# Image set of a function?

• Nov 19th 2011, 02:36 AM
Orlando
Image set of a function?
How do u find the image set for a function with no domain?
• Nov 19th 2011, 02:40 AM
Deveno
Re: Image set of a function?
you don't. a function without a domain is an empty function, with a "blank" graph.

put it this way, how many pairs (x,f(x)) can occur in such a function, when there is no possible "first coordinate"?
• Nov 19th 2011, 02:47 AM
Orlando
Re: Image set of a function?
I thought it was the lowest possible value for y and the highest possible vaule foe y??
• Nov 19th 2011, 03:51 AM
Plato
Re: Image set of a function?
Quote:

Originally Posted by Orlando
How do u find the image set for a function with no domain?

Please show us what you mean by "a function with no domain".
At least, describe it in detail.
• Nov 19th 2011, 05:30 AM
Orlando
Re: Image set of a function?
Sorry I am tring to find the image set of :

f(x)=1/4(x-1)^2 -9

In interval notation.

There is no domain given so I persume the answear would be

[-9, to infinity) Sorry don`t know how to us the symbol for infinity

Or am I completly of the mark
• Nov 19th 2011, 05:40 AM
Plato
Re: Image set of a function?
Quote:

Originally Posted by Orlando
f(x)={1/4(x-1)^2} -9
In interval notation.

Assuming that the function is $f(x)=\frac{1}{4(x-1)^2}-9$ then the domain is all $x\ne 1$.

The range is $(-9,\infty)$. Because $f(x)\ne -9$ for any $x$.
• Nov 19th 2011, 05:43 AM
Orlando
Re: Image set of a function?
sorry the function is 1/4*(x-1)^2-9

sorry for how I have presented it still trying to get use to typing maths in a computer

• Nov 19th 2011, 05:48 AM
HallsofIvy
Re: Image set of a function?
Then that has "natural domain" of all real numbers. It's graph is a parabola with vertex at (1, 9), vertical axis, opening upward.

If a function is given without an explicit domain, unless stated to the contrary, the domain is taken to be the "natural domain"- the set of all real numbers for which it is possible to calculate the value. The natural domain of a polynomial is all real numbers. The natural domain of [1/(4(x-1)^2]- 9] is all real number except x= 1. That is because x= 1 would make the denominator of that fraction 0.
• Nov 19th 2011, 05:49 AM
Plato
Re: Image set of a function?
Quote:

Originally Posted by Orlando
sorry the function is [TEX]f(x)=\frac{1/4}(x-1)^2-9[TEX]

Is the function $f(x)=\frac{1}{4}(x-1)^2-9~???$
• Nov 19th 2011, 05:53 AM
Orlando
Re: Image set of a function?
yes it is. Sorry for messing everyone about still getting my head round La Tex
• Nov 19th 2011, 05:55 AM
Plato
Re: Image set of a function?
Quote:

Originally Posted by Orlando
yes it is. Sorry for messing everyone about still getting my head round La Tex

Then the domain is $\mathbb{R}$ and range $[-9,\infty)$
• Nov 19th 2011, 06:02 AM
Orlando
Re: Image set of a function?
Thanks for everyones help. Going to learn how to use La Tex looks a great bit of software.