1. Domain and Range

Can anyone explain domain and range to me. I know what it is and I can sort of tell from look at a graph what the domain is but i never get it 100% right. An example would be 5x+4/x^(1/2)+3x-2. Really anything you can tell me about domain and range and how to find it and put it in the correct intervals would be great.
Thanks
AC

2. Originally Posted by Casas4
Can anyone explain domain and range to me. I know what it is and I can sort of tell from look at a graph what the domain is but i never get it 100% right. An example would be 5x+4/x^(1/2)+3x-2. Really anything you can tell me about domain and range and how to find it and put it in the correct intervals would be great.
Thanks
AC
see the explanations of domain and range in the following threads:

http://www.mathhelpforum.com/math-he...lus-notes.html (post #2)

http://www.mathhelpforum.com/math-he...ain-range.html

http://www.mathhelpforum.com/math-he...ions-help.html

http://www.mathhelpforum.com/math-he...-function.html

http://www.mathhelpforum.com/math-he...ain-range.html

you should know how to find the domain and range for several classes of functions: polynomials, logs, radical, rationals, trigonometric functions. i think the threads i gave cover them all. if not, ask about them

3. Originally Posted by Casas4
Can anyone explain domain and range to me. I know what it is and I can sort of tell from look at a graph what the domain is but i never get it 100% right. An example would be 5x+4/x^(1/2)+3x-2. Really anything you can tell me about domain and range and how to find it and put it in the correct intervals would be great.
Thanks
AC
Domain is the values of "x" for which a function f(x) function has some defined answer. Range is the values of the function f(x) for a defined domain.
$\displaystyle y\;=\;5x + \frac{4}{\sqrt {x}} + 3x - 2$
For domain, the term inside the root should be positive
$\displaystyle x>0$
Also demoninator should not be equal to zero,
x not = 0
Therefore, Domain = {x belongs to real number R / x > 0}
Range = {y belongs to real number R / x > 0}

See another example:
$\displaystyle y\;=\;5x + \sqrt {3x - 2}$
Here, for domain, the term inside the root should be positive

$\displaystyle \;3x - 2 > 0$

$\displaystyle \;x > \frac{2}{3}$
So, Domain = {$\displaystyle {x\;belongs\;R/x > \frac{2}{3}}$}

Calculate y for $\displaystyle \;x = \frac{2}{3}$

we got $\displaystyle \;y = \frac{10}{3}$

So Range = {$\displaystyle {y\;belongs\;R/y > \frac{10}{3}}$}

OK, See one more example
$\displaystyle y\;=\;(3x - 1)^{2}-5$
Here for domain, we can put any value of x, either positive, zero or negative, in the equation, we will get an answer for y.
therefore Domain = {x belongs to real number R}
For Range, We know that $\displaystyle y\;=\;(3x - 1)^{2}>0$ for any value of x. It is zero if $\displaystyle \;x = \frac{1}{3}$

Therefore the minimum value of y will be 0 - 5
i. e. y= -5

So Range= {$\displaystyle {y\;belongs\;R/y > -\;5}$}