Hi would i be correct in saying that the image set for the following is correct?
Closed Interval [-3,9] ?
This is a parabola, yes? It just has been shifted horizontally and vertically, as well as stretched. Still just a parabola.
I don't see why crossing the x-axis would cause any problems. As long as the domain remains real, then all values of x are part of the pre-image set and correspond to another value after passing through F.
I'm trying to make sure I'm not missing some simple idea, but I can't think of it. Take a look at the Wiki page on images:
Image (mathematics) - Wikipedia, the free encyclopedia
I am just trying to be careful because it's easy for me to think you mean one thing when you don't. I don't want to give you wrong advice.
Is that the whole problem? What did your instructor tell you an image was?
This is how I view this problem:
Like I said this is a parabola, so it has a domain of all real numbers. The image of F is the set of outputs over the domain of F, which is all read numbers. So now all we have to do is figure out what y-values make up F. A parabola that opens upwards has it's lowest point at its vertex. Above the vertex the parabola keeps increasing to positive infinity. Agree?
I think the image set over the domain of F is from the vertex to positive infinity. Does this make sense?
Is this similar to what you've been hearing in class? If not, can you give me some examples of the info in your book. Maybe it's asking something other than the image of the entire function. It could be asking for an image of some subset.
You are viewing the problem correctly, it was just the way i presented it.
So taking your advice to agree; i agree the parabola opening upwards can continue to positive infinatly. So yes that makes sense.
So from your advice is it a half open interval, [a,b) and so it would be [-4,∞)?