1 Answer

Rubiela · Answer 1 · 2023-11-30T07:52:50+0000

20. A rational function
(p(x))/(q(x)) (where
p,q are polynomials in
) has a removable discontinuity at
x=a if for some positive integer
we can factorize

p(x) = (x-a)^n p^*(x)

q(x) = (x-a)^n q^*(x)

$\implies (p(x))/(q(x)) = ((x-a)^n p^*(x))/((x-a)^n q^*(x)) = (p^*(x))/(q^*(x))$

That is, we "remove" the discontinuity at
x=a because while
$x\\eq a$ , we have
(x-a)/(x-a)=1 . The discontinuity is still there, since
(p(a))/(q(a)) is undefined, but in the limit sense we can essentially ignore the factors of
x-a . The graph of
(p(x))/(q(x)) will have all the features of
(p^*(x))/(q^*(x)) aside from a hole at
x=a .

In this case, we have

f(x) = (x+4)/(x^2-x-20) = (x+4)/((x+4)(x-5))

and when
$x\\eq-4$ , we can cancel the factor of
x+4 so that

$f(x) = \begin{cases}\frac1{x-5} & \text{if }x \\eq -4 \\\\ \text{DNE} & \text{if }x = -4\end{cases}$

("DNE" = "does not exist", i.e. is undefined. For some reason "undefined" wouldn't render...)

This means
x=-4 is a removable discontinuity.

We cannot do the same with the factor of
x-5 , so in contrast
x=5 is a non-removable discontinuity.

21. The pieces of
f(x) defined on
x<2 and
x>2 are themselves continuous since they are polynomials. Then the continuity of
f(x) over the entire real line depends on the point where the pieces meet.