Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

if an equation has x^2 or y^2 will it always be not a function?

OpenStudy (freckles):

is x^2+y^2=0 a function?

OpenStudy (anonymous):

is that the question exactly as it is written?

OpenStudy (anonymous):

in general

OpenStudy (anonymous):

like x^2+1=y or y^2+5=x anything that's squared of x or y.

OpenStudy (anonymous):

"will it always not be a function"?!

OpenStudy (anonymous):

\[y=x^2+1\] is not a function, it is an equation but the function \[f(x)=x^2+1\] is a function

OpenStudy (freckles):

y=x^2+1 is a function y^2+5=x isn't a function x^2+y^2=0 is a function x^2+y^2=1 isn't a function

OpenStudy (anonymous):

i must disagree with my learned colleague @freckles

OpenStudy (freckles):

well a function of x that is

OpenStudy (anonymous):

and also y is a function of x

OpenStudy (freckles):

disagree with me? on which one?

OpenStudy (anonymous):

y=x^2+1 is a function

OpenStudy (freckles):

i mean is a function of x

OpenStudy (anonymous):

i think that's a function

OpenStudy (freckles):

oh because i put y instead of f(x)

OpenStudy (anonymous):

\[\{(x,y):y=x^2+1\}\] is a function \[f(x)=x^2+1\] is a function \[f:=x\to x^2+1\] is a function

OpenStudy (anonymous):

because anything squared is positive?

OpenStudy (anonymous):

\[y=x^2+1\] is an equation

OpenStudy (anonymous):

no it has nothing to do with positive or negative

OpenStudy (anonymous):

y can represent f(x)

OpenStudy (anonymous):

a function is a rule that assigns to each element in one set (the domain) a unique element in another set (range, or codmain)

OpenStudy (anonymous):

make that codomain in any case the equation given by \[x^2+y^2=1\] does not represent a function as @freckles said, because there can be two different y values for the same x value

OpenStudy (freckles):

http://www.mathsisfun.com/sets/function.html

OpenStudy (anonymous):

wait so there can't be 2 or more x values that can give the same y?

OpenStudy (freckles):

both of these links say equations can be functions

OpenStudy (freckles):

and i like paul a lot

OpenStudy (anonymous):

im confused with the definition of functions

OpenStudy (anonymous):

the definition i got is that a relation for which every element in the domain has a unique corresponding element in the range? but a range can be the same?

OpenStudy (anonymous):

if there are multiple ranges that are the same, wouldn't that be not unique?

OpenStudy (freckles):

Well this is the way Paul and I would define a function: A function is an equation for which any x that can be plugged into the equation will yield exactly one y out of the equation.

OpenStudy (calculusfunctions):

In general, just remember that when looking at an equation in two variables x (independent variable) and y (dependent variable), the equation always represents a function as long as i). the exponent of y is 1 and ii). any term is either "+" ve or "-"ve but not both +/-

OpenStudy (anonymous):

can you give an example of ii)? @calculusfunctions

OpenStudy (anonymous):

@freckles when wouldn't a function yield one y?

OpenStudy (freckles):

Pretend we have x^2+y^2=1 This is not a function. For example if x=0 then y^2=1 which means y=1 or y=-1 So we have (0,1) and (0,-1) are on the circle. But look we have an x value yielding two different y-values So this x^2+y^2=1 is definitely not a function.

OpenStudy (anonymous):

ok, so whenever i see y^2, it will never be a function? if there is x

OpenStudy (freckles):

y^2=x-5 is not a function. Example, if x=6, we have y^2=6-5=1 which means y=1 or y=-1 so we have (6,1) and (6,-1) on the parabola. So we have an x yielding two different y-values.

OpenStudy (freckles):

Well x^2+y^2=0 is a function

OpenStudy (freckles):

And I say this because over the reals (0,0) is the only point on the graph of x^2+y^2=0

OpenStudy (calculusfunctions):

Some examples of functions:\[y =x ^{2}-3\] \[y=\sqrt{x}\] \[x ^{3}-2xy +3=0\] \[y =\frac{ 1 }{ x }\]Examples of some equations that are not functions:\[4x ^{2}-y ^{2}=36\] \[y =\pm \sqrt{x}\] \[x ^{2}+y ^{2}\]

OpenStudy (anonymous):

so can i check if an equation is a function by solving for x and see if it comes out to 2 or more y values?

OpenStudy (calculusfunctions):

Oops, I forgot to finish the last one\[x ^{2}+y ^{2}=25\]

OpenStudy (freckles):

You have to pick an x such that happens to show it is not a function not just any x

OpenStudy (calculusfunctions):

@polarpenguin did you understand what I said?

OpenStudy (anonymous):

so a function can't have +/- together?

OpenStudy (calculusfunctions):

Yes

OpenStudy (calculusfunctions):

That is correct

OpenStudy (freckles):

|dw:1414288286955:dw| Do you think g(x) is a function?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!