Master Intermediate Value Theorem Problems

Consider the function:
$f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$
$f(x)=\frac{x}{|x|+1}$
I already showed that this function is continuous and injective. Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b $\in <!--\; \in \; -->$ $\mathbb{R}$. I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits.

Show that if $f$ is continuous on [0,1] with $f(0)=f(1)$, there must exist $x,y\in <!--\; \in \; -->[0,1]$ with $|x-<!--\; -\; -->y|=\frac{1}{2}$ and $f(x)=f(y)$
I've been working on this for a while, and can't seem to figure out where to start. Any hints would be appreciated

State and prove an intermediate value theorem for functions mapping into the digital line.


The digital line is defined as $\mathbb{Z}$ with the topology given by the basis elements $B(n)=\{n\},$ if $n$ odd, and $B(n)=\{n-<!--\; -\; -->1,n,n+1\}$ if $n$ even.

I really have no clue, where to start. I have tried just using the regular intermediate value theorem, but I cannot prove it in this case. Can anyone help me?

How would I solve the following two questions.
Using the intermediate value theorem to show that there is a solution of the equation $\frac{si{n}^{2}x}{2}-<!--\; -\; -->x+1=0$ in the interval $[0,pi]$
I showed by the IVT there is a c in $[0,pi]$ give that c is zero because $0<1$ $0>-<!--\; -\; -->pi+1$ but I am not sure if it did this correctly.
2.My second question asked me to sketch a function that satisfies the following conditions or prove that it is impossible $f(x)$ is a continous function on [0,2] its minimum value is −3 but it does not have a max value.
I said this is impossible because if $f(x)$ is continous on a bounded interval it must have both a max and min value.

$f$ is continuous function in $[0,2]$ and $f(2)=1$
proof that there is a point $x$ in $[0,2]$ so that:
$f(x)=\frac{1}{x}$
nobody in my class solved this exercise. our teacher give us a hint: $p(x)=x\ast <!--\; \ast \; -->f(x)$ and to do Intermediate value theorem on this function. so I tried:
$p(0)=0\ast <!--\; \ast \; -->f(0)=0,P(2)=2\ast <!--\; \ast \; -->f(2)=2$
$p(0)<h<p(2)$ by Intermediate value theorem, there is a point $c$ so that $a<c<b$ and $f(c)=h$
please solve

Show that $\tan <!--\; \; -->x-<!--\; -\; -->2x$ has a root in (0,1.4).
I got: $f(x)=\tan <!--\; \; -->x-<!--\; -\; -->2x$
$f(0)=0$, $f(1.4)=-<!--\; -\; -->2.775$
Does this mean there is no root in (0,1.4)? Because I got −2.775 and it did not fall in (0,1.4).

Let $X$ be a compact metric space (feel free to impose more conditions as long as they're also satisfied by spheres) and $F:X\times <!--\; \times \; -->[0,1]\to <!--\; \to \; -->\mathbb{R}$ a continuous function such that

1. $F(x,0)>0$ for all $x\in <!--\; \in \; -->X$
2. $F(x,1)<0$ for all $x\in <!--\; \in \; -->X$

Then, for each $x\in <!--\; \in \; -->X$, let $t_0(x)\in <!--\; \in \; -->[0,1]$ be the smallest such that $F(x,t_0(x))=0$. Is $t_0:X\to <!--\; \to \; -->[0,1]$ a continuous function?

A friend suggested that I applied the Maximum theorem, but to show that the relevant correspondence is lower semicontinuous I need to prove the following statement:

If $x_n\to <!--\; \to \; -->x$ and $F(x,t)=0$, there is a subsequence $x_{n_k}$ and a sequence $t_k$ such that $F(x_{n_k},t_k)=0$ and $t_k\to <!--\; \to \; -->t$. This doesn't seem very obvious or even true, but I'm not sure.

Any suggestions?

$f(x)=\frac{1}{\sqrt{1-<!--\; -\; -->|x|}}+\frac{1}{x^4}$
I was asked to show though the intermediate value theorem that $f(x)$ has at least one solution when $f(x)=314$. I found that $f(x)$ is continuous when $-<!--\; -\; -->1<x<0$ and $0<x<1$. The problem I'm having is that I thought that the theorem only worked for closed intervalls. Any tips are greatly appreciated!

Let $f$ be measurable in $\mathbb{R}$ and ${\int <!--\; \int \; -->}_{\mathbb{R}}fdx=1$ Show that for any $r\in <!--\; \in \; -->(0,1)$, there is a Lebesgue measurable set $E$ such that ${\int <!--\; \int \; -->}_{E}fdx=r$.

I have tried the interval $(-<!--\; -\; -->R,R)$. If I can show that $R\mapsto <!--\; \mapsto \; -->{\int <!--\; \int \; -->}_{-<!--\; -\; -->R}^{R}fdx$ is continuous and I can use the intermediate value theorem to show that there is a $R$ such that ${\int <!--\; \int \; -->}_{-<!--\; -\; -->R}^{R}fdx=r$. If I consider $|\int <!--\; \int \; -->f({\mathrm{\chi <!--\; \chi \; -->}}_{(-<!--\; -\; -->R_1,R_1)}-<!--\; -\; -->{\mathrm{\chi <!--\; \chi \; -->}}_{(-<!--\; -\; -->R_2,R_2)})dx|$, I do not know to show the continuous since I think that ${\int <!--\; \int \; -->}_{-<!--\; -\; -->R}^{R}fdx$ may not be finite for some $R$ or $\int <!--\; \int \; -->|f|dx$ may not be finite.

I've an homework question where i need to prove that the following equation contains at-least three roots $\frac{x^4}{10}=\frac{x^4-<!--\; -\; -->100}{x-<!--\; -\; -->1}$.
I was able to find three roots after blackefining the equation as function: $f(x)=\frac{x^4-<!--\; -\; -->100}{x-<!--\; -\; -->1}-<!--\; -\; -->\frac{x^4}{10}$ where the segments:
1. [3.3,3.4] because $f(3.3)>0$ and $f(3.4)<0$
2. [10.9,11] because $f(10.9)>0$ and $f(11)<0$
3. [−3,−2] because $f(-<!--\; -\; -->3)<0$ and $f(-<!--\; -\; -->2)>0$
I know the Intermediate-Value theorem state that for each mentioned segment there has to be a 0 (a root).
I would like to know if my answer is correct and if so how can i prove that the function is continuous at the mentioned segments.
UPDATE: The roots are indeed in the above mentioned segments but in the open segment like:
1. (3.3,3.4)
2. (10.9,11)
3. (−3,−2)

Is the Intermediate Value Theorem basically saying that if a function is continuous on an interval, then the function is surjective?
The formal definition states something to the effect of "any value in the domain will map to a value in the range", unless I misunderstand it. So, since we're mapping from $[a,b]\to <!--\; \to \; -->[f(a),f(b)]$, the range is made up of the images of the values in the domain, that is, the range is the codomain.
My understanding of a surjective function is one in which $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->D,f(x)\in <!--\; \in \; -->C$, where $f:D\to <!--\; \to \; -->C$.
Or an I horribly mistaken?

Suppose that $(A,\mathrm{\Sigma <!--\; \Sigma \; -->},m)$ is a measure space and $H$ is a linear functional on $L^{\mathrm{\infty <!--\; \infty \; -->}}(A,\mathrm{\Sigma <!--\; \Sigma \; -->},m)$. If
$\mathcal{U}:=\{u:A\to <!--\; \to \; -->\mathbb{R}|u\text{ is measurable, bounded and }{\int <!--\; \int \; -->}_{A}udm=1\}$
and there are functions $u_1,u_2\in <!--\; \in \; -->\mathcal{U}$ such that
$H(u_1)\le <!--\; \le \; -->0\text{and}H(u_2)\ge <!--\; \ge \; -->0,$
then my question is:

Is there a function $u_3\in <!--\; \in \; -->\mathcal{U}$ such that $H(u_3)=0$?

Suppose $f:[1,2]\to <!--\; \to \; -->[5,7]$ is continuous. Show that $f(c)=2c+3$ for some $c\in <!--\; \in \; -->[1,2]$.
First note $f(1)=5$ and $f(2)=7$. By the IVT, all values $c\in <!--\; \in \; -->[1,2]$ are hit. I'm just wondering how to put all of these facts together to arrive at f(c)=2c+3 for all $c\in <!--\; \in \; -->[1,2]$.

Note: I know how to prove this for $f(a)=f(a+100)$, but no clue how to do it for the value that's asked of me.
There's a continuous function $f(x):\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$
With these properties:
1. $f(x)\le 2\text{for}x\ge 200$
2. $f(x)\le 1\text{for}x\le 100$
3. $f(x)=3\text{for}x=150$
I'm supposed to show that there exists an $a$ for which the property $f(a)=(a+200)$ is true. I've tried applying the intermediate value theorem every which way, but at this point I don't know if that's the right way to go.

$f(x)=x^2\exp <!--\; \; -->(\sin <!--\; \; -->(x))-<!--\; -\; -->\cos <!--\; \; -->(x)$ on the interval $[0,\mathrm{\pi <!--\; \pi \; -->}/2]$, I have shown the function is continuous and that there is at least one solution on the interval via using IVT, I know I have to find another solution in the interval such that $f(x)<0$, $f(x)>0$ and then $f(x)<0$, just wondering what is the best way to approach, Im sure there must be a more effective way than just to number crunch, could we consider MVT on the interval, many thanks in advance.

Use the Intermediate Value Theorem to prove $f:[0,1]\to <!--\; \to \; -->[0,1]$ continuous and $C\in <!--\; \in \; -->[0,1]$, there is some $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=C$.
Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an $f(x)=C$, but I am having trouble proving that a $f(c)=C$.
This is what I have:
Since $f$ is continuous on [0,1] there exists $f(a)=0$ and $f(b)=1$.
Let $g(x)=F(x)-<!--\; -\; -->C$. We assume that $f(a)<C<f(b)$ $\to <!--\; \to \; -->$ $0<C<1$
when $x=a$, $g(a)$ is negative when $x=b$, $g(b)$ is positive
Therefore, $g(a)<0<g(b)$, and since $f$ is continuous on [0,1], so is $g$.
Therefore there exists a $g(x)=0$, and $f(x)=C$.
How can I prove there is a $f(c)=C$ ? Is this a rule for IVT?

How can I prove
if $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous and $\stackrel{¯<!--\; ¯\; -->}{f(\mathbb{N})}=\mathbb{R}$, then $f$ is onto (surjective)
by the intermediate value theorem?

I want to give an informal proof of the intermediate value theorem to my calculus students. They have just learned about continuity and they don't know yet about derivatives among more other advanced topics.
Do you think it's possible?

Show that a path going from one vertex of the unit square (in $\mathbb{C}$) to its opposite and a second path going between the other pair of opposite vertices must intersect. By path I mean continuous function $f:[0,1]\to <!--\; \to \; -->\mathbb{C}$ with $f(0)$ one vertex and $f(1)$ the other. [EDIT] Both are paths in the unit square (no going outside)
I looked through other questions and didn't quite find an answer to this. I know that the intermediate value theorem says that $\text{Im}((f-<!--\; -\; -->g)(t))$ and $\text{Re}((f-<!--\; -\; -->g)(t))$ must be 0 at some point, but I'm not sure how to show that this must occur at the same point.

For example, to prove that the function $x^8+x-1=0$ has exactly two roots, we first prove that $f$ has at most 2 real roots by using differentiation. $f^{\prime }(x)=8x^7+1=0$, by using that the function has two roots and therefore it should have at least one point such that $f\prime (c)=0$ and then we use the IVT to prove that it has exactly two roots.
However, the derivative of equation $x^4-6x^2-8x+1=0$ has 2 roots but still, I should try to prove that it has exactly two roots. If its derivative had one solution, then I could first prove that it has two roots at most as I did but it has two roots, so how should we approach the problem?