Master Intermediate Value Theorem Problems

I am having this equation:
$\frac{1}{x-<!--\; -\; -->a_1}+\frac{1}{x-<!--\; -\; -->a_2}+\cdots <!--\; \cdots \; -->+\frac{1}{x-<!--\; -\; -->a_n}=0$
where $a_1<a_2<\cdots <!--\; \cdots \; --><a_n$ are real numbers.
Now I want to prove with the intermediate value theorem that this equation has $n-<!--\; -\; -->1$ solutions in the real numbers.
My thoughts:
With $a_1<a_2<\cdots <!--\; \cdots \; --><a_n$, you can see that every summand gets smaller than the summand before.
My other thought was that about the $n$-summands, with the intermediate value theorem you know that every zero (point) is in the interval and is located between the $n$-summands. So there are $n-<!--\; -\; -->1$ solutions for this equation!
Questions:
How can I prove my thoughts in a formal correct way? (Are my thoughts generally correct?)

The theorem is stated as follows:

Assume that $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous and that $f(a)$ and $f(b)$ have opposite signs. Then there is a point $c\in <!--\; \in \; -->(a,b)$ such that $f(c)=0$.

The beginning of the proof is as follows:

Consider the case where $f(a)<0<f(b)$. Let $A=\{x\in <!--\; \in \; -->[a,b]:f(x)<0\}$ and set $c=supA$. Observe first that since $f$ is continuous and $f(b)>0$, then $c<b$.

I don't see why $c<b$ follows from the fact that f is continuous and $f(b)>0$, but I would really like to understand this, especially since the author seems to think it needs no further explanation. I did try looking at other proofs for the IMT, on this site and other sites, but none of them phrased the proof in this way and they didn't really further my understanding.

I am struggling with the following questions:

Let $g(x)$ be continuous in $[a,b]$ and $0\le <!--\; \le \; -->r\le <!--\; \le \; -->1$ I need to prove that there is $x_0\in <!--\; \in \; -->[a,b]$ that for him exists: $g(x_0)=rg(a)+(1-<!--\; -\; -->r)g(b)$

I know that for $r=0$ and $r=1$. I am getting immediate proof of that(it is the easy part). Where I am struggling is finding and understanding the values of $r$ I need to use. I've tried defining a helping function $g(x)=rg(a)+(1-<!--\; -\; -->r)g(b)-<!--\; -\; -->g(x_0)$ but no success on that end. I need to show that for 2 different values in the range of $0<<!--\; <\; -->r<<!--\; <\; -->1$ one time I am getting a positive result and for the other a negative.
I am not asking for the solution, asking for guidance on how it can be done.

$a,b>0$ and $c<d$
Show that $g(x)={\displaystyle \frac{a}{(x-<!--\; -\; -->c{)}^4}+\frac{b}{(x-<!--\; -\; -->d{)}^7}=0}$ has at least a solution in $(c,d)$
I defined
$\underset{x\searrow <!--\; \searrow \; -->c}{lim}g(x)=+\mathrm{\infty <!--\; \infty \; -->}$
$\underset{x\nearrow <!--\; \nearrow \; -->d}{lim}g(x)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$
I don't know if the limits are just enough for the proof, or is there something else?

A corollary to the Intermediate Value Theorem is that if $f(x)$ is a continuous real-valued function on an interval $I$, then the set $f(I)$ is also an interval or a single point.
Is the converse true? Suppose $f(x)$ is defined on an interval $I$ and that $f(I)$ is an interval. Is $f(x)$ continuous on I?
Would the answer change if $f(x)$ is one-to-one?

Show that there is a number $x\in <!--\; \in \; -->[\mathrm{\pi <!--\; \pi \; -->}/2,\mathrm{\pi <!--\; \pi \; -->}]$ such that $\tan <!--\; \; -->(x)=-<!--\; -\; -->x$
Can I solve this by the intermediate value theorem?

The function $g:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$, is continuous, $g(1)>0$ and
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}g(x)=0$
Show that for every $y$ between 0 and $g(1)$ the function takes on a value in $(y,g(1))$

Let $f:[0,1]\to <!--\; \to \; -->\mathbb{R}$ be continuous with $f(1)=f(0)$. Prove that if $h\in <!--\; \in \; -->(0,\frac{1}{2})$ is not of the form $\frac{1}{n}$, then there does not necessarily exist $|x-<!--\; -\; -->y|=h$ satisfying $f(x)=f(y)$. Provide an example that illustrates this using $h=\frac{2}{5}$.
So I was given the hint that I can use a modified $sin$ function, however I'm not really sure how I would go about that. Preferably, an example not using that would be great.
My thoughts so far are that I use a proof by contradiction saying that $\mathrm{\exists <!--\; \exists \; -->}$ $x,y\in <!--\; \in \; -->[0,1]$ such that $f(x)=f(0)$ and $|x-<!--\; -\; -->y|=h$. And I need to get to the point where $h=\frac{1}{n}$, which would create the contradiction (I assume that's the endpoint?). However, how would I use $h=\frac{2}{5}$ to prove that? Is it trivial in that $h=\frac{1}{n}$ is false? Also, it isn't given that $n\in <!--\; \in \; -->\mathbb{N}$, so maybe I shouldn't assume that?

Use the intermediate value theorem to show that the equation,
$tan(x)=2x$
has an infinite amount of real solutions.
So far I have used the IVT to show that for $f(x)=tan(x)$ in the interval $(-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ there is a $L$ between $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ and $\mathrm{\infty <!--\; \infty \; -->}$ such that there is a value $c$ where $f(c)=L$
I also have shown that for $g(x)=2x$ in the interval $[-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$ there is a $K$ between $-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}$ and $\mathrm{\pi <!--\; \pi \; -->}$ such that there is a value $d$ where $f(d)=K$
Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f(x)$ consists of all reals that $f(x)$ must intersect with $g(x)$ or there is some value $x$ in the interval such that $2x=tan(x)$? If this is correct then how would I extend this result to all intervals $[\mathrm{\pi <!--\; \pi \; -->}(n-<!--\; -\; -->\frac{1}{2}),\mathrm{\pi <!--\; \pi \; -->}(n+\frac{1}{2})]$?

For each $x\in <!--\; \in \; -->[0,10)$, write its decimal expansion $x=a_0.a_1{a}_2{a}_3\dots <!--\; \dots \; -->a_n\dots <!--\; \dots \; -->$. Take into account the number $x^{\prime }$ formed by the digits of odd rank of this decimal expansion; i. e., $x^{\prime }=0.a_1{a}_3{a}_5\dots <!--\; \dots \; -->a_{2n-<!--\; -\; -->1}\dots <!--\; \dots \; -->$ We have two cases:
(1) $x^{\prime }$ is not periodic
(2) $x^{\prime }$ is periodic after the digit $a_{(2n_0-<!--\; -\; -->1)}$
Now let $f$ be the function from the half-open interval [0,10) to $\mathbb{R}$ defined by
$f(x)=\{\begin{array}{ll}0& \text{in case (1) for }{x}^{\prime }\\ a_{(2n_0)}.a_{2(n_0+1)}\dots <!--\; \dots \; -->a_{2(n_0+k)}\dots <!--\; \dots \; -->& \text{in case (2) for }{x}^{\prime }\end{array}$
(Note that in case (2), the integer part $[f(x)]=a_{(2n_0)}$ of $f(x)$ can be nonzero.)

QUESTION
Prove that for all $x_1,x_2$, the function $f$ takes on the open interval $(x_1,x_2)$ all values between $f(x_1)$ and $f(x_2)$ despite $f$ being discontinuous in all points of its domain.

My textbook Elementary Classical Analysis claims that by Darboux's theorem (the intermediate value theorem for derivatives), if a function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ has a nonzero derivative on $\mathbb{R}$, then is $f$ strictly monotonic (i.e., either $f^{\prime }(x)>0$ on $\mathbb{R}$ or $f^{\prime }(x)<0$ on $\mathbb{R}$).
The claim is definitely true if $f^{\prime }s$ domain were instead a closed interval, but since $\mathbb{R}$ is open, I don't understand why Marsden's claim should be true.

Suppose that I have a polynomial of degree 4 like $x^4+4x+c=0$ where $c\in <!--\; \in \; -->\mathbb{R}$.
What is the difference when the question says
Show that it has at most two real roots
and
Show that it has exactly two real roots
?
Are approaches to solve the problem in these two cases different?

Let $f:[a,b]\to <!--\; \to \; -->\text{R}$ be a continuous function such that for each $x\in <!--\; \in \; -->[a,b]$ there exists $y\in <!--\; \in \; -->[a,b]$ such that $|f(y)|\le <!--\; \le \; -->\frac{1}{2}|f(x)|.$ Prove that there exists $c\in <!--\; \in \; -->[a,b]$ such that $f(c)=0.$

I am trying to use intermediate value theorem for continuous function. Here $|f(x)|$ is maximum with respect to $|f(y)|.$ Hence
$|f(y)|\le <!--\; \le \; -->1/2(|f(x)|+|f(y)|)\le <!--\; \le \; -->|f(x)|.$
Now I can say $f(c)=1/2(|f(x)|+|f(y)|$ for some $c\in <!--\; \in \; -->(a,b).$ But how to show $f(c)=0?$

Use the Intermediate Value Theorem and Rolle's Theorem to show the that the polynomial
$p(x)=x^5+x^3+7x-<!--\; -\; -->2$
has a unique real root.

Can someone please give some hints on how to do this question.

How would I go about showing this question:
Suppose that the function $f:[0,1]\to <!--\; \to \; -->[0,1]$ is continuous. Use the Intermediate Value Theorem to prove that there exists $c\in <!--\; \in \; -->[0,1]$ such that
$f(c)=c(2-<!--\; -\; -->c^2)$

My thoughts are we can define a function say $g(x):=f(x)-<!--\; -\; -->x(2-<!--\; -\; -->x^2)$ and then $g(0)=f(0)$ and then $g(1)=f(1)-<!--\; -\; -->1$ and then I'm stuck on how to complete it to get to the point where
$g(0)<0<g(1)$
so there would exist a c s.t. $g(c)=0$ [by IVT] and thus $f(c)=c(2-<!--\; -\; -->c^2)$
If anyone could help, would be much appreciated thank you!

What steps would I take or use in order to use the intermediate value theorem to show that $\cos <!--\; \; -->x=x$ has a solution between $x=0$ and $x=1$?

Given continuous function $g:[a,b{]}^n\subset <!--\; \subset \; -->{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$. By Weistress $g$ has a max and a min.

Can I also conclude its image contains all values in-between this maximum and minimum?

I need this result to complete a proof but cannot seem to find a generalisation of the Intermediate Value Theorem to ${\mathbb{R}}^n$.

What steps would I take or use in order to use the intermediate value theorem to show that $\cos <!--\; \; -->x=x$ has a solution between $x=0$ and $x=1$?

I'm doing a review packet for Calculus and I'm not really sure what it is asking for the answer?
The question is: Let f be a continuous function on the closed interval [-3, 6]. If f(-3)=-2 and f(6)=3, what does the Intermediate Value Theorem guarantee? I get that the intermediate value theorem basically means but not really sure how to explain it?

The Intermediate Value Theorem states that, for a continuous function $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$, if $f(a)<d<f(b)$, then there exists a $c\in <!--\; \in \; -->(a,b)$ such that $f(c)=d.$

I wonder if I change the hypothesis of $f(a)<d<f(b)$ to $f(a)>d>f(b)$, the result still holds. I believe so, since $f$ assumes a fixed point $f(x)=x$ in $[a,b]$, so we would have $c=d$, although I'm not completely sure.

I need this result in order to prove the set $X=\{x\in <!--\; \in \; -->[a,b]s.t.f|[a,x]\text{is bounded}\}$, with a continuous $f$, is not empty.