Wize AP Calculus (BC) Textbook > Limits & Continuity

The Intermediate Value Theorem

Popular Courses

AP Calculus (BC) Exam Prep Course

AP Exam Prep

MATH 116

University of Waterloo

Find My Course

0:00 / 0:00

The Intermediate Value Theorem
Continuity is very powerful since it can tell us if a function attains certain values. In particular, continuity tells us function outputs between certain values.

Intermediate Value Theorem (IVT)
 If f(x)f(x)f(x) is continuous on [a,b],[a, b],[a,b], then for any LLL between f(a)f(a)f(a)and f(b)f(b)f(b) there exists a point c∈(a,b)c\in(a, b)c∈(a,b) for which f(c)=L.f(c)=L.f(c)=L. 

Wize Concept
In other words, since fff is continuous, f(x)f(x)f(x) , covers all the values between f(a)f(a)f(a) and f(b)f(b)f(b).

Wize Tip
Whenever the question asks about roots or solutions of an equation L=0L=0L=0 in IVT.

Procedure for Problem Solving 
Determine the domain of significance.
Evaluate the function at the endpoints of the domain. 
Determine if the points of interest are within the end points. If not, consider choosing a new domain.

0:00 / 0:00

Example: Intermediate Value Theorem

Determine if the functionf(x)=x3−6x+4f(x)=x^3-6x+4f(x)=x3−6x+4has a root in the domain [0,1].[0,1].[0,1]. 

f(0)=0−0+4=4f(1)=1−6+4=−1}f(1)<0<f(0)\left.\begin{array}{r}   f(0) = 0-0+4 = 4\\   f(1) = 1-6+4 = -1  \end{array}\right\} f(1)<0<f(0)f(0)=0−0+4=4f(1)=1−6+4=−1​}f(1)<0<f(0)
  Therefore, f(c)=0f(c)=0f(c)=0 for some c, 0<c<1,0 < c < 1,0<c<1, so at least one root exists as f(x) is continuous on [0,1].
 

Mark Yourself Question

Grab a piece of paper and try this problem yourself.
When you're done, check the "I have answered this question" box below.
View the solution and report whether you got it right or wrong.

Use the Intermediate Value Theorem to show that x2+x−3=2\sqrt{x^2+x-3}=2x2+x−3​=2 has a solution on the interval (2,3).

I have answered this question