 # Master Intermediate Value Theorem Problems

Recent questions in Intermediate Value Theorem Mollie Wise 2022-11-27

## What is its worth ${15}^{\circ }$? hEorpaigh3tR 2022-11-25

## What is the value of $\frac{100}{0}$? 1) 02) 13) 64) Not defined Cherish Rivera 2022-11-24

## 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, Lorena Becker 2022-11-24

## 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 Nicholas Hunter 2022-11-23

## Suppose $f:\mathbb{R}\to \mathbb{R}$ is continuous and periodic with period 2a for some $a>0$; that is, $f\left(x\right)=f\left(x+2a\right)$ for all $x\in R$. Show there is some $c\in \left[0,a\right]$ such that $f\left(c\right)=f\left(c+a\right)$.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 $f\left(x\right)=f\left(x+2a\right)$. Therefore, if i can show that $f\left(c+a\right)=f\left(c+2a\right)$ or $f\left(c+a\right)-f\left(c+2a\right)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f\left(2a\right)-f\left(3a\right)$, but I can't show that's less than, equal to, or greater than 0. Any ideas? Barrett Osborn 2022-11-23

## Use the intermediate value theorem to prove that if $f:\left[1,2\right]\to R$ is a continuous function, that there is at least one number $c$ in the interval $\left(1,2\right)$ such that $f\left(c\right)=1/\left(1-c\right)+1/\left(2-c\right)$This is a question for my intro calc class that I am having a hard time understanding. Jared Lowe 2022-11-20

## The intermediate value theorem states the following: Consider an interval $I=\left[a,b\right]$ in the real numbers $ℝ$ and a continuous function $f:I\to ℝ$. Then, If $u$ is a number between $f\left(a\right)$ and $f\left(b\right)$, $f\left(a\right) (or $f\left(a\right)>u>f\left(b\right)$ ),then there is a $c\in \left(a,b\right)$ such that $f\left(c\right)=u$.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 $f:I\to \mathbb{R}$ with $I=\left[a,b\right]$, but here it is different. ajakanvao 2022-11-17

## The statement for Intermediate Value Theorem is thus:Let $f$ denote a function that is continuous on the closed interval $\left[a,b\right]$ and suppose $f\left(a\right)\ne f\left(b\right)$. If $N$ is any number between $f\left(a\right)$ and $f\left(b\right)$, then there is at least one number $c$ in $\left(a,b\right)$ so that $f\left(c\right)=N$.But who is to stop me if I put it like this?Let $f$ denote a function that is continuous on the open interval $\left(a,b\right)$. Now, $\underset{x\to {a}^{+}}{lim}f\left(a\right)=L$ and $\underset{x\to {b}^{-}}{lim}f\left(b\right)=M$. Suppose, $L\ne M$. If N is any number between $L$ and $M$, then there is at least one number $c$ in $\left(a,b\right)$ so that $f\left(c\right)=N$. Uroskopieulm 2022-11-17

## let $f:\left[0,a\right]\to \mathbb{R}$ of class ${C}^{1}$ such that $f\left(0\right)=0$ show that$\mathrm{\exists }c\in \right]0,a\left[,{f}^{\prime }\left(c\right)=\frac{2f\left(a\right)+a{f}^{\prime }\left(a\right)}{3a}$First I apply mean value theorem then $\mathrm{\exists }b\in \right]0,a\left[,{f}^{\prime }\left(b\right)=\frac{f\left(a\right)-f\left(0\right)}{a-0}=\frac{f\left(a\right)}{a}$ then $f\left(a\right)=a{f}^{\prime }\left(b\right)$so$\frac{2f\left(a\right)+a{f}^{\prime }\left(a\right)}{3a}=\frac{2}{3}{f}^{\prime }\left(b\right)+\frac{1}{3}{f}^{\prime }\left(a\right)$How to prove that $\frac{2}{3}{f}^{\prime }\left(b\right)+\frac{1}{3}{f}^{\prime }\left(a\right)$ is between ${f}^{\prime }\left(a\right)$ and ${f}^{\prime }\left(b\right)$? sbrigynt7b 2022-11-16

## Let $f$ be continuous on $\left[0,1\right]$ and $f\left(0\right)=f\left(1\right)$. Also, $n\in \mathbb{N}$. Prove that there exists some $x$ s.t. $f\left(x\right)=f\left(x+\frac{1}{n}\right)$.I think I need to assume towards a contradiction that if $g\left(x\right)=f\left(x\right)-f\left(x+\frac{1}{n}\right)$, then $g\left(x\right)\ne 0$ $\mathrm{\forall }x$.So, $g\left(0\right)=f\left(0\right)-f\left(\frac{1}{n}\right)$, $g\left(1\right)=f\left(1\right)-f\left(1+\frac{1}{n}\right)$. Since $f\left(0\right)=f\left(1\right)$,$g\left(0\right)-g\left(1\right)=f\left(1+\frac{1}{n}\right)-f\left(\frac{1}{n}\right)$. 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. Jared Lowe 2022-11-15

## I'm trying to solve the below exercise.Let $f:\left[0,1\right]\to \left[0,1\right]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a\in A$ such that $f\left(a\right)=a$. Is the same true for discontinuous functions?Here is my attempt.Notice that if $f\left(0\right)=0$ or $f\left(1\right)=1$, the theorem is proved. Suppose not. Define $g\left(x\right)=f\left(x\right)-x$, which is a continuous function from $\left[0,1\right]$ to $\left[0,1\right]$. Then since $f\left(0\right)\ne 0$, $g\left(0\right)\ne 0$, so $g\left(0\right)>0$. Since $f\left(1\right)\ne 1$, $g\left(1\right)<0$. By the intermediate value theorem, there exists $a\in \left[0,1\right]$ such that $g\left(a\right)=0$. So $f\left(a\right)-a=0$, so $f\left(a\right)=a$.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? Jaiden Elliott 2022-11-12

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

Calculus 2Open question deal4markaz 2022-11-07

## Q:   State And Prove Rolles Theorem With Diagram? Taniya Melton 2022-10-31

## As the title suggests I am trying to prove that for all $c\in \left[0,\mathrm{\infty }\right)$ there exists $x\in \mathbb{R}$ 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 $x{e}^{x}=f\left(x\right)$, but this function is defined on $\mathbb{R}\to \mathbb{R}$ and the IVT requires that our domain be $\left[a,b\right]$ with $a,b\in \mathbb{R}$. 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. Josiah Owens 2022-10-26

## Prove that there exists a real number $x$ such that ${x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}\left(x\right)}=125$ using Intermediate Value Theorem. Paola Mayer 2022-10-22

## Let $f:\left[0,2\right]\to \mathbb{R}$ be a continuous function such that $f\left(0\right)=f\left(2\right)$. Use the intermediate value theorem to prove that there exist numbers $x,y\in \left[0,2\right]$ such that $f\left(x\right)=f\left(y\right)$ and $|x-y|=1$.Hint: Introduce the auxiliary function $g:\left[0,1\right]\to \mathbb{R}$ defined by $g\left(x\right)=f\left(x+1\right)-f\left(x\right)$. Antwan Perez 2022-10-21

## Suppose that $f,g:\left[a,b\right]\to \mathbb{R}$ are continuous functions with $f\left(a\right) and $f\left(b\right)>g\left(b\right)$. Prove that there exists an $x\in \left(a,b\right)$ with $f\left(x\right)=g\left(x\right)$. I believe this uses the Intermediate Value Theorem. Eliza Gregory 2022-10-20

## I know every continuous function from $\left[0,1\right]$ to $\left[0,1\right]$ has fixed point. I proved it using Intermediate value theorem. But I can not see why every continuous function from $\left[0,1\right]$ to $\mathbb{R}$ has fixed point.Is this result even true? ajanlr 2022-10-15
## $f:\left[0;1\right]\to \mathbb{R}$ is continuous andProof that there exists an $c\in \left[0,1\right]$ such that $f\left(c\right)=c$.Idea: So i would define a function $g\left(x\right)=f\left(x\right)-x$ and then find points so that $g\left(x\right)$ would be $g\left(y\right)<0$ and $g\left(z\right)>0$. 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 $f\left(x\right)$ for which the integral would work like $f\left(x\right)=\frac{1}{2}$? Aydin Jarvis 2022-10-13