# 2.4 Continuity  (Page 5/16)

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## The intermediate value theorem

Let f be continuous over a closed, bounded interval $\left[a,b\right].$ If z is any real number between $f\left(a\right)$ and $f\left(b\right),$ then there is a number c in $\left[a,b\right]$ satisfying $f\left(c\right)=z$ in [link] .

## Application of the intermediate value theorem

Show that $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ has at least one zero.

Since $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ is continuous over $\left(\text{−}\infty ,\text{+}\infty \right),$ it is continuous over any closed interval of the form $\left[a,b\right].$ If you can find an interval $\left[a,b\right]$ such that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number c in $\left(a,b\right)$ that satisfies $f\left(c\right)=0.$ Note that

$f\left(0\right)=0-\text{cos}\phantom{\rule{0.1em}{0ex}}\left(0\right)=-1<0$

and

$f\left(\frac{\pi }{2}\right)=\frac{\pi }{2}-\text{cos}\frac{\pi }{2}=\frac{\pi }{2}>0.$

Using the Intermediate Value Theorem, we can see that there must be a real number c in $\left[0,\pi \text{/}2\right]$ that satisfies $f\left(c\right)=0.$ Therefore, $f\left(x\right)=x-\text{cos}\phantom{\rule{0.1em}{0ex}}x$ has at least one zero.

## When can you apply the intermediate value theorem?

If $f\left(x\right)$ is continuous over $\left[0,2\right],f\left(0\right)>0$ and $f\left(2\right)>0,$ can we use the Intermediate Value Theorem to conclude that $f\left(x\right)$ has no zeros in the interval $\left[0,2\text{]?}$ Explain.

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between $f\left(0\right)$ and $f\left(2\right);$ it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the function $f\left(x\right)={\left(x-1\right)}^{2}.$ It satisfies $f\left(0\right)=1>0,f\left(2\right)=1>0,$ and $f\left(1\right)=0.$

## When can you apply the intermediate value theorem?

For $f\left(x\right)=1\text{/}x,f\left(-1\right)=-1<0$ and $f\left(1\right)=1>0.$ Can we conclude that $f\left(x\right)$ has a zero in the interval $\left[-1,1\right]?$

No. The function is not continuous over $\left[-1,1\right].$ The Intermediate Value Theorem does not apply here.

Show that $f\left(x\right)={x}^{3}-{x}^{2}-3x+1$ has a zero over the interval $\left[0,1\right].$

$f\left(0\right)=1>0,f\left(1\right)=-2<0;f\left(x\right)$ is continuous over $\left[0,1\right].$ It must have a zero on this interval.

## Key concepts

• For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
• Discontinuities may be classified as removable, jump, or infinite.
• A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
• The composite function theorem states: If $f\left(x\right)$ is continuous at L and $\underset{x\to a}{\text{lim}}g\left(x\right)=L,$ then $\underset{x\to a}{\text{lim}}f\left(g\left(x\right)\right)=f\left(\underset{x\to a}{\text{lim}}g\left(x\right)\right)=f\left(L\right).$
• The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

$f\left(x\right)=\frac{1}{\sqrt{x}}$

The function is defined for all x in the interval $\left(0,\infty \right).$

$f\left(x\right)=\frac{2}{{x}^{2}+1}$

$f\left(x\right)=\frac{x}{{x}^{2}-x}$

Removable discontinuity at $x=0;$ infinite discontinuity at $x=1$

$g\left(t\right)={t}^{-1}+1$

$f\left(x\right)=\frac{5}{{e}^{x}-2}$

Infinite discontinuity at $x=\text{ln}\phantom{\rule{0.1em}{0ex}}2$

$f\left(x\right)=\frac{|x-2|}{x-2}$

$H\left(x\right)=\text{tan}\phantom{\rule{0.1em}{0ex}}2x$

Infinite discontinuities at $x=\frac{\left(2k+1\right)\pi }{4},$ for $k=0,±1,±2,±3\text{,…}$

$f\left(t\right)=\frac{t+3}{{t}^{2}+5t+6}$

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

#### Questions & Answers

Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
how do i evaluate integral of x^1/2 In x
ayo Reply
first you simplify the given expression, which gives (x^2/2). Then you now integrate the above simplified expression which finally gives( lnx^2).
Ahmad
by using integration product formula
Roha
find derivative f(x)=1/x
Mul Reply
-1/x^2, use the chain rule
Andrew
f(x)=x^3-2x
Mul
what is domin in this question
noman
all real numbers . except zero
Roha
please try to guide me how?
Meher
what do u want to ask
Roha
?
Roha
the domain of the function is all real number excluding zero, because the rational function 1/x is a representation of a fractional equation (precisely inverse function). As in elementary mathematics the concept of dividing by zero is nonexistence, so zero will not make the fractional statement
Mckenzie
a function's answer/range should not be in the form of 1/0 and there should be no imaginary no. say square root of any negative no. (-1)^1/2
Roha
domain means everywhere along the x axis. since this function is not discontinuous anywhere along the x axis, then the domain is said to be all values of x.
Andrew
Derivative of a function
Waqar
right andrew ... this function is only discontinuous at 0
Roha
of sorry, I didn't realize he was taking about the function 1/x ...I thought he was referring to the function x^3-2x.
Andrew
yep...it's 1/x...!!!
Roha
true and cannot be apart of the domain that makes up the relation of the graph y = 1/x. The value of the denominator of the rational function can never be zero, because the result of the output value (range value of the graph when x =0) is undefined.
Mckenzie
👍
Roha
Therefore, when x = 0 the image of the rational function does not exist at this domain value, but exist at all other x values (domain) that makes the equation functional, and the graph drawable.
Mckenzie
👍
Roha
Roha are u A Student
Lutf
yes
Roha
What is the first fundermental theory of Calculus?
ZIMBA Reply
do u mean fundamental theorem ?
Roha
I want simple integral
aparna Reply
for MSc chemistry... simple formulas of integration
aparna
hello?
funny
how are you
funny
I don't understand integration
aparna
r u insane
aparna
integration is so simple not typical..
funny
tell me any questions about integration then i will solve.
funny
we use integration for whole values or for sum of values any there are some basic rule for integration..
funny
I just formulas
aparna
I just want formulas of integration
aparna
value of log ax cot-x cos-x
aparna
there are many formulas about integration
funny
more then one formula are exist about integration..
funny
so I want simple formulas Because I'm studying MSc chem...Nd have done bsc from bio...
aparna
I am M.sc physics now i am studying in m.phil
funny
so what can i do
aparna
I will send you basic formula for integration after two mint first of all i write then i will send you.
funny
send me your messenger id where i can send you formulas about integration because there is no option for image sending..
funny
integration f(X) dx this is basic formula of integration sign is not there you can look integration sign in methematics form... and f(X) my be any function any values
funny
you send me your any ID where i can send you information about integration
funny
send me SMS at this ID Adnan sathi Adnan sathi
funny
Hi
RIZWAN
I don't understand the formula
Adaeze Reply
who's formula
funny
which formula?
Roha
what is the advantages of mathematical economics
Mubarak
What is a independent variable
Sifiso Reply
a variable that does not depend on another.
Andrew
which can be any no... does not need to find its value by any other variable.. often x is independent and y is dependent
Roha
solve number one step by step
bil Reply
x-xcosx/sinsq.3x
Hasnain
x-xcosx/sin^23x
Hasnain
how to prove 1-sinx/cos x= cos x/-1+sin x?
Rochel Reply
1-sin x/cos x= cos x/-1+sin x
Rochel
how to prove 1-sun x/cos x= cos x / -1+sin x?
Rochel

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