b.)f(x) = 2/3 x+2 (there were no brackets the x+2 is just beside the fraction)

Rule: If f(g(x)) = x and f(f(x)) = x, then f(x) and g(x) are inverse of each other.

This is my exercise question I don't understand what to do I get the rule but how am getting an answer of x? I have the solutions also, but I don't know what to do with the numbers am I solving for something?

Correction : If f(g(x)) = x and g(f(x)) = x, then f(x) and g(x) are inverse of each other, or g = f^-1. That should give you the answer.

 

[aisha]

Correction : If f(g(x)) = x and g(f(x)) = x, then f(x) and g(x) are inverse of each other, or g = f^-1. That should give you the answer.

I understand that but I don't know what to do? lol Omg I don't even know what to do with the numbers given. Show me please The solutions are both are inverses of each other.

 

[Skomatth]

Well you have to find the inverse of f(x) first. Then show $$f(f^{-1}(x)) = x$$. Can you find an inverse? An inverse g(x) is the graph of f(x) reflected across the line y=x .

For example:

$$f(x) = 1/x + 5$$

Defined implicitly the inverse is therefore:

$$x= 1/y + 5$$

where

$$<br /> y= f^{-1}$$

solve for y to get the inverse defined explicitly:

$$y = \frac{1}{x-5}$$

 

[aisha]

Yes I know how to find the inverse that much makes sense but what does the stuff in these brackets mean? f(f^(-1)(x))=x what do I do after I find the inverse? multiply f? by the inverse? Thats the part I don't get how to =x

 

[Skomatth]

$$f(f^{-1}(x)) = x$$
is a composite function . Do you know about these? BTW what math level are you in, what circumstances are you in? It seems like you have an awful lot of questions that could be answered by reading the textbook.

 

[learningphysics]

Suppose you have two different functions:
f(x)=3x+1
g(x)=x^2+1

You get a composite function f(g(x)) by taking g(x) and plugging into f(x) where x is:
So f(g(x))=3(x^2+1)+1

And g(f(x))=(3x+1)^2+1

Now let's look at a function and its inverse:
f(x)=3x+1
f-1(x)=(x-1)/3

I form the composite of f(x) with f-1(x)
f(f-1(x))=3[(x-1)/3]+1=(x-1)+1=x


f-1(f(x))=[(3x+1)-1]/3=(3x)/3=x

So we find f(f-1(x))=f-1(f(x))=x

This is always true for any relation and its inverse. So you should notice the same thing in your exercise.

 

[aisha]

Suppose you have two different functions:
f(x)=3x+1
g(x)=x^2+1

You get a composite function f(g(x)) by taking g(x) and plugging into f(x) where x is:
So f(g(x))=3(x^2+1)+1

And g(f(x))=(3x+1)^2+1

Now let's look at a function and its inverse:
f(x)=3x+1
f-1(x)=(x-1)/3

I form the composite of f(x) with f-1(x)
f(f-1(x))=3[(x-1)/3]+1=(x-1)+1=x


f-1(f(x))=[(3x+1)-1]/3=(3x)/3=x

So we find f(f-1(x))=f-1(f(x))=x

This is always true for any relation and its inverse. So you should notice the same thing in your exercise.

WOW THANKS SOOOO MUCH what a great explanation and example! So easy to understand, Thankyou.