Recent questions in Intermediate Value Theorem

Calculus 2Answered question

hEorpaigh3tR 2022-11-25

What is the value of $\frac{100}{0}$?

1) 0

2) 1

3) 6

4) Not defined

1) 0

2) 1

3) 6

4) Not defined

Calculus 2Answered question

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,

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,

Calculus 2Answered question

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

Calculus 2Answered question

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, $f(x)=f(x+2a)$ for all $x\in R$. Show there is some $c\in [0,a]$ such that $f(c)=f(c+a)$.

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 $f(x)=f(x+2a)$. Therefore, if i can show that $f(c+a)=f(c+2a)$ or $f(c+a)-f(c+2a)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f(2a)-f(3a)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

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 $f(x)=f(x+2a)$. Therefore, if i can show that $f(c+a)=f(c+2a)$ or $f(c+a)-f(c+2a)=0$, I'd be done. But I just don't know where to go from there. I tried substituting $x=a$, getting $f(2a)-f(3a)$, but I can't show that's less than, equal to, or greater than 0. Any ideas?

Calculus 2Answered question

Barrett Osborn 2022-11-23

Use the intermediate value theorem to prove that if $f:[1,2]\to R$ is a continuous function, that there is at least one number $c$ in the interval $(1,2)$ such that $f(c)=1/(1-c)+1/(2-c)$

This is a question for my intro calc class that I am having a hard time understanding.

This is a question for my intro calc class that I am having a hard time understanding.

Calculus 2Answered question

Jared Lowe 2022-11-20

The intermediate value theorem states the following: Consider an interval $I=[a,b]$ in the real numbers $\mathbb{R}$ and a continuous function $f:I\to \mathbb{R}$. Then, If $u$ is a number between $f(a)$ and $f(b)$, $f(a)<u<f(b)$ (or $f(a)>u>f(b)$ ),

then there is a $c\in (a,b)$ such that $f(c)=u$.

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 $I=[a,b]$, but here it is different.

then there is a $c\in (a,b)$ such that $f(c)=u$.

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 $I=[a,b]$, but here it is different.

Calculus 2Answered question

ajakanvao 2022-11-17

The statement for Intermediate Value Theorem is thus:

Let $f$ denote a function that is continuous on the closed interval $[a,b]$ and suppose $f(a)\ne f(b)$. If $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

But who is to stop me if I put it like this?

Let $f$ denote a function that is continuous on the open interval $(a,b)$. Now, $\underset{x\to {a}^{+}}{lim}f(a)=L$ and $\underset{x\to {b}^{-}}{lim}f(b)=M$. Suppose, $L\ne M$. If N is any number between $L$ and $M$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

Let $f$ denote a function that is continuous on the closed interval $[a,b]$ and suppose $f(a)\ne f(b)$. If $N$ is any number between $f(a)$ and $f(b)$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

But who is to stop me if I put it like this?

Let $f$ denote a function that is continuous on the open interval $(a,b)$. Now, $\underset{x\to {a}^{+}}{lim}f(a)=L$ and $\underset{x\to {b}^{-}}{lim}f(b)=M$. Suppose, $L\ne M$. If N is any number between $L$ and $M$, then there is at least one number $c$ in $(a,b)$ so that $f(c)=N$.

Calculus 2Answered question

Uroskopieulm 2022-11-17

let $f:[0,a]\to \mathbb{R}$ of class ${C}^{1}$ such that $f(0)=0$ show that

$\mathrm{\exists}c\in ]0,a[,{f}^{\prime}(c)=\frac{2f(a)+a{f}^{\prime}(a)}{3a}$

First I apply mean value theorem then $\mathrm{\exists}b\in ]0,a[,{f}^{\prime}(b)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)}{a}$ then $f(a)=a{f}^{\prime}(b)$

so

$\frac{2f(a)+a{f}^{\prime}(a)}{3a}=\frac{2}{3}{f}^{\prime}(b)+\frac{1}{3}{f}^{\prime}(a)$

How to prove that $\frac{2}{3}{f}^{\prime}(b)+\frac{1}{3}{f}^{\prime}(a)$ is between ${f}^{\prime}(a)$ and ${f}^{\prime}(b)$?

$\mathrm{\exists}c\in ]0,a[,{f}^{\prime}(c)=\frac{2f(a)+a{f}^{\prime}(a)}{3a}$

First I apply mean value theorem then $\mathrm{\exists}b\in ]0,a[,{f}^{\prime}(b)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)}{a}$ then $f(a)=a{f}^{\prime}(b)$

so

$\frac{2f(a)+a{f}^{\prime}(a)}{3a}=\frac{2}{3}{f}^{\prime}(b)+\frac{1}{3}{f}^{\prime}(a)$

How to prove that $\frac{2}{3}{f}^{\prime}(b)+\frac{1}{3}{f}^{\prime}(a)$ is between ${f}^{\prime}(a)$ and ${f}^{\prime}(b)$?

Calculus 2Answered question

sbrigynt7b 2022-11-16

Let $f$ be continuous on $[0,1]$ and $f(0)=f(1)$. Also, $n\in \mathbb{N}$. Prove that there exists some $x$ s.t. $f(x)=f(x+\frac{1}{n})$.

I think I need to assume towards a contradiction that if $g(x)=f(x)-f(x+\frac{1}{n})$, then $g(x)\ne 0$ $\mathrm{\forall}x$.

So, $g(0)=f(0)-f(\frac{1}{n})$, $g(1)=f(1)-f(1+\frac{1}{n})$. Since $f(0)=f(1)$,

$g(0)-g(1)=f(1+\frac{1}{n})-f(\frac{1}{n})$. 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.

I think I need to assume towards a contradiction that if $g(x)=f(x)-f(x+\frac{1}{n})$, then $g(x)\ne 0$ $\mathrm{\forall}x$.

So, $g(0)=f(0)-f(\frac{1}{n})$, $g(1)=f(1)-f(1+\frac{1}{n})$. Since $f(0)=f(1)$,

$g(0)-g(1)=f(1+\frac{1}{n})-f(\frac{1}{n})$. 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.

Calculus 2Answered question

Jared Lowe 2022-11-15

I'm trying to solve the below exercise.

Let $f:[0,1]\to [0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a\in A$ such that $f(a)=a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0)=0$ or $f(1)=1$, the theorem is proved. Suppose not. Define $g(x)=f(x)-x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0)\ne 0$, $g(0)\ne 0$, so $g(0)>0$. Since $f(1)\ne 1$, $g(1)<0$. By the intermediate value theorem, there exists $a\in [0,1]$ such that $g(a)=0$. So $f(a)-a=0$, so $f(a)=a$.

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?

Let $f:[0,1]\to [0,1]$ be a continuous function. Prove that $f$ has at least one fixed point: an $a\in A$ such that $f(a)=a$. Is the same true for discontinuous functions?

Here is my attempt.

Notice that if $f(0)=0$ or $f(1)=1$, the theorem is proved. Suppose not. Define $g(x)=f(x)-x$, which is a continuous function from $[0,1]$ to $[0,1]$. Then since $f(0)\ne 0$, $g(0)\ne 0$, so $g(0)>0$. Since $f(1)\ne 1$, $g(1)<0$. By the intermediate value theorem, there exists $a\in [0,1]$ such that $g(a)=0$. So $f(a)-a=0$, so $f(a)=a$.

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?

Calculus 2Answered question

Jaiden Elliott 2022-11-12

I found that $f(x)={x}^{2}$ is continuous on $[1,2]$ but does not assume the values 2 intermediate between the value 1 and 4. I do not understand what they mean. For me, $\sqrt{2}\in [1,2]$ and $f(\sqrt{2})=2$.

Also, using the the intermediate value theorem, since $f$ is continuous on $[1,2]$ and $f(1)=1<2<4=f(2)$, there exists $c\in [1,2]$ such that $f(c)=2$.

Can I get some explanation about this?

Also, using the the intermediate value theorem, since $f$ is continuous on $[1,2]$ and $f(1)=1<2<4=f(2)$, there exists $c\in [1,2]$ such that $f(c)=2$.

Can I get some explanation about this?

Calculus 2Answered question

Taniya Melton 2022-10-31

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

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 $x{e}^{x}=f(x)$, but this function is defined on $\mathbb{R}\to \mathbb{R}$ and the IVT requires that our domain be $[a,b]$ 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.

Calculus 2Answered question

Josiah Owens 2022-10-26

Prove that there exists a real number $x$ such that ${x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}(x)}=125$ using Intermediate Value Theorem.

Calculus 2Answered question

Paola Mayer 2022-10-22

Let $f:[0,2]\to \mathbb{R}$ be a continuous function such that $f(0)=f(2)$. Use the intermediate value theorem to prove that there exist numbers $x,y\in [0,2]$ such that $f(x)=f(y)$ and $|x-y|=1$.

Hint: Introduce the auxiliary function $g:[0,1]\to \mathbb{R}$ defined by $g(x)=f(x+1)-f(x)$.

Hint: Introduce the auxiliary function $g:[0,1]\to \mathbb{R}$ defined by $g(x)=f(x+1)-f(x)$.

Calculus 2Answered question

Antwan Perez 2022-10-21

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

Calculus 2Answered question

Eliza Gregory 2022-10-20

I know every continuous function from $[0,1]$ to $[0,1]$ has fixed point. I proved it using Intermediate value theorem. But I can not see why every continuous function from $[0,1]$ to $\mathbb{R}$ has fixed point.

Is this result even true?

Is this result even true?

Calculus 2Answered question

ajanlr 2022-10-15

$f:[0;1]\to \mathbb{R}$ is continuous and

${\int}_{0}^{1}\text{}f(x)dx=\frac{1}{2}$

Proof that there exists an $c\in [0,1]$ such that $f(c)=c$.

Idea: So i would define a function $g(x)=f(x)-x$ and then find points so that $g(x)$ would be $g(y)<0$ and $g(z)>0$. 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 $f(x)$ for which the integral would work like $f(x)=\frac{1}{2}$?

${\int}_{0}^{1}\text{}f(x)dx=\frac{1}{2}$

Proof that there exists an $c\in [0,1]$ such that $f(c)=c$.

Idea: So i would define a function $g(x)=f(x)-x$ and then find points so that $g(x)$ would be $g(y)<0$ and $g(z)>0$. 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 $f(x)$ for which the integral would work like $f(x)=\frac{1}{2}$?

Calculus 2Answered question

Aydin Jarvis 2022-10-13

I'm taking differential calculus, and I've read that is actually a pretty interesting function that just has one solution. I'm a little confused since most proofs that involve the Intermediate value theorem give a closed interval. But I need to prove that it has a solution in the real numbers. Thanks in advance.

As you might assume, the majority of intermediate value theorem problems are quite challenging even for the brightest students because you have to use equations and explain what you know with the help of a verbal solution. If it sounds like rocket science, start with the questions on answers that provide intermediate value theorem problems with solutions. Basically, they stand for the real-valued functions that are continuous on an interval, as long as you have a value between f(a) and you have the variables where f(x)=y (as you will see from the answers we have provided below).