+0  
 
0
116
4
avatar+72 

https://prnt.sc/BXzDgql7NOR3

 

I think it should start with a closed dot because of the brackets,

but im not sure if it's the first left or the second left? 

 Feb 18, 2022
 #1
avatar+117834 
+1

 

\(\lfloor x \rfloor \)     this is the floor of x. 

It means that if x is written as a decimal you should just truncate it leaving just the integer number.

so

 

\(\lfloor 2 \rfloor =2\\ \lfloor 2.3 \rfloor =2\\ \lfloor 2.9 \rfloor =2\\ \lfloor 2.9999999 \rfloor =2\\ but\\ \lfloor 3 \rfloor =3\\~\\ \text{So if you were going to graph } f(x)=\lfloor x \rfloor \text{over the interval [2,3]}\\ \text{ what will your graph look like. }\)

 

Plese describe it for me.  Where are the open circles, the closed circles etc

 

 

 Feb 18, 2022
 #2
avatar+117834 
+1

Also remember that an easy thing to do is to test a point/s

 

\(f(x)=\lfloor x \rfloor -2\)

 

 

If x=0  what is f(0)    equal to?

 Feb 18, 2022
 #3
avatar+72 
+1

Ok i tested points 1, 1.5 and 3

and got (1,-1) (1.5, -0.5) and (3,1)

 

And it seems like its the top-left graph 

 Feb 18, 2022
 #4
avatar+117834 
+1

You got the right answer but what you have written is not altogether correct.

 

You have to treat the floor funtion like you would brackets.

So you work out what is inside and THEN take the flor of it befrome moving on to the rest of the expression.

 

I also desctribed what the floor funtion does rather badly.

The floor funtion simply ROUNDS A NUMBER DOWN to the nearest INTEGER.   

(Thanks for the guidence EP) 

 

I know you already worked out much of this but I wanted to reiterate the whole lot.  ;)

 

\(f(1)=\lfloor 1 \rfloor -2 = 1-2=-1\\ f(1.5)=\lfloor 1.5 \rfloor -2 = 1-2 = -1\\ f(3)=\lfloor 3 \rfloor -2 =3-2=1\)

 

giving    (1, -1)  (1.5, -1) and  (3, 1)

Melody  Feb 18, 2022
edited by Melody  Feb 18, 2022

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