Let f(x)=[{√-x, if x<0}{3-x, if 0≤x<3}{(x-3)^2, if x>3}]. (i) Check whether f is discontinuous. If yes, find where? (ii) Give a rough sketch of the graph of f.

Answer 1

For x<0, f(x) = sqrt(x)

for 0<=x<3, f(x) = 3-x

But sqrt(0) = 0 and 3-0 = 3, so there is a jump discontinuity at x=0

That is NOT the case at the other mesh point x=3, because

3-3 = 0 and (3-3)^2 = 0^2 = 0