Log in

View Full Version : Determine if the relation is a function


MANDMO 91
Sep 2, 2008, 01:33 PM
I am doing some functions review but I forget how to determin if it is a relation when given an equation
help me please!!

y=-3x +5

y=square root of (3x - 1)

Unknown008
Sep 2, 2008, 10:24 PM
Don't quite understand what you want here...

l_byl
Sep 10, 2008, 10:13 AM
the relation is a function if when graphed it passes the vertical line test aka no two points have the same x value

ebaines
Sep 10, 2008, 10:28 AM
An equation represents a function if for any x in the function's domain there is preciely one value of y that emerges. If there is more than 1 possible value of y for any x , then the equation is not a function.

Example: consider the equation y=2x. The domain for x is - \infty < x < + \infty. For any x, there is only one value of y that comes out of the equation. For example y(2) = 4, y(-10) = -20, etc. Therefore y= 2x is a function.

Another example: consider y = \sqrt x . Here x is in the domain 0 \le x < \infty . For x equals zero there is one value of y: zero. But for any positive x there are two values of y, since the square root of a positive number number is either plus or minus. For example, \sqrt 4 = \pm 2. Hence y= \sqrt x is not a function.

Hope this helps.

Unknown008
Sep 11, 2008, 03:46 AM
You mean that functions are the equations having an inverse, that is, are one one? Ok, but I thought that all equations could be represented as functions, but some of which had inverse, and others don't. K, thanks for the specification ebaines.

ebaines
Sep 11, 2008, 05:36 AM
You mean that functions are the equations having an inverse, that is, are one one? .

Not quite. Consider the equation f(x)= sin(x). This is defintely a fuction since for any value of x there is precisely one value of f(x). Its inverse is g(x) = arcsin(x), which is not a function, as for any value of x there are an infinite number of values for g(x), for example arcsin(0) = 0, pi, 2*pi, 3*pi, etc.

Unknown008
Sep 12, 2008, 03:37 AM
Oh, OK, thanks then, your really a help ebaines. Btw, my school is ebene, having the same pronunciation as your username, tee hee!