Explore the Intermediate Value Theorem Equation

High School
Calculus and Analysis
Calculus 2
Intermediate Value Theorem

Recent questions in Intermediate Value Theorem

glycleWogry 2022-06-22

I have a problem with the Intermediate value theorem. For example if I have the function $(x) = - 4 x^{2} + 12 x$ , I can get for example all the values from $x = 0$ to $x = 2$ , so $f (0) = 0$ and $f (2) = 8$ , with the Intermediate value theorem, I know that the function takes all the values from 0 to 8. But it also takes the value 9 when x goes from 0 to 2, so with the Intermediate value theorem I can´t know all the value that the function takes.

Can anyone explain me why this happens in this example, and obviously in other example.

Semaj Christian 2022-06-21

Using the Intermediate Value Theorem and a calculator, find an interval of length 0.01 that contains a root of $x^{5} - x^{2} + 2 x + 3 = 0$ , rounding off interval endpoints to the nearest hundredth.

I've done a few things like entering values into the given equation until I get two values who are 0.01 apart and results are negative and positive ( $- 1.15$ & $- 1.16$ ), but these answers were incorrect.

I'm at the point where I'm thinking there is not enough information to solve. Any ideas?

Jaqueline Kirby 2022-06-19

Suppose that a function $f (x)$ is differentiable $\forall x \in [a, b]$ . Prove that $f^{'} (x)$ takes on every value between $f^{'} (a)$ and $f^{'} (b)$ .

If the above question is a misprint and wants to say "prove that $f (x)$ takes on every value between $f (a)$ and $f (b)$ ", then I have no problem using the intermediate value theorem here.

If, on the other hand, it is not a misprint, then it seems to me that I can't use the Intermediate value theorem, as I can't see how I am authorised to assume that $f^{'} (x)$ is continuous on $[a, b]$ .

Or perhaps there is another way to look at the problem?

Sarai Davenport 2022-06-19

I am thinking of a proposition analogous to the intermediate value theorem for a scalar function on reals:

In a topological space $(X, τ)$ where X is a connected set, consider a set $S \subseteq X$ and a continuous function $f : R \to X$ . For some $a, b \in R$ with $a < b$ , the function satisfies
$f (a) \in S^{\circ} \land f (b) \in X ∖ \bar{S}$
Then, there exists $c \in (a, b)$ such that
$f (c) \in \partial S$
Is there any theorem useful for proving this proposition? Or any advice will be appreciated.

Dwllane4 2022-06-16

The following claim seems true for me:

For any continuous function over reals $f (\vec{x})$ , if $f (\vec{x}) = c$ has no zero, then either $f (\vec{x}) > c$ for all $\vec{x}$ or $f (\vec{x}) < c$ for all $\vec{x}$ .

However, my calculus is a bit rusty, and I am wondering from which this claim follows? (For univariate functions, this follows from the intermediate value theorem?)

Bailee Short 2022-06-15

let $f$ defined and continuous function in $[a, b]$ with $[a, b] \subseteq f ([a, b])$ .Prove that $f (x) = x$ admits at least one solution in $[a, b]$

I am not sure how to deal with this theorem so I have no idea where to start. if someone could walk me through the process that would be great.

Gabriella Sellers 2022-06-15

Is there a $p$ -adic analogue to the intermediate value theorem? I know there is a notion of convex sets in the $p$ -adic context but can we hope for an intermediate value theorem in this context?

dourtuntellorvl 2022-06-15

If i have 2 functions f and g, both with the domain [a,b] interval. Where both are continuous. Also, f(a)>g(a) and f(b)<g(b), how would i use intermediate value theorem to show there exists a c in [a,b] for the fucntion h(x)=f-g such that f(c)=g(c)

Gybrisysmemiau7 2022-06-14

I need help please with the following question. (Do I need to use radians in dealing with the sin part?))
(a) Use the Intermediate Value Theorem to show that the equation $x + \sin (x) = 1$ , has at least one solution in the interval [0,1].
(b) Find an approximation of the solution with an error of at most 0.05.

Finley Mckinney 2022-06-14

$f : R^{2} \to R$ is continuous and $f (t, x) < 0$ holds for $t x > 0$ and $f (t, x) > 0$ holds for $t x < 0$

I want to show, that $x (t) = 0$ is the only solution of $x^{'} = f (t, x)$

You can say, that there is an solution, because f is continuous. By the intermediate value theorem exists $ξ \in [- a, a]$ , such that $f (t, ξ) = 0$ . I only have to show, that $x (t) = 0$ is unique. How can I do that?

Leland Morrow 2022-06-13

Let $f$ be a continuously differentiable function on $[a, b]$ with $f (a) = f (b)$ and $f^{'} (a) = f^{'} (b)$ . Then, do there exist $x_{1}, x_{2} \in (a, b)$ such that $f^{'} (x_{1}) = f^{'} (x_{2})$ for $x_{1} \neq x_{2}$ ?

I think the answer is yes by intermediate value theorem, but am not getting the sufficient rigour to prove it. Any hints. Thanks beforehand.

excluderho 2022-06-13

Let $f : [a, b] \to R$ be continuous, $p, q > 0$ . We want to show that there is a $c \in [a, b]$ s.t. $p f (a) + q f (b) = (p + q) f (c)$ . $\frac{p}{p + q}, \frac{q}{p + q}$ are both in (0,1). Then $\frac{p}{p + q} f (a) + \frac{q}{p + q} f (b)$ is in $(0, f (a) + f (b))$ . If we want to apply the intermediate value theorem we better would need it to be in [ $f (u)$ , $f (w)$ ], where we know that $f (u)$ is the Minimum and $f (w)$ is the Maximum. But we have not any info about what the Min/Max would be?

Jordan Victorio2022-06-11

Find a counterexample to the statement that every positive integer can be written as the sum of the squares of

kincirrboh7 2022-06-03

Let $f : [a, b] \to R$ be a continuous function and let $f (a) < f (b)$ . Then by intermediate value theorem
$(1) f ([a, b]) = [f (a), f (b)]$
$(2) f ([a, b]) \supseteq [f (a), f (b)]$
$(3) f ([a, b]) \subseteq [f (a), f (b)]$
$(4) f ([a, b]) \neq [f (a), f (b)]$
For example if I take the function $f (x) = x^{2}$ on $[- 1, 2]$ then $f ([- 1, 2]) = [0, 4]$ which is not equal to $[f (- 1), f (2)] = [1, 4]$ , also $[1, 4] \subseteq [0, 4]$ , so options 1 is incorrect. But how to use IVT here? Please help.

Monserrat Monroe 2022-06-03

Let $f$ be continuous on $[a, b]$ , where $a < b$ , such that $f (x) \neq 0$ for all $x \in [a, b]$ . Prove that there is a $c > 0$ , either $f (x) > c$ or $f (x) < - c$ for all $x \in [a, b]$ .

What I did,

Proving by contradiction, Let $f (x) > c$ for some $x_{1} \in [a, b]$ and $f (x) < - c$ for some $x_{2} \in [a, b]$ . By the Intermediate Value Theorem, there exists exists a number $z \in (a, b)$ such that $f (z) = 0$ , however this contradicts the fact that $f (x) \neq 0$ for all $x \in [a, b]$ . Hence $f (x) > c$ or $f (x) < - c$ for all $x \in [a, b]$ .

Hi, was wondering whether my proof is sound. Do help me verify! Thanks!

Andy Erickson 2022-05-30

Let $f$ be a continuous function on $[0, 1]$ and differentiable on $(0, 1)$ with $f (1) = f (0) = 0$ and $f (\frac{1}{2})$ =1 .Show that

(1) there exist some a $\in (\frac{1}{2}, 1)$ such that $f (a) = a$

while this just can be proved by using the intermediate value theorem and I have solved it. But the second part made me confuse,

(2) for any real number $λ$ , there exist some point $ε \in (0, a)$ such that $f^{'} (ε) - 1 = λ (f (ε) - ε))$

For the second part, why there's a $f^{'} (ε)$ there? and how I can approach the proof of second part?

Jazmine Bruce 2022-05-28

Let $g \geq 0$ on $[a, \infty)$ , and $\int_{a}^{\infty} g (x) d x < \infty$ . Also, $f$ is continuous, bounded on $[a, \infty)$ . Show that there exists a $ξ \in [a, \infty)$ such that
$\int_{a}^{\infty} f (x) g (x) d x = f (ξ) \int_{a}^{\infty} g (x) d x .$
This is extension of Intermediate value theorem.

Pitrellais 2022-05-27

I'm starting to study the basics of measure theory and found interesting this problem: Let us define $| E | = \int_{- \infty}^{\infty} 1_{E} (x) d x$ , where $E \subset R$ and $1_{E}$ its characteristic function. Given $A \subset B \subset R$ , with $| A | = a < b = | B |$ and $x \in (a, b)$ , is it possible to find $A \subset X \subset B$ such that |X|=x? Any ideas? Can I say this is an analogue of the Intermediate value theorem for the function $| • |$ ?

cyfwelestoi 2022-05-26

I'm going to give a lecture in a public contest (like a civil service examination to be a university public teacher in my country) about the intermediate value theorem in 45 minutes. I'm going to give this lecture to three professors (the examiners) but it's like I'm going to give it to undergraduate students.

Since the proof of the intermediate value theorem is too hard to show to someone who's never learned real analysis, what should I do in this class? I'm open to suggestions.

skottyrottenmf 2022-05-26

Prove that $f (x) = 3$ has a solution on the interval $[a, b]$

And Intermediate Value Theorem says that

if $f (a) * f (b) < 0$ , then it has a solution on that interval

So instead of seeing if 3 is between the interval and stuff like that. Can't I just do this:
$f (x) = 3$
$f (x) - 3 = 0$
And then we consider $f (x) - 3$ a completely new whole function called $g (x) = 0$

According to the theorem, I can say that since $g (a) * g (b) < 0$ , it has a solution on the interval.

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).

Calculus 2

Master Intermediate Value Theorem Problems