Chapters

Exercise 1
Exercise 2
Exercise 3
Exercise 4
Exercise 5
Exercise 6
Exercise 7
Exercise 8
Exercise 9
Exercise 10
Exercise 11
Exercise 12
Solution of exercise 1
Solution of exercise 2
Solution of exercise 3
Solution of exercise 4
Solution of exercise 5
Solution of exercise 6
Solution of exercise 7
Solution of exercise 8
Solution of exercise 9
Solution of exercise 10
Solution of exercise 11
Solution of exercise 12

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Exercise 1

Prove that the function $\text{[math]}$ intersects the x-axis on the interval $\text{[math]}$ . Can the same be said for the function: $\text{[math]}$ ?

Exercise 2

Given the function:

$\text{[math]}$

Can it be said that f(x) is bounded in the interval $\text{[math]}$ ?

Exercise 3

Given the function $\text{[math]}$ . Can it be said that the function exists for all values in the interval $\text{[math]}$ ?

Exercise 4

Prove that the equation: $\text{[math]}$ , has at least one solution $\text{[math]}$ such that $\text{[math]}$ .

Exercise 5

Given the function $\text{[math]}$ . Can it be said that there is at least one point, c, inside the interval $\text{[math]}$ which verifies that $\text{[math]}$ ?

Exercise 6

Prove that the polynomial function $\text{[math]}$ has a value of zero between $\text{[math]}$ and $\text{[math]}$ .

Exercise 7

Prove that the equation $\text{[math]}$ has at least one real solution.

Exercise 8

Prove that there is a real number, x, such that $\text{[math]}$ .

Exercise 9

Given the function:

$\text{[math]}$

Prove that there is a point in the open interval $\text{[math]}$ in which the function $\text{[math]}$ has a value of $\text{[math]}$ .

Exercise 10

Given the function $\text{[math]}$ , determine if it is bounded superiorly and inferiorly in the interval $\text{[math]}$ and indicate if it reaches its maximum and minimum values within this interval.

Exercise 11

Prove that the function $\text{[math]}$ is continuous at $\text{[math]}$ and prove that there exists at least one real root of the equation $\text{[math]}$ .

Exercise 12

f and g are two continuous functions in $\text{[math]}$ and such that $\text{[math]}$ and $\text{[math]}$ . Prove the existance of c withinin $\text{[math]}$ such that $\text{[math]}$ .

Solution of exercise 1

Prove that the function $\text{[math]}$ intersects the x-axis on the interval $\text{[math]}$ . Can the same be said for the function: $\text{[math]}$ ?

The first function is continuous at $\text{[math]}$ .

$\text{[math]}$ .

Since it verifies the intermediate value theorem, at least one c belongs to the interval $\text{[math]}$ and intersects the x-axis.

We cannot confirm the same of the second function because it is not continuous at $\text{[math]}$ .

Solution of exercise 2

Given the function:

$\text{[math]}$

Can it be said that f(x) is bounded in the interval $\text{[math]}$ ?

Since f(x) is not continuous at $\text{[math]}$ , the function is not continuous in the closed interval $\text{[math]}$ , as a result, it cannot be said that the function is bounded in that interval.

Solution of exercise 3

Given the function $\text{[math]}$ . Can it be said that the function exists for all values in the interval $\text{[math]}$ ?

$\text{[math]}$ $\text{[math]}$

The function is continuous at $\text{[math]}$ since it is a polynomial function.

It is in the interval $\text{[math]}$ as it is verified that $\text{[math]}$ and $\text{[math]}$ .

Since it verifies the intermediate value theorem, the function exists at all values in the interval $\text{[math]}$ .

Solution of exercise 4

Prove that the equation: $\text{[math]}$ , has at least one solution $\text{[math]}$ such that $\text{[math]}$ .

f(x) is continuous in $\text{[math]}$

$\text{[math]}$

Since it verifies the Bolzano's Theorem, there is c $\text{[math]}$ such that:

$\text{[math]}$

Therefore there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 5

Given the function $\text{[math]}$ . Can it be said that there is at least one point, c, inside the interval $\text{[math]}$ which verifies that $\text{[math]}$ ?

f(x) is continuous in $\text{[math]}$ .

$\text{[math]}$

The Bolzano theorem cannot be applied because it does not change sign.

Solution of exercise 6

Prove that the polynomial function $\text{[math]}$ has a value of zero between $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$ is a polynomial and therefore is continuous in the interval $\text{[math]}$ .

$\text{[math]}$

$\text{[math]}$ .

There is a $\text{[math]}$ such that $\text{[math]}$

Solution of exercise 7

Prove that the equation $\text{[math]}$ has at least one real solution.

The function is continuous in the interval $\text{[math]}$ .

$\text{[math]}$ .

Since it verifies Bolzano's theorem, there is $\text{[math]}$ such that:

$\text{[math]}$

Therefore there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 8

Prove that there is a real number, x, such that $\text{[math]}$ .

Consider the function $\text{[math]}$ .

It is continuous at $\text{[math]}$ .

$\text{[math]}$

There is a $\text{[math]}$ such that:

$\text{[math]}$

Therefore, there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 9

Given the function:

$\text{[math]}$

Prove that there is a point in the open interval $\text{[math]}$ in which the function $\text{[math]}$ has a value of $\text{[math]}$ .

The exponential function is positive at $\text{[math]}$ , therefore the denominator of the function cannot be annulled.

There is only doubt of the continuity at $\text{[math]}$ , which is out of the interval being studied, therefore $\text{[math]}$ is continuous in $\text{[math]}$ .

Consider the function g defined by $\text{[math]}$ .

g is continuous on the interval $\text{[math]}$ .

$\text{[math]}$ $\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

Solution of exercise 10

The function is continuous in the interval $\text{[math]}$ , as a result, it can be affirmed that it is bounded in that interval.

As well as being continuous in the interval $\text{[math]}$ , it has fulfilled the extreme value theorem, which affirms that it attains at least one maximum and absolute minimum in the interval $\text{[math]}$ .

Solution of exercise 11

Prove that the function $\text{[math]}$ is continuous at $\text{[math]}$ and prove that there exists at least one real root of the equation $\text{[math]}$ .

The function is continuous since it is the sum of continuous functions.

$\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

Therefore, there is at least one real solution to the equation $\text{[math]}$ .

Solution of exercise 12

f and g are two continuous functions in $\text{[math]}$ and such that $\text{[math]}$ and $\text{[math]}$ . Prove the existance of c withinin $\text{[math]}$ such that $\text{[math]}$ .

h is the function defined by $\text{[math]}$ .

Since f and g are continuous in $\text{[math]}$ , the function h also is.

$\text{[math]}$ $\text{[math]}$

Since it verifies the intermediate value theorem, there is a $\text{[math]}$ such that:

$\text{[math]}$

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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.

Theory

One to the Power of Infinity

Zero Over Zero

Calculating Limits

Continuity

Continuous Function

Continuity on a Closed Interval

Discontinuity

Division by Zero

Essential Discontinuity

Intermediate Value Theorem

Jump Discontinuity

Limit

One-Sided Limit

Indeterminate Forms

Zero Times Infinity

Infinity Minus Infinity

Infinity over Infinity

Properties of Infinity

Extreme Value Theorem

Bolzano’s Theorem

Limits at Infinity

Infinite Limit

Limit of a Logarithmic Function

Removable Discontinuity

Limit of an Exponential Function

Limit Rules

Formulas

Continuity Formulas

Limit Formulas

Exercises

Limits Worksheet

Continuity Problems

Limit Problems

Intermediate Value Theorem Problems

Continuity Worksheet

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Laura

February 2024

Correct equals infinity equals 16 but not true. It’s it’s six it’s infinity.

TonyL

September 2023

And this, my friends, is why humans will be conquered by AI… so many logic holes it’s… well, infinite. lol

Raj

May 2023

Is 0^infinity ( zero to the power infinity) indeterminate form? How?

Tara Kapali

May 2023

Didn’t Cantor proved that there are a group of infinities (the Aleph zero & Aleph one sets, for example)? And these are grouped around the concept of infinite to the power infinity, if I remember correctly .

Dhayq

July 2022

Exercise 3
I think it is discontinuous at 0

rohit

September 2020

exercise 2 : the function is not defined for x= 0.

rohit

September 2020

exercise 1 q 5 The function has a jump discontinuity at x = 1, should be x=0.