Investigating Properties of Inverse Functions

AI Thread Summary
The discussion centers around the properties of the function g(x) = 4(x-3)^2 + 1 and its inverse. The user successfully completed parts a-d but struggles with part e, which questions whether the inverse is a function given a specific domain. The conclusion is that the inverse is not a function because it produces two different y-values for the same x-value (specifically at x=5). The algebraic justification involves evaluating the inverse function at x=5, yielding two results, which confirms the violation of the function definition. The conversation highlights the importance of understanding domain restrictions for inverse functions.
msimard8
Messages
58
Reaction score
0
Ok here is teh question, there is parts a-e, i have a-d answered correctly, but am having trouble on e.

Q: A function g is g(x)=4(x-3)^2 + 1

a) Graph g and the inverse of g. (Already completed)

b) At what points do g and the inverse of g intersect. (completed)

c) Determine an equation that defines the inverse of g.
the answer is y=3 +/- (root of (x-1/4) )

d) State restrictions on the domain or range of g so that its inverse is a function.

i got x is greater or equal to 3 and x is less than or equal to 3


then finally

e) Suppose teh domain of g is {x|2 < or equal to x < or equal to 5, XER}
Would the inverse be a function? Justify your answer.

Ok. The answers say the inverse is not a function because the

inverse of g (5) = 2 and
inverse of g (5) =4

i guess that makes sense because there is two y coordinates for that x. My question how do you algabraically solve that. Help would be appreciated. Thanks
 
Physics news on Phys.org
msimard8 said:
Ok here is teh question, there is parts a-e, i have a-d answered correctly, but am having trouble on e.

Q: A function g is g(x)=4(x-3)^2 + 1

a) Graph g and the inverse of g. (Already completed)

b) At what points do g and the inverse of g intersect. (completed)

c) Determine an equation that defines the inverse of g.
the answer is y=3 +/- (root of (x-1/4) )

d) State restrictions on the domain or range of g so that its inverse is a function.

i got x is greater or equal to 3 and x is less than or equal to 3


then finally

e) Suppose teh domain of g is {x|2 < or equal to x < or equal to 5, XER}
Would the inverse be a function? Justify your answer.

Ok. The answers say the inverse is not a function because the

inverse of g (5) = 2 and
inverse of g (5) =4

i guess that makes sense because there is two y coordinates for that x. My question how do you algabraically solve that. Help would be appreciated. Thanks

Simple. Hopefully your inverse function was supposed to be:
y=3 \pm \sqrt{\frac{x-1}{4}}

So using the + sign: y(5) = 3 + 1 = 4.
Using the - sign: y(5) = 3 -1 = 2.

Since we may use either sign, that means that y(5) is double valued, which violates the definition of a function.

-Dan
 
Last edited:
thanks

Thanks a lot Dan.

One question. Why are you using f(5). Why did you chose the number 5. Is it because it is the highest possible x coordinate.

Thanks
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Replies
3
Views
2K
Replies
15
Views
2K
Replies
6
Views
2K
Replies
3
Views
1K
Replies
10
Views
2K
Replies
19
Views
3K
Replies
2
Views
4K
Back
Top