Where are these examples discontinuous?

Please explain step by step. Thank you.enter image source here

1 Answer
Mar 28, 2018

a. Removable discontinuity at #x=0,# infinite discontinuity at #x=1.# b. Removable at #x=-3,# infinite at #x=-2.# c. Jump at #x=2.#

Explanation:

Discontinuities tend to happen at three distinct points:

  1. At vertical asymptotes (infinite discontinuities), which can be found by setting the denominator of the rational function equal to zero and solving for #x.#

  2. At holes (removable discontinuities), which come to exist by canceling out a common factor in both the numerator and dominator.

  3. At jumps, which generally happen in the graphs of piecewise functions, usually resulting in a notable gap between two curves in the graph.

Keeping these in mind, let's proceed:

a. Let's simplify a little. The denominator can be factored, yielding:

#f(x)=x/(x(x-1))#

#x# cancels out:

#f(x)=cancelx/(cancelx(x-1))#, which tells us there's a removable discontinuity at #x=0#. We now have #f(x)=1/(x-1)#.

Setting the denominator equal to zero and solving yields

#x-1=0#

#x=1#

So, there's an infinite discontinuity (vertical asymptote) at #x=1#

b. Again, the first thing we do is simplify. There's not much that can be done with the numerator; however, we factor the denominator (it's a quadratic), yielding

#f(x)=(x+3)/((x+3)(x+2))#. #x+3# cancels out:

#f(x)=cancel(x+3)/(cancel(x+3)(x+2))#.

This tells us there is a removable continuity at #x+3=0,# or #x=-3.#

The simplified form is #f(x)=1/(x+2).# Once again, set the denominator equal to zero and solve for infinite discontinuities:

#x+2=0#

#x=-2#

c. This function is much unlike the other two. The first objective should be to get rid of the absolute value and get a piecewise function instead. We do this by realizing that #|x-2|=0# for #x=2,# and seeing that for

#x<2, |x-2|<0=-(x-2)=2-x#

#x>2, |x-2|>0=x-2#

So, we really have a piecewise function in the form

#f(x)=-(x-2)/(x-2)=-1, x<2, f(x)=(x-2)/(x-2)=1, x>2#

It appears that we have removable discontinuities, but upon closer inspection, this is not the case. Yes, #x-2# cancels out in both obtained functions, but as evident by the simplification, at #x=2,# the function jumps from being #-1# to #1.# So, there is a jump discontinuity at #x=2.#