1. Continuity!

f(x) = { -x+b if x<1 or 1 if x=1 or (-5/(x-b))-4 if x>1 and x doesnot equal b}

(a) For what value(s) of b is f continuous at 1 ?

(b) For what value(s) of b does f have a removable discontinuity at 1 ?

(c) For what value(s) of b does f have an infinite discontinuity at 1 ?

(d) For what value(s) of b does f have a (finite) jump discontinuity at 1 ? Write your answer in interval notation.

I literally have no idea where to start. If someone could just atleast get me started it would be a huge help. Thanks.

2. Originally Posted by demonmath
f(x) = { -x+b if x<1 or 1 if x=1 or (-5/(x-b))-4 if x>1 and x doesnot equal b}

(a) For what value(s) of b is f continuous at 1 ?

(b) For what value(s) of b does f have a removable discontinuity at 1 ?

(c) For what value(s) of b does f have an infinite discontinuity at 1 ?

(d) For what value(s) of b does f have a (finite) jump discontinuity at 1 ? Write your answer in interval notation.

I literally have no idea where to start. If someone could just atleast get me started it would be a huge help. Thanks.
$f(x)=\left\{\begin{array}{lr}x+b&:x<1\\1&:x=1\\ \frac{-5}{x-b}-4&:x>1\end{array}\right\}$

(a) $f(x)$ is continuous at $1$ if $\lim_{x\to1^-}f(x)=f(1)=\lim_{x\to1^+}f(x)$

So you want both limits to equal $f(1)$, which is $1$.

For the left limit, we use $f(x)=-x+b$. The limit as $x\to1^-$ is $-1+b$, so $b$ needs to be $2$. Now, since the other equation for $x>1$ also depends on $b$, we have to hope that $b=2$ works for it as well. And it does: $\lim_{x\to1^+}f(x)=\frac{-5}{1-2}-4=5-4=1$, so $b=2$ implies $f$ is continuous at $x=1$.

(b) Here you want $\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)\neq 1$ (and also $\neq\pm\infty$).

(c) Here you want the function to blow up near $x=1$, so what value of $b$ can you choose so that $f\to\infty$ or $f\to-\infty$ as $x\to1$?

(d) Here you want $\lim_{x\to1^-}f(x)\neq\lim_{x\to1^+}f(x)\neq\pm\infty$. This is basically a "what's left over" question. Any value of $b$ that was not an answer to one of the above parts will work here.

3. Wow I definitely understood that first one when you walked me through it thanks a ton. I don't know how to continue on though, I thought I just needed a starting point but clearly that's not enough because I am still struggling with the basic concept there

4. Originally Posted by demonmath
Wow I definitely understood that first one when you walked me through it thanks a ton. I don't know how to continue on though, I thought I just needed a starting point but clearly that's not enough because I am still struggling with the basic concept there
For (b), we want to solve

$\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)$ i.e. $-1+b=\frac{-5}{1-b}-4$

$-1+b=\frac{-5}{1-b}-4\implies b+3=\frac{-5}{1-b}\implies (b+3)(1-b)=$ $-5\implies -b^2-2b+3=-5\implies b^2+2b-8\implies(b+4)(b-2)=0\implies b=\{-4,2\}$

However, we have stated that for this part these limits cannot equal $1$, and from part (a) we know that $b=2$ means that they do, so $b\neq2$. Thus, only $b=-4$ will give you a removable discontinuity (a "hole"), because $\lim_{x\to1^-}(-x-4)=-5=\lim_{x\to1^+}\frac{-5}{x+4}-4$.

For part (c) we just want the denominator approach $0$ as $x\to1^+$, so $b=1$ will do the trick by making $f(x)\to-\infty$ as $x\to1^+$.

For (d), every other value of $b$ will give a finite discontinuity, so the set $B=\{b\in\mathbb{R}:b\neq-4,b\neq1,b\neq2\}$ contains the values you're looking for.

5. I really appreciate all the help it was really good for my understanding of the material.

6. You're welcome.