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Mathematics 22 Online
OpenStudy (anonymous):

Graph the equation written:

OpenStudy (anonymous):

\[f(x)=\frac{ 1+|3-x| if x is \neq 3 }{ 2 if x =3 }\]

OpenStudy (austinl):

What? \(f(x)=\dfrac{1+|3-x|}{2},~\text{if x}\ne3,~ \text{if x}=3\) Is this the question?

OpenStudy (anonymous):

Yes but the x unequal to 3 goes for the top and the x=3 is for the denominator

OpenStudy (austinl):

I have never seen an equation in that form where the location of.... just.... um. @nincompoop Would you mind looking at this?

OpenStudy (anonymous):

Alright, Well see what this person says. Lol

OpenStudy (anonymous):

Its a part of limits with calculus

OpenStudy (austinl):

I guess nin is busy.

OpenStudy (anonymous):

Its alright. Wanna help with another one?

OpenStudy (anonymous):

the function doesnt have a denominator right??

OpenStudy (anonymous):

its not a function. its for a limit to be graphed. Look at the very first one @hantenks

OpenStudy (anonymous):

The fraction was to keep them separated.

OpenStudy (anonymous):

\[F(x)= 1+|3-x|; if x \neq 3\]

OpenStudy (anonymous):

\[Then 2 if x=3\]

OpenStudy (austinl):

THAT is what confused me.

OpenStudy (anonymous):

Ohhhh, I'm sorry. I didn't know.

OpenStudy (anonymous):

But its asking for it to be graphed.

OpenStudy (anonymous):

ohkzz.. its simple if you can graph the 1st part of f(x) at x not equal to 0.. for it you need to know how to draw |x| graph m posting it..

OpenStudy (debbieg):

So graph y=1+|3-x| which is just a transformation of an abs value function. But, at x=3, you need an "open circle" for the value you get for y, and then a closed circle at the point (3,2).

OpenStudy (anonymous):

So what exactly would I want to do first. |dw:1378749273650:dw|

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