Investigating Properties of Inverse Functions

[msimard8]

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

 

[topsquark]

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:
[tex]y=3 \pm \sqrt{\frac{x-1}{4}}[/tex]

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

 

[msimard8]

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