Algebra
posted by Jimmy .
Use the function below to answer the following parts.
1/3x  4 if 6 </= x <  3
f(x) = 2 if 3 < x < 3
x^2  4 if x > 3
I found out the domain, range, and points of disontinuity. I think i did them right:
Domain= [6,infinity]
Range= [6,infinity]
and points of discontinuity = x=3,x=3
How do i graph these functions?

Your first segment runs from (6,6) to (3,5)
for 6 ≤ x < 3
so draw a solid dot for (6,6) and an "open" dot for (3,5)
the 2nd segment runs from (3,2) to (3,2).
draw that line leaving the end points as open dots.
the 3rd graph is part of a parabola, which has its vertex at (0,4) and which would open upwards.
I would pass through (1,3), (2,0) and (3,5)
draw it only from (3,5) on, leaving the (3,5) open.
you are correct with your domain and range, and your value of x that have a discontinuity. 
thanks
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