Master Intermediate Value Theorem Problems

Trying to take a different approach in proving if $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous and injective, then the maximum value of $f$ must occur at $a$ or $b$. trying to prove this by using the intermediate value theorem.
Here is what I have so far:
Let $b\in <!--\; \in \; -->\mathbb{R}$ Assume to the contrary that the max value occurs at an interior point of the interval $[a,b]$. Let $x_{\circ <!--\; \circ \; -->}$ equal this interior point.
The first case would occur when $f(a)<f(b)$ then $f(a)<f(b)<f(x_{\circ <!--\; \circ \; -->})$ If we consider $y_{\circ <!--\; \circ \; -->}\in <!--\; \in \; -->(f(b),f(x_{\circ <!--\; \circ \; -->}))$ is there a way to show that $y_{\circ <!--\; \circ \; -->}$ has at least 2 preimages by applying the intermediate value theorem?
Of course the second case would be $f(b)<f(a)$...

Show that the following equation has exactly one real root:
$3x+2\cos <!--\; \; -->x+5=0$

So I think the way to approach this problem is to use a combination of the intermediate value theorem and the mean value theorem.
But since we are not given a specific interval, I'm not entirely sure what values of $f(a)$ and $f(b)$ to choose. Am I suppose to choose $f(0)$ and $f(1)$ and see that 0 lies in between them. If so, how exactly can we proceed with the mean value theorem after using the intermediate value theorem. Do we have to check for continuity and the number c, or is there some other way?

Found it:
Let $I$ be a real interval. Let $a,b\in <!--\; \in \; -->I$ such that $(a..b)$ is an open interval. Let $f:I\to <!--\; \to \; -->\mathbb{R}$ be a real function which is continuous on $(a..b)$. Let $k\in <!--\; \in \; -->\mathbb{R}$ lie between $f(a)$ and $f(b)$. That is, either $f(a)<k<f(b)$ or $f(b)<k<f(a)$. Then $\mathrm{\exists <!--\; \exists \; -->}c\in <!--\; \in \; -->(a..b)$ such that $f(c)=k$.

Shouldn't this suppose continuity on the closed interval $[a,b]$? Otherwise consider $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->\mathbb{R}$ such that
$f(x)=\{\begin{array}{ll}\arcsin <!--\; \; -->(x)& \text{-1<x<1}\\ 23& \text{if x=1}\\ -<!--\; -\; -->23& \text{if x= -1}\end{array}$
where $a=-<!--\; -\; -->1$ and $b=1$.

Let $f:[0,1]\to <!--\; \to \; -->\mathbb{R}$ continuous and let $f([0,1])\subset <!--\; \subset \; -->[0,1]$. By using intermediate value theorem, show that $f$ has at least got one fixed point in $[0,1]$ (aka there is one solution $x\in <!--\; \in \; -->[0,1]$ of the equation $f(x)=x$).

I'm not sure at all if I did it right but I have taken the function:
$f(x)=x$
Then take the given intervals and insert for $x$ the beginning of interval:
$f(0)=0$
And the end of the interval:
$f(1)=1>0$
Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution $x_1=[0,1]$.
Did I do it correctly? Did I explain correctly?

The function $g:(0,\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->\mathbb{R}$, is continuous, $g(1)>0$ and
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}g(x)=0$
It is a fact that for every $y$ between 0 and $g(1)$ the function takes on a value in $(y,g(1))$
How would one show that if:
$\underset{e\to <!--\; \to \; -->0+}{lim}sup\{g(x):0<x<e\}=0$
then $g$ attains a maximum value on 0 to infinity. Does $g$ necessarily have to be bounded below for this to hold? Why?

If $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous over the entire domain Ms is not bounded above or below, show that $f(\mathbb{R})=\mathbb{R}.$

My approach would be to say that since $f$ is continuous on $\mathbb{R}$ it must be continuous on $[a,b],a,b\in <!--\; \in \; -->\mathbb{R}$ and using the intermediate value theorem this tells us that $f$ must take every value between $f(a)$ and $f(b)$. Now I want to argue that we can make this closed bounded interval as large as we want and the result still holds and I want to show that this implies that we can make the values of $f(a)$ as small as we like and $f(b)$ as large as we like (or the other way round) since $f$ is neither bounded below or above but I'm not sure if this argument can used to show that the image of $f$ is $\mathbb{R}$ since we switch from closed bounded intervals to an unbounded interval.

Does my argument hold to show the result?

Show that $\sqrt{1+x}<1+\frac{x}{2}$ if $x>0$

Let $f$ be continuous on $[0,1]$ and such that $f(0)<0$ and $f(1)>1$. Prove that there exists $c\in <!--\; \in \; -->(0,1)$ such that $f(c)=c^4$

The Intermediate Value Theorem states that over a closed interval $[a,b]$ for line $L$, that there exists a value $c$ in that interval such that $f(c)=L$.
We know both functions require $x>0$, however this is not a closed interval.
However, I went ahead on the problem anyway. I decided to solve for x.
$2^x=\frac{10}{x}$
$x{2}^x=10$
$x^{2}log2=log10$
$x^2=\frac{1}{log2}$
$x=\sqrt{\frac{1}{log2}}$
However, when graphing these two functions I found that they both met up at $\sim <!--\; \sim \; -->2.236$, which is not equal to $x=\sqrt{\frac{1}{log2}}$
I am very new to Calculus and the Intermediate Value theorem and would appreciate a simple explanation. Thank you.

A number $a$ is called a fixed point of a function $f$ if $f(a)=a$. Consider the function $f(x)=x^{87}+4x+2$, $x\in <!--\; \in \; -->\mathbb{R}$.
(a) Use the Mean Value Theorem to show that $f(x)$ cannot have more than one fixed point.
(b) Use the Intermediate Value Theorem and the result in (a) to show that $f(x)$ has exactly one fixed point.

Is there an intermediate value like theorem for $\mathbb{C}$? I know $\mathbb{C}$ isn't ordeblack, but if we have a function $f:\mathbb{C}\to <!--\; \to \; -->\mathbb{C}$ that's continuous, what can we conclude about it?
Also, if we have a function, $g:\mathbb{C}\to <!--\; \to \; -->\mathbb{R}$ ,continuous with $g(x)>0>g(y)$ does that imply a z "between" them satisfying $g(z)=0$.
Edit: I apologize if the question is vague and confusing. I really want to ask for which definition of between, (for example maybe the part of the plane dividing the points), and with relaxed definition on intermediateness for the first part, can we prove any such results?

Let $f\in <!--\; \in \; -->L^1({\mathbb{R}}^2)$ with respect to the Lebesgue measure $m\times <!--\; \times \; -->m$ on ${\mathbb{R}}^2$. Prove that if
${\iint <!--\; \iint \; -->}_{{\mathbb{R}}^2}f(x,y)dxdy=0,$
then there exits a square $S_{a,b}=\{(x,y)\mid <!--\; \mid \; -->a\le <!--\; \le \; -->x\le <!--\; \le \; -->a+1,b\le <!--\; \le \; -->y\le <!--\; \le \; -->b+1\}$, such that
${\iint <!--\; \iint \; -->}_{S_{a,b}}f(x,y)dxdy=0.$
I tried to show that the integral
${\iint <!--\; \iint \; -->}_{[a,a+x]\times <!--\; \times \; -->[b,b+y]}f(s,t)dsdt$
is absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.
Is there any theorem for the two-dimensional case?

I am currently doing a problem that asks me to use the intermediate value theorem to show that $x^{1/3}=1-<!--\; -\; -->x$ lies between 0 and 1. I want to start by evaluating the function at 0 and 1, but it seems the function is undefined because if you plug in 1, you get $1=0$. This doesn't seem right. Is my interpretation correct?

Use the intermediate value theorem to prove that if $f:[0,1]\to <!--\; \to \; -->[0,1]$ is continuous, then there exists $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=\sqrt{c}$
Suppose that $f(0)<f(1)$. Consider now, $f(0)<\sqrt{c}<f(1)$. By the intermediate value theorem, $\mathrm{\exists <!--\; \exists \; -->}b\in <!--\; \in \; -->[0,1]:f(b)=\sqrt{c}$.
Now we need to show that its possible that $b=c$, how?

Spivak saves the proof of the intermediate value theorem for his chapter on supremums and infemums.
His proof makes clear use of the completeness of reals, but why I struggle to see why it is dependent on completeness and not just continuity.

Let $h:[0,1]\to <!--\; \to \; -->\mathbb{R}$ be continuous. Prove that there exists $w\in <!--\; \in \; -->[0,1]$ such that
$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$
I tried several things with this problem. I first tried a new function
$\begin{array}{rl}g(x)& =h(x)-<!--\; -\; -->\frac{x+1}{2}h(0)+\frac{2x+2}{9}h\left(\frac{1}{2}\right)+\frac{x+1}{12}h(1)\\ & =h(x)-<!--\; -\; -->(x+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)).\end{array}$
Evaluating $g(0)$ and $g(1)$ didn't really work.

Since I want to find a point $x_0$ where $g(x_0)>0$ and another $x_1$ where $g(x_1)<0$, I started thinking how I can make $g(x_0)>0$,or
$h(x_0)-<!--\; -\; -->(x_0+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1))>0,$
or
$\frac{h(x_0)}{x_0+1}>\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1).$
Since the LHS is the gradient from the point $(-<!--\; -\; -->1,0)$ to $(x_0,h(x_0))$, I need to somehow prove that that
$\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)$
is always less than the steepest gradient from $(-<!--\; -\; -->1,0)$ to $(x_0,h(x_0))$. However, I am stuck here. Any help?

I know I have to use the intermediate value theorem but how?

I think that, since the since $g(x)=\cos <!--\; \; -->(x)$ and $g(x)=x^3$ are continuous everywhere, the function $f(x)=\cos <!--\; \; -->(x)-<!--\; -\; -->x^3$ must also be continuous everywhere, and therefore, according to the Intermediate Value Theorem, $\cos <!--\; \; -->(x)=x^3$ must have a solution. However, I'm not sure if that's true.
How can I show that $\cos <!--\; \; -->(x)=x^3$ has a solution?

Question: Sketch all the continuous functions $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ which satisfy
$(f(x){)}^2=(x-<!--\; -\; -->1{)}^2(x-<!--\; -\; -->2{)}^2.$
Justify your answers.
I have found the eight possible continuous functions as follows:
$\begin{array}{rl}f(x)& =(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)\\ f(x)& =-<!--\; -\; -->(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)\\ f(x)& =|x-<!--\; -\; -->1|(x-<!--\; -\; -->2)\\ f(x)& =-<!--\; -\; -->|x-<!--\; -\; -->1|(x-<!--\; -\; -->2)\\ f(x)& =(x-<!--\; -\; -->1)|x-<!--\; -\; -->2|\\ f(x)& =-<!--\; -\; -->(x-<!--\; -\; -->1)|x-<!--\; -\; -->2|\\ f(x)& =|x-<!--\; -\; -->1||x-<!--\; -\; -->2|\\ f(x)& =-<!--\; -\; -->|x-<!--\; -\; -->1||x-<!--\; -\; -->2|.\end{array}$
and have got the conclusion of this question which has at most eight possible such functions, but how to proof the "at most" statement.

I received a question on a previous exam, but I had no clue how to go about doing it. I know I'm supposed to use the MVT, IVT and FTC, but I'm not sure where. The question is
Suppose $f(x)$ is integrable on $[a,b]$, with $f(x)\ge <!--\; \ge \; -->0$ on $[a,b]$, and that $g(x)$ is continuous on $[a,b]$. Assuming that $f(x)g(x)$ is integrable on $[a,b]$, show that $\mathrm{\exists <!--\; \exists \; -->}c\in <!--\; \in \; -->[a,b]$ so that
${\int <!--\; \int \; -->}_a^{b}f(x)g(x)dx=g(c){\int <!--\; \int \; -->}_a^{b}f(x)dx.$
Thank you in advance.