Master Intermediate Value Theorem Problems

What is its worth ${15}^{\circ <!--\; \circ \; -->}$?

What is the value of $\frac{100}{0}$?
1) 0
2) 1
3) 6
4) Not defined

What is the integral of log(z) over the unit circle?
I tried three times, but in the end I am concluding that it equals infinity, after parametrizing, making a substitution and integrating directly (since the Residue Theorem is not applicable, because the unit circle encloses an infinite amount of non-isolated singularities.)
Any ideas are welcome.
Thanks,

My final for my introductory analysis course is tomorrow and my teacher gave us a list of possible theorems to prove. If anyone could please show me a proof for The Intermediate Value Theorem that is short and easy to follow, so even if I still cannot understand it I can at least memorize it. Also, I have looked through numerous texts and the internet, but they all seem to confuse me. I know that itis an insult to all you math experts to memorize proofs, but I am desperate at this point. Thank you

Suppose $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous and periodic with period 2a for some $a>0$; that is, $f(x)=f(x+2a)$ for all $x\in <!--\; \in \; -->R$. Show there is some $c\in <!--\; \in \; -->[0,a]$ such that $f(c)=f(c+a)$.
The only way I see to do this question is to apply the intermediate value theorem, but I just don't know how to apply it to this question. I know that $f(x)=f(x+2a)$. Therefore, if i can show that $f(c+a)=f(c+2a)$ or $f(c+a)-<!--\; -\; -->f(c+2a)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f(2a)-<!--\; -\; -->f(3a)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

Use the intermediate value theorem to prove that if $f:[1,2]\to <!--\; \to \; -->R$ is a continuous function, that there is at least one number $c$ in the interval $(1,2)$ such that $f(c)=1/(1-<!--\; -\; -->c)+1/(2-<!--\; -\; -->c)$
This is a question for my intro calc class that I am having a hard time understanding.

The intermediate value theorem states the following: Consider an interval $I=[a,b]$ in the real numbers $ℝ$ and a continuous function $f:I\to ℝ$. Then, If $u$ is a number between $f(a)$ and $f(b)$, $f(a)<u<f(b)$ (or $f(a)>u>f(b)$ ),

then there is a $c\in (a,b)$ such that $f(c)=u$.
What happens if we consider $\mathbb{R}$ instead of $I$ ?
Is there a simple method to prove the theorem in this case ? I know the method for $f:I\to <!--\; \to \; -->\mathbb{R}$ with $I=[a,b]$, but here it is different.

The statement for Intermediate Value Theorem is thus:

Let $f$ denote a function that is continuous on the closed interval $[a,b]$ and suppose $f(a)\ne <!--\; \ne \; -->f(b)$. If $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

But who is to stop me if I put it like this?

Let $f$ denote a function that is continuous on the open interval $(a,b)$. Now, $\underset{x\to <!--\; \to \; -->a^{+}}{lim}f(a)=L$ and $\underset{x\to <!--\; \to \; -->b^{-<!--\; -\; -->}}{lim}f(b)=M$. Suppose, $L\ne <!--\; \ne \; -->M$. If N is any number between $L$ and $M$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

let $f:[0,a]\to <!--\; \to \; -->\mathbb{R}$ of class $C^1$ such that $f(0)=0$ show that
$\mathrm{\exists <!--\; \exists \; -->}c\in <!--\; \in \; -->]0,a[,f^{\prime }(c)=\frac{2f(a)+a{f}^{\prime }(a)}{3a}$
First I apply mean value theorem then $\mathrm{\exists <!--\; \exists \; -->}b\in <!--\; \in \; -->]0,a[,f^{\prime }(b)=\frac{f(a)-<!--\; -\; -->f(0)}{a-<!--\; -\; -->0}=\frac{f(a)}{a}$ then $f(a)=a{f}^{\prime }(b)$
so
$\frac{2f(a)+a{f}^{\prime }(a)}{3a}=\frac{2}{3}{f}^{\prime }(b)+\frac{1}{3}{f}^{\prime }(a)$
How to prove that $\frac{2}{3}{f}^{\prime }(b)+\frac{1}{3}{f}^{\prime }(a)$ is between $f^{\prime }(a)$ and $f^{\prime }(b)$?

Let $f$ be continuous on $[0,1]$ and $f(0)=f(1)$. Also, $n\in <!--\; \in \; -->\mathbb{N}$. Prove that there exists some $x$ s.t. $f(x)=f(x+\frac{1}{n})$.
I think I need to assume towards a contradiction that if $g(x)=f(x)-<!--\; -\; -->f(x+\frac{1}{n})$, then $g(x)\ne <!--\; \ne \; -->0$ $\mathrm{\forall <!--\; \forall \; -->}x$.
So, $g(0)=f(0)-<!--\; -\; -->f(\frac{1}{n})$, $g(1)=f(1)-<!--\; -\; -->f(1+\frac{1}{n})$. Since $f(0)=f(1)$,
$g(0)-<!--\; -\; -->g(1)=f(1+\frac{1}{n})-<!--\; -\; -->f(\frac{1}{n})$. For context, I am only up to Ch. 8 in Spivak, so I cannot use any derivative-based theorems which is why I am stuck here.

I'm trying to solve the below exercise.

Let $f:[0,1]\to <!--\; \to \; -->[0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a\in <!--\; \in \; -->A$ such that $f(a)=a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0)=0$ or $f(1)=1$, the theorem is proved. Suppose not. Define $g(x)=f(x)-<!--\; -\; -->x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0)\ne <!--\; \ne \; -->0$, $g(0)\ne <!--\; \ne \; -->0$, so $g(0)>0$. Since $f(1)\ne <!--\; \ne \; -->1$, $g(1)<0$. By the intermediate value theorem, there exists $a\in <!--\; \in \; -->[0,1]$ such that $g(a)=0$. So $f(a)-<!--\; -\; -->a=0$, so $f(a)=a$.
I am fairly sure that the result is not true for discontinuous functions, largely because I requiblack continuity of $g$ to invoke the intermediate value theorem. I am having trouble finding a counterexample, however. Do I define a function piece-wise, with jumps at $x=0$ or $x=1$, to try to break the intermediate value theorem?

I found that $f(x)=x^2$ is continuous on $[1,2]$ but does not assume the values 2 intermediate between the value 1 and 4. I do not understand what they mean. For me, $\sqrt{2}\in <!--\; \in \; -->[1,2]$ and $f(\sqrt{2})=2$.
Also, using the the intermediate value theorem, since $f$ is continuous on $[1,2]$ and $f(1)=1<2<4=f(2)$, there exists $c\in <!--\; \in \; -->[1,2]$ such that $f(c)=2$.
Can I get some explanation about this?

As the title suggests I am trying to prove that for all $c\in <!--\; \in \; -->[0,\mathrm{\infty <!--\; \infty \; -->})$ there exists $x\in <!--\; \in \; -->\mathbb{R}$ such that $x{e}^x=c$.
Now, the previous parts of this question imply that the way we are meant to do this is by using the intermediate value theorem. Then we can consider $x{e}^x=f(x)$, but this function is defined on $\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ and the IVT requires that our domain be $[a,b]$ with $a,b\in <!--\; \in \; -->\mathbb{R}$. So how exactly can we do this? Please note that this comes from an analysis course so calc I or II methods won't work here.

Prove that there exists a real number $x$ such that $x^{177}+\frac{165}{1+x^8+{\sin }^2<!--\; \; -->(x)}=125$ using Intermediate Value Theorem.

Let $f:<!--\; :\; -->[0,2]\to <!--\; \to \; -->\mathbb{R}$ be a continuous function such that $f(0)=f(2)$. Use the intermediate value theorem to prove that there exist numbers $x,y\in <!--\; \in \; -->[0,2]$ such that $f(x)=f(y)$ and $|x-y|=1$.
Hint: Introduce the auxiliary function $g:<!--\; :\; -->[0,1]\to <!--\; \to \; -->\mathbb{R}$ defined by $g(x)=f(x+1)-f(x)$.

Suppose that $f,g:<!--\; :\; -->[a,b]\to <!--\; \to \; -->\mathbb{R}$ are continuous functions with $f(a)<g(a)$ and $f(b)>g(b)$. Prove that there exists an $x\in <!--\; \in \; -->(a,b)$ with $f(x)=g(x)$. I believe this uses the Intermediate Value Theorem.

I know every continuous function from $[0,1]$ to $[0,1]$ has fixed point. I proved it using Intermediate value theorem. But I can not see why every continuous function from $[0,1]$ to $\mathbb{R}$ has fixed point.
Is this result even true?

$f:[0;1]\to <!--\; \to \; -->\mathbb{R}$ is continuous and
${\int <!--\; \int \; -->}_0^{1}f(x)dx=\frac{1}{2}$
Proof that there exists an $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=c$.
Idea: So i would define a function $g(x)=f(x)-<!--\; -\; -->x$ and then find points so that $g(x)$ would be $g(y)<0$ and $g(z)>0$. And proceed with my proof and use the intermediate value theorem.
However, i am confused how to do that since i don't have the exact function but only the integral.
Can i just define an $f(x)$ for which the integral would work like $f(x)=\frac{1}{2}$?

I'm taking differential calculus, and I've read that is actually a pretty interesting function that just has one solution. I'm a little confused since most proofs that involve the Intermediate value theorem give a closed interval. But I need to prove that it has a solution in the real numbers. Thanks in advance.