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# Week 6 Problem Set B.
## Task to do.
Catch up on all the past homework if you haven't done so. Make sure you know how to do them, and/or seek help or work together so that you understand them. The problem section for this week's homework is at the end of this page.
## Reading.
Read chapter 2.5 and finish reading chapter 2 of Thomas. And preview chapter 3 for next week.
## Brief review.
Last week we discussed continuity of a function at a point, where we say a function $f(x)$ is continuous at the point $x=a$ if $\displaystyle \lim_{x\to a}f(x)=f(a)$. Intuitively this corresponds to being able to draw the graph of $f(x)$ without lifting the pen near the point $x=a$. And we say a function $f$ is continuous to mean it is continuous at each point of the domain of $f$.
Since continuity at a point can be defined by limits, the precise definition of continuity at a point will look similar to the limit definition:
> **Definition.** Continuity of $f$ at the $x=a$.
> We say $f$ is continuous at $x=a$ if for every $\epsilon > 0$, there exist $\delta > 0$ such that whenever $|x-a| < \delta$, we have $|f(x)-f(a)| < \epsilon$.
If a function has a domain with one-sided end points, then we use one-sided limit to speak of one-sided continuity.
Continuous functions are nice, that composition and arithmetic combinations of continuous functions remain continuous (on their domain). In particular, every **elementary function** as well as **absolute value function** are continuous **on their domain**. This is large class of functions!
Continuity also plays well with limits, namely if $f(y)$ is continuous at $y=b$, and that $g$ is another function such that $\displaystyle\lim_{x\to a}g(x)=b$, then we have the following **limit exchange property**: $$
\lim_{x\to a}f(g(x)) = f(\lim_{x\to a}g(x))=f(b).
$$
An important property of continuous is the **intermediate value theorem**, which states the following:
> **Intermediate value theorem.**
> If $f(x)$ is continuous on the interval $[a,b]$, and $y_{0}$ is any number between $f(a)$ and $f(b)$, then there exists some point $c\in[a,b]$ such that $f(c)=y_{0}$. In other words, if $f$ is a continuous over the closed interval $[a,b]$, then the graph of $f(x)$ from $f(a)$ to $f(b)$ must go through every number between $f(a)$ and $f(b)$ as $x$ go from $a$ to $b$.
Now, $f$ may also hit numbers outside of $f(a)$ and $f(b)$, but that is not guaranteed and the theorem does not say anything about that nor prevent it.
Intermediate value theorem is useful in showing the existence of solutions to some equation. It is a key result in the sense that intermediate value theorem is equivalent to the axiomatic definition of the real numbers! In any case, let us see a quick example:
**Example.** Show that the equation $x^{3}-2x+1 = 0$ has at least a solution.
$\blacktriangleright$ Note that the function $f(x)=x^{3}-2x+1$ is **continuous** (as it is polynomial, hence elementary). Now by testing and trying some points into $f$, we see that $f(-2)=-3$ and $f(0)=1$. So as $$
f(-2) < 0 < f(0)
$$by **intermediate value theorem**, there exists some point $c\in[-2,0]$ such that $f(c) = 0$, namely there is a number $c$ in $[-2,0]$ satisfying $c^{3}-2c+1=0$. Amazing! $\blacklozenge$
## Problems.
### Using intermediate value theorem.
1. Use the intermediate value theorem in each of the following to prove that each equation has a solution. **Give a closed interval on where the solution is**. Then use DESMOS or a graphing calculator to verify your claim.
1. $x^{3}-3x-1=0$
2. $2x^{3}-2x^{2}-2x+1=0$
3. $x(x-1)^{2} = 1$
4. $x^{x} = 2$
5. $\sqrt{x} + \sqrt{1+x} = 4$
6. $x^{3}- 15x +1 = 0$, there are three roots. For each root, give an interval containing that root.
7. $\cos(x)=x$.
8. $2\sin(x)=x$. There are three roots. For each root, give an interval containing that root.
2. A continuous function $y=f(x)$ is known to be negative at $x=0$ and positive at $x=1$. Why does the equation $f(x)=0$ have at least one solution between $x=0$ and $x=1$? Illustrate it with a sketch.
3. Explain why the equation $\cos(x)=x$ has at least one solution.
4. Show that the equation $x^{3} - 15x +1 =0$ has three solutions in the interval $[-4,4]$. (Hint: You need to find several places where the sign changes.)
5. Show the function $F(x) = (x-a)^{2}(x-b)^{2}+x$ takes on the value $\frac{a+b}{2}$ for some value of $x$.
6. If $f(x) = x^{3}-8x+10$, show that there are values $c$ for which $f(c)$ equals (a) $\pi$; (b) $-\sqrt{3}$; (c) $5000000$.
### Continuity and limits.
1. For each of the following, at which points is the function continuous? Hint: These are all elementary functions.
1. $\displaystyle y= \frac{1}{x-2} - 3x$
2. $\displaystyle y= \frac{1}{(x+2)^{2}} + 4$
3. $\displaystyle y = \frac{x+1}{x^{2}-4x+3}$
4. $\displaystyle y = \frac{x+3}{x^{2}-3x-10}$
5. $\displaystyle y = |x-1| + \sin(x)$
6. $\displaystyle y = \frac{1}{|x| + 1} - \frac{x^{2}}{2}$
7. $\displaystyle y = \frac{\cos(x)}{x}$
8. $\displaystyle y = \frac{x+2}{\cos(x)}$
9. $g(x)=\begin{cases}\displaystyle\frac{x^{2}-x-6}{x-3}, & x\neq3 \\ 5, & x=3\end{cases}$
10. $f(x)=\begin{cases}\displaystyle\frac{x^{3}-8}{x^{2}-4}, & x\neq 2, x\neq-2 \\ 3, & x = 2 \\ 4, & x = -2\end{cases}$
2. For each of the following, compute the limit, if exists:
1. $\displaystyle\lim_{x\to\pi} \sin(x-\sin(x))$
2. $\displaystyle\lim_{t \to 0} \sin\left( \frac{\pi}{2} \cos(\tan(t))\right)$
3. $\displaystyle\lim_{y \to 1} \sec(y \sec^{2} y - \tan^{2}y -1)$
4. $\displaystyle\lim_{x\to 0} \tan\left( \frac{\pi}{4} \cos(\sin x^{1/3}) \right)$
5. $\displaystyle \lim_{t\to 0} \cos\left( \frac{\pi}{\sqrt{19-3\sec(2t)}} \right)$
6. $\displaystyle\lim_{x\to \frac{\pi}{6}} \sqrt{\csc^{2}(x)+5\sqrt{3}\tan(x)}$
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