# Understanding inverse functions

#### Casio

##### Member
I have a graph f(x) = 3(x + 1)^2 - 12 , I have sketched this graph (Not shown) hand it is a parabola with a y- intercept at - 9. the vertex being - 12.

The image set is a closed interval {- 12, infinity} Sorry no square brackets and no sign for infinity.

I am asked to explain why the function of f does not have an inverse?

Given that the graph is a parabola I would have thought that the graph did have an inverse?

The domain {-3 < x < 1}

I am missing something here in the understanding if anyone can advise It would be much appreciated.

Sorry I didn't include a sketch the paint package won't allow me to draw a curve #### CaptainBlack

##### Well-known member
I have a graph f(x) = 3(x + 1)^2 - 12 , I have sketched this graph (Not shown) hand it is a parabola with a y- intercept at - 9. the vertex being - 12.

The image set is a closed interval {- 12, infinity} Sorry no square brackets and no sign for infinity.

I am asked to explain why the function of f does not have an inverse?

Given that the graph is a parabola I would have thought that the graph did have an inverse?

The domain {-3 < x < 1}

I am missing something here in the understanding if anyone can advise It would be much appreciated.

Sorry I didn't include a sketch the paint package won't allow me to draw a curve A function is single valued, so for your f(x) to have an inverse for every y in its range there must be one and only one x in its domain such that y=f(x).

That is every horizontal line y=u that cuts the curve y=f(x) cuts it in one point only.

CB

• Casio

#### Casio

##### Member
A function is single valued, so for your f(x) to have an inverse for every y in its range there must be one and only one x in its domain such that y=f(x).

That is every horizontal line y=u that cuts the curve y=f(x) cuts it in one point only.

CB
Thanks CB