Master Intermediate Value Theorem Problems

Let $f(x)$ be a continuous function on $[a,b]$ and $f^{\mathrm{\prime <!--\; \prime \; -->}}(a)$ exists. Let $\mathrm{\xi <!--\; \xi \; -->}$ be a number such that
$f^{\mathrm{\prime <!--\; \prime \; -->}}(a)>\mathrm{\xi <!--\; \xi \; -->}>\frac{f(b)-<!--\; -\; -->f(a)}{b-<!--\; -\; -->a}$
Prove that there is a $c\in <!--\; \in \; -->(a,b)$ such that
$\frac{f(c)-<!--\; -\; -->f(a)}{c-<!--\; -\; -->a}=\mathrm{\xi <!--\; \xi \; -->}$
I try to use Intermediate Value Theorem to show this. I let $g(x)=\frac{f(x)-<!--\; -\; -->f(a)}{x-<!--\; -\; -->a}$. I try to show this function is continuous on $[a,b]$ but I don know how to show it continuous at endpoint.

I have a question regarding Intermediate Value theorem. I feel like I understand the concept but I'm having trouble understanding a particular problem. I'm not what this problem is asking me to do, perhaps someone could clear it up for me. It reads as follows:
Show that if f is continuous on [0, 1] and $0\le <!--\; \le \; -->f(x)\le <!--\; \le \; -->1$ for all x on [0, 1], then there exists at least one point c in [0, 1] at which f(c) = c. HINT: Apply the Intermediate Value theorem to the function $g(x)=x–f(x)$.
Obviously the last part is important, but I don't know how it all fits. This seems fairly different than proving that there lies a root of a function between two points.

I'm aware that $\mathbb{R}$ is not the only set that satisfies the least upper bound property, the $p$-adics do also. Does the intermediate value theorem also hold in the $p$-adics then?

What are the spaces where the intermediate value theorem hold?

In the context of a measure space $(X,M,\mu )$, suppose $f$ is a bounded measurable function with $a\le <!--\; \le \; -->f(x)\le <!--\; \le \; -->b$ for $\mathrm{\mu <!--\; \mu \; -->}$-a.e. $x\in <!--\; \in \; -->X$. Prove that for each integrable function $g$, there exists a number $c\in <!--\; \in \; -->[a,b]$ such that ${\int <!--\; \int \; -->}_{X}f|g|d\mathrm{\mu <!--\; \mu \; -->}=c{\int <!--\; \int \; -->}_X|g|d\mathrm{\mu <!--\; \mu \; -->}$

I tried to use the Intermediate value theorem for integral of Riemann but i had no idea. Somebody have any tip?

I am trying to understand and prove the fundamental theorem of calculus and I ran into some confusion understanding the intermediate value theorem . several sources online claim that if a function f(x) is continuous on [a,b] let s be a number such that $f(a)<s<f(b)$ then there exists a number k in the open interval (a,b) such that f(k)=s my question is why do we only assume the open interval shouldn't it also include the closed interval [a,b] and also why does s have to be less than both $f(a)$ and $f(b)$?

I want to prove that for a continuous function mapping a connected space to ℝ such that f(p) never equals s, it follows that f(p) < s for all p or f(p) > s for all s.
So here's what I know so far:
-Because f is continuous and the metric space M is connected, f(M) is connected.
-f(M) is connected in the real numbers, so if x,$y\mathrm{ϵ<!--\; ϵ\; -->}f(M)$ and x < z < y, then $z\mathrm{ϵ<!--\; ϵ\; -->}f(M)$
-I have a feeling the intermediate value theorem will play a part in this, but I'm not sure how.
I'm having trouble relating these three points.

Use the Intermediate Value Theorem to show that the equation $x^3+x+1=0$ has a solution.
How to do this?

I have given real numbers $x_1,x_2,y_1,y_2$ such that $x_1>x_2$ and $y_1<y_2$. The the claim is that there exists some $\mathrm{\lambda <!--\; \lambda \; -->}\in <!--\; \in \; -->(0,1)$ such that $\mathrm{\lambda <!--\; \lambda \; -->}(x_1-<!--\; -\; -->x_2)+(1-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->})(y_1-<!--\; -\; -->y_2)=0$. In order to proof this, one needs ( at least in my opinion) the intermediate value theorem. But the intermediate value theorem does not hold in constructive mathematics (that is without the law of excluded middle; or constructive mathematics acts in intuitionistic logic). Is there any constructive way to show the above equation?

Let $f$ be a continuous function on $\mathbb{R}$ which is periodic with period $2\mathrm{\pi <!--\; \pi \; -->}$. This means $f(t+2\mathrm{\pi <!--\; \pi \; -->})=f(t)$ for all $t$. Show that there exists $x\in <!--\; \in \; -->[0,\mathrm{\pi <!--\; \pi \; -->}]$ such that $f(x)=f(x+\mathrm{\pi <!--\; \pi \; -->})$.
I know it's an intermediate-value theorem problem. I think I have to take the difference of both sides, but not quite sure. Can anyone help?

Suppose that $f$ is a continuous function and that $f(-<!--\; -\; -->1)=f(1)=0$. Show that there is $c\in <!--\; \in \; -->(-<!--\; -\; -->1,1)$ such that
$f(c)=\frac{c}{1-<!--\; -\; -->c^2}$
I am not sure if this is a new question as I set it this morning, after solved a similar question. I wanted to prove it using the same idea (Intermediate Value Theorem) but it was not that nice...

Can anyone help me on this?

By the way, the 'similar question' I mentioned above is as follow:

Suppose that $f$ is a continuous function and that $f(0)=1$ and $f(1)=2$. Show that there is $c\in <!--\; \in \; -->(0,1)$ such that
$f(c)=\frac{1}{c}$
The hint given was to let $g(x)=xf(x)$ and use the Intermediate Value Theorem.

Let $a,b\in <!--\; \in \; -->\mathbb{R}$, $a<b$ and let $f$ be a differentiable real-valued function on an open subset of $\mathbb{R}$ that contains [a,b]. Show that if $\mathrm{\gamma <!--\; \gamma \; -->}$ is any real number between $f^{\prime }(a)$ and $f^{\prime }(b)$ then there exists a number $c\in <!--\; \in \; -->(a,b)$ such that $\mathrm{\gamma <!--\; \gamma \; -->}=f^{\prime }(c)$.

Hint: Combine mean value theorem with the intermediate value theorem for the function $\frac{(f(x_1)-<!--\; -\; -->f(x_2))}{x_1-<!--\; -\; -->x_2}$ on the set $\{(x_1,x_2)\in <!--\; \in \; -->E^2:a\le <!--\; \le \; -->x_1<x_2\le <!--\; \le \; -->b\}$.

I am having a lot of trouble trying to start on this problem.

Given equation $x^{100}+x^{99}+x^{98}+...+x+1=5$ I need to determine whether there exists one solution $x$ such that $x>0$. Also, whether I have one solution $x<0$.

First, I let $a=0$ such that $f(a)=1$. Let $b=1$ such that $f(b)=101$. Since $f(a)<5<f(b)$, thus, by intermediate value theorem, there exits one solution $x\in <!--\; \in \; -->(0,1)$. Similarly, I can conclude that there exits one solution smaller than 0. Is my process correct?

I have been stuck on this Real Analysis problem for hours and am just totally clueless- I am sure it is some application of the Intermediate Value Theorem- suppose $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous at every point. Prove that the equation $f(x)=c$ cannot have exactly two solutions for every value of $c.$

Would appreciate some help

I have just recently covered the Intermediate Value Theorem, and I wanted to practice solving problems involving this theorem. However, I encountered a problem that I am not exactly sure how to tackle (it's a question involving a periodic function).

I thought about splitting the proofs into 3 cases, but I don't think it would be applicable here?

The question is:

If f is periodic with a period of 2a for some $a>0$, then $f(x)=f(x+2a)$ for all $x\in R$. Show that if f is continous, there exists some $c\in [0,a]$ such that $f(c)=f(c+a)$.

Any help would be greatly appreciated. Thanks a lot!

suppose we have following function
$f(x)=x^2+10\sin <!--\; \; -->(x)$
we should show that there exist number $c$ such that $f(c)=1000$

clearly we can solve this problem using intermediate value theorem for instance
$f(0)=0$
$f(90)={90}^2+10\sin <!--\; \; -->(90)=8100+10\ast <!--\; \ast \; -->1=8110$
and because 1000 is between these two number we can see that there exist such c so that $f(c)=1000$

am i right? thanks in advance

Let $f(x)$ be a monic polynomial of odd degree. Prove that $\mathrm{\exists <!--\; \exists \; -->}A\in <!--\; \in \; -->\mathbb{R}$ s.t. $f(A)<0$ and there exists $B\in <!--\; \in \; -->\mathbb{R}$ such that $f(B)>0$.

Deduce that every polynomial of odd degree has a real root.

There are questions that answer the final part, but they do not do so by proving the first part. I am fairly sure that this involves the intermediate value theorem, but not sure how to implement it in this case.

Let $I=[a,b]$ with $a<b$ and let $u:I\to <!--\; \to \; -->\mathbb{R}$ be a function with bounded pointwise variation, i.e.
$Va{r}_{I}u=sup\{\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}|u(x_i)-<!--\; -\; -->u(x_{i-<!--\; -\; -->1})|\}<\mathrm{\infty <!--\; \infty \; -->}$
where the supremum is taken over all partition $P=\{a=x_0<x_1<...<x_{n-<!--\; -\; -->1}<b=x_n\}$. How can I prove that if $u$ satisfies the intermediate value theorem (IVT), then $u$ is continuous?

My try: $u$ can be written as a difference of two increasing functions $f_1,f_2$. I know that a increasing function that satisfies the (ITV) is continuous, hence, if I prove that $f_1,f_2$ satisfies the (ITV) the assertion follows. But, is this true? I mean, $f_1,f_2$ satisfies (ITV)?

Suppose the function $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ is continuous. For a natural number $k$, let $x_1,x_2,...,x_k$ be points in $[a,b]$. Prove there is a point $z$ in $[a,b]$ at which
$f(z)=(f(x_1)+f(x_2)+...+f(x_k))/k$
So I'm thinking about applying the intermediate value theorem:

If
$a<x_1<b,a<x_2<b,...,a<x_k<b$
then
$f(a)<f(x_1)<f(b),...,f(a)<f(x_k)<f(b)$
or
$k.f(a)<f(x_1)+...+f(x_k)<k.f(b)$
$f(a)<(f(x_1)+f(x_2)+...+f(x_k))/k<f(b)$
But I couldn't think of any way to prove that $f(a)<f(x_k)<f(b)$ or is it even true?

EDIT: Thanks everyone for your effort.

Use the Intermediate Value Theorem to show that there is a root of equation
$x^3+2x-<!--\; -\; -->\frac{1}{x}=0$
in $(\frac{1}{4},1)$. Find this root to two decimal places.