Question

Please help me with this 2 Calc questions

Answer 1

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

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

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

$\implies \dfrac{p(x)}{q(x)} = \dfrac{(x-a)^n p^*(x)}{(x-a)^n q^*(x)} = \dfrac{p^*(x)}{q^*(x)}$

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

In this case, we have

$f(x) = \dfrac{x+4}{x^2-x-20} = \dfrac{x+4}{(x+4)(x-5)}$

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

$f(x) = \begin{cases}\dfrac1{x-5} & \text{if }x \neq -4 \\\\ \text{DNE} & \text{if }x = -4\end{cases}$

("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 $x$ 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.

Here we have

$f(x) = \begin{cases}x+3 & \text{if }x\le2 \\ cx+6 & \text{if }x>2\end{cases}$

so the pieces meet at $x=2$. Continuity at this point requires that the both limits from either side of $x=2$ be the same. This means

$\displaystyle \lim_{x\to2^-} f(x) = \lim_{x\to2} (x+3) = 2+3 = 5$

$\displaystyle \lim_{x\to2^+} f(x) = \lim_{x\to2} (cx+6) = 2c + 6$

Solve for $c$.

$2c + 6 = 5 \implies 2c = -1 \implies \boxed{c = -\dfrac12}$