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Find all values of t such that \lfloor t\rfloor = 2t + 3. If you find more than one value, then list the values you find in increasing order, separated by commas.

 Jan 8, 2015

Best Answer 

 #2
avatar+27537 
+10

I don't think a graphical solution is "cheating" - it seems perfectly valid to me.  However, if you want a non-graphical method then here is a possibility:

 

Non graphical method:

(That last d should be a δ, of course!)

.

 Jan 9, 2015
 #1
avatar+98091 
+5

This is cheating...but have a look at the graph of y =  floor(x) and y = 2x + 3...seen here....https://www.desmos.com/calculator/8kenredv3g

Notice that there appear to be only two points of interesection in the two graphs.... one at (-3, -3) and one at  (-3.5, -4)

There is also another "near" miss at (-4, -5).....but  2(-4) + 3 = -5 ....but floor (-4) = -4  ...so they do not intersect when x = -4....

So the x (or, t, if you prefer) values that make this true is when t = -3 and t = -3.5

Note the oddity here.....if the graphs were both just "linear," and because they have different "slopes,"  there would be- at most - only one point of intersection, ....but since the "floor" function isn't truly a linear function, it actually gives rise to an additional intersection point...!!!

 

 Jan 8, 2015
 #2
avatar+27537 
+10
Best Answer

I don't think a graphical solution is "cheating" - it seems perfectly valid to me.  However, if you want a non-graphical method then here is a possibility:

 

Non graphical method:

(That last d should be a δ, of course!)

.

Alan Jan 9, 2015
 #3
avatar+98091 
0

Thanks, Alan, for that answer....

 

 Jan 9, 2015

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