Hi would i be correct in saying that the image set for the following is correct?

$\displaystyle f(x)= \frac{1}{9}(x - 3)^2 - 4$

Closed Interval [-3,9] ?

Thanks

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- Oct 25th 2009, 07:49 AMADYimage set
**Hi would i be correct in saying that the image set for the following is correct?**

$\displaystyle f(x)= \frac{1}{9}(x - 3)^2 - 4$

**Closed Interval [-3,9] ?**

Thanks - Oct 25th 2009, 08:36 AMJameson
- Oct 25th 2009, 08:39 AMADY
Hi Jameson,

You're right, but i thought it would be restrictive due to the points crossing the x axis at -3 and 9. What do you suggest, that there could be multiple outputs and should be so restrictive?

Any help wouldnt go a miss please - Oct 25th 2009, 08:54 AMJameson
I'm not quite following you.

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 - Oct 25th 2009, 09:26 AMADY
Sorry for being confusing...i'm probably just confused my self, so could f be closed and infinate?

- Oct 25th 2009, 09:36 AMJameson
Don't be sorry!

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. - Oct 25th 2009, 10:09 AMADY
Well, firstly you are very sincere and i'm humbled by that as you are the admin of the forum!

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,∞)? - Oct 25th 2009, 11:15 AMJameson
That looks great to me :)

- Oct 26th 2009, 12:09 PMADY
Just to confirm, are my brackets ok?

- Oct 26th 2009, 02:59 PMJameson
Yes. The vertex is at (3,4) so you should include 4. The reason I was hesitant about this question is that usually you would just be asked for the "range", not the "image". Image is a more generalized idea than range that usually isn't discussed until Linear Algebra in 3rd year college math.