Master Intermediate Value Theorem Problems

Let $f$ be a function such that $f:U\subseteq <!--\; \subseteq \; -->{\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$. $U$ is open in ${\mathbb{R}}^n$ and path connected and $f$ is continuous. Let $x_1,x_2\in <!--\; \in \; -->U$. Proof that for all $c\in <!--\; \in \; -->[f(x_1),f(x_2)]$ there exists an $x\in <!--\; \in \; -->U$ such that $f(x)=c$. I'm supposed to use one-dimensional intermediate value theorem to proof this. There is a hint stating that I should look out for a function $\mathrm{\phi <!--\; \phi \; -->}:[0,1]\to <!--\; \to \; -->U$ such that we use a "useful" composition of $f$ and $\mathrm{\phi <!--\; \phi \; -->}$. I really don't know how to do this proof I would appreciate help a lot!

How is intermediate value theorem valid for $\sin <!--\; \; -->x$ in $[0,\mathrm{\pi <!--\; \pi \; -->}]$?

It has max value 1 in the interval $[0,\mathrm{\pi <!--\; \pi \; -->}]$ which doesn't lie between values given by $\sin <!--\; \; -->0$ and $\sin <!--\; \; -->\mathrm{\pi <!--\; \pi \; -->}$.

By using the Intermediate value theorem.

Show that f is continuos on $[-<!--\; -\; -->1,1]$, then there exist n in the natural numbers such that the equation $f(x)+n=n(e^x)$ has a solution in $[-<!--\; -\; -->1,1]$.

I'm having trouble solving this problem. Any help is appreciated.

Let $f:[0,1{]}^d⟶<!--\; ⟶\; -->{\mathbb{R}}^d$ with $d\ge <!--\; \ge \; -->2$. $f$ is continuous and let $c\in <!--\; \in \; -->(0,1)$. If we have that $f(0,...,0)<<(c,...,c)$ and $f(1,...,1)>>(c,...,c)$, is there an extension of the Intermediate Value Theorem for vector-valued functions that would help me prove that there indeed exists $x\in <!--\; \in \; -->(0,1{)}^d$ such that $f(x)=(c,...,c)$ ?

I have a really general question. Luzin's theorem implies that Lebesgue measurable functions are continuous almost everywhere. So is there an analogous version of the intermediate value theorem for measurable functions, with maybe some extra conditions?

How do you use the intermediate value theorem to prove the existence of numbers? For example, with $f(x)=c^2=2$, how can I prove that $\sqrt{2}$ or a positive number $"c"$ such that $f(x)$ is true exists?

I'm guessing that through the definition of the intermediate value theorem, which defines intervals of continuity, I'll have to guess with intervals around my desired value. Is that the right way to approach this?

I was thinking about this. My intuition is that there is a counterexample. Suppose $f:[0,1]⟶<!--\; ⟶\; -->\mathbb{R}$ is continuous. Also suppose

1. If $q\in <!--\; \in \; -->\mathbb{Q}$, then $f(q)\in <!--\; \in \; -->\mathbb{Q}$.
2. $f(0)<0$
3. $f(1)>1$

By the Intermediate Value Theorem, for any rational number $r$ with $0<r<1$ there is an x with $0<x<1$ such that $f(x)=r$. Is this true of the restriction $f:\mathbb{Q}⟶<!--\; ⟶\; -->\mathbb{Q}$.

Suppose that $f:[0,1]\to \mathbb{R}$ is a continuous function on the interval $[0,1]$, and that $f(0)=f(1)$. Use the Intermediate Value Theorem to show that there exists $c\in [0,1/2]$ such that $f(c)=f(c+1/2)$.

I understand the Intermediate Value Theorem and I know how it is used to prove an equation has a root within an interval. I just don't understand how to approach this question

I realized I was confused by this concept (while preparing for my exam).

If a function f has no jump discontinuities, does it have the intermediate value theorem property?

Facts, I know: I know that continuity implies intermediate value theorem property. However, intermediate value theorem property does not imply continuity. All derivatives have the intermediate value theorem property, but we can have discontinuous derivatives. Derivatives do not have jump discontinuities, or discontinuities of the first kind.

I don't know if a function has no jump discontinuities then it necessarily has the intermediate value theorem property. I know this works for derivatives. I was trying to construct a discontinuous function with discontinuities of the second kind, (one of left or right does not exist), with no jump discontinuities, that doesn't have intermediate value theorem property, but I couldn't think of any.

Let
$f(x)=\frac{x^6-<!--\; -\; -->1}{3x-<!--\; -\; -->1}$
Prove that the range of $f$ is $\mathbb{R}$.( Hint: use the Intermediate Value Theorem.)

I thought IVT was meant to show that the function has a root? Please help, I don't know how I can use IVT to prove the range.

Let's define $f$ as a continuous function with $f:[0;2]\to <!--\; \to \; -->\mathbb{R}$ and $f(0)=f(2)$.

Now, I want to show that:
$\mathrm{\exists <!--\; \exists \; -->}{x}_0\in <!--\; \in \; -->[0;1]:f(x_0)=f(x_0+1)$
I tried to plot a few functions in order to construct a counterexample, but it seems that this statement really is true.

Unfortunately, I don't think I'm entirely sure why this works, yet. My current guess is that it has something to do with the Intermediate Value Theorem, as f is continuous. In other words: If our function value 'goes up', it'll have to 'come back down' eventually (and vice versa), since it still needs to fulfill $f(0)=f(2)$.

Can someone help me prove this statement?

I have a continious function $f$ that is strictly increasing. And a continious function $g$ that is strictly decreasing. How to I rigorously prove that $f(x)=g(x)$ has a unique solution?

Intuitively, I understand that if I take limits to infinity, then $f$ grows really large and $g$ grows very small so the difference is less than zero. If I take the limits to negative infinity then the opposite happens. Using the intermediate value theorem, there must be an intersection, and since they are strictly increasing/decreasing, only one intersection happens.

My question is how do I apply the intermediate value theorem here? I don't have the interval to apply it on. I don't know when $f$ crosses $g$ and therefore can't take any interval. Or is there some sort of axiom applied here that I am missing?

The Intermediate Value Theorem has been proved already: a continuous function on an interval $[a,b]$ attains all values between $f(a)$ and $f(b)$. Now I have this problem:

Verify the Intermediate Value Theorem if $f(x)=\sqrt{x+1}$ in the interval is $[8,35]$.

I know that the given function is continuous throughout that interval. But, mathematically, I do not know how to verify the theorem. What should be done here?

I cant understand how to prove this question. We learned about intermediate value theorem but this makes no sense because 120 km isn't in bounds of either upper or lower limit. Here is the question

At 2:00 PM, a car's speedometer reads 30km/h. At 2:10 PM, it reads 50 km/h. Show that at some time between 2:00 and 2:10 the acceleration was exactly 120 km/${\text{h}}^2$ (YES she wrote the question are ${\text{h}}^2$. I hope its a typo). Indicate which theorem you must use in your explanation.

I'm new to mathematical proofs, and I have just covered the Intermediate Value Theorem. I have tried to practice my understanding of the theorem, but I have encountered a question that I'm not sure how to approach.

The question is:

Assume that $f$ and $g$ are continuous on the interval [0,1] and $0\le f(x)\le 1$ for all $x\in <!--\; \in \; -->[0,1]$. Show that if $g(0)=0$ and $g(1)=1$, then there exists a $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=g(c)$.

What I have tried doing:

Introducing another function like such $h(x)=f(x)-<!--\; -\; -->g(x)$ and applying the theorem. However, I'm not sure if that is possible, or how I should go about doing it.

I am asked to find an example of a discontinuous function $f:[0,1]\to \mathbb{R}$ where the intermediate value theorem fails. I went over the intermediate value theorem today

Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Suppose that there exists a y such that $f(a)<y<f(b)$ or $f(a)>y>f(b).$ Then there exists a $c\in [a,b]$ such that $f(c)=y$.

I understand the theory behind it, however, we did not go over many example of how to use it to solve such problems so I do not really know where to begin

If $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous on $[a,b]$, show that $\mathrm{\exists <!--\; \exists \; -->}c\in <!--\; \in \; -->[a,b]$ so that
${\int <!--\; \int \; -->}_{c-<!--\; -\; -->a}^{b-<!--\; -\; -->c}f(x)dx=0$
I think it can be solved using the intermediate value theorem but I can't find a suitable function.

Let $f(x)$ be a continues function for all $x$, and $|f(x)|\le <!--\; \le \; -->7$ for all $x$.

Prove the equation $2x+f(x)=3$ has one solution.

I think the intermediate value theorem is key in this, but I'm not sure of the proper usage.

II want to use the Intermediate Value Theorem and Rolle’s theorem to show that the graph of $f(x)=x^3+2x+k$ crosses the x-axis exactly once, regardless of the value of the constant k.

I know I can use the intermediate value theorem, but I don't necessarily know how to show the change in a sign for two select inputs. any hero would be appreciated in that regard.

I also know the derivative of $x^3+2x+k$ is greater than zero, but what does that mean?

Suppose $f$ is continuous on [0,2] and that $f(0)=f(2)$. Then $\mathrm{\exists <!--\; \exists \; -->}$$x,$$y\in <!--\; \in \; -->[0,2]$ such that $|y-<!--\; -\; -->x|=1$ and that $f(x)=f(y)$.


Let $g(x)=f(x+1)-<!--\; -\; -->f(x)$ on [0,1]. Then $g$ is continuous on [0,1], and hence $g$ enjoys the intermediate value property! Now notice
$g(0)=f(1)-<!--\; -\; -->f(0)$
$g(1)=f(2)-<!--\; -\; -->f(1)$
Therefore
$g(0)g(1)=-<!--\; -\; -->(f(0)-<!--\; -\; -->f(1){)}^2<0$
since $f(0)=f(2)$. Therefore, there exists a point $x$ in [0,1] such that $g(x)=0$ by the intermediate value theorem. Now, if we pick $y=x+1$, i think the problem is solved.

I would like to ask you guys for feedback. Is this solution correct? Is there a better way to solve this problem?