1. Table of Contents
8. Section 2.7 Intermediate Value Theorem

2. Essential Question
What is the IVT used for?

3. Intermediate value theorem
A function that is continuous on a closed interval takes on every value between f(a) and f(b).

4. In other words…..
A functions that is continuous can’t skip values

5. Example Prove that sin x = 0.3 has at least one solution.
We can apply the IVT because sin x is continuous
Choose an interval (0, 𝜋 2 )
Sin(0) = 0, sin( 𝜋 2 ) = 1
0.3 is between these 2 values
Therefore, the IVT tells us that sin x = 0.3 has a solution in (0, 𝜋 2 )
Could choose other intervals as well

6. Existence of zeros The IVT can be used to show the existence of zeros
If one value of the function is negative and another is positive, there must be a zero somewhere in between

7. Bisection Method
Cut intervals in half successively to narrow down where zero is

8. Example Show that 𝑐𝑜𝑠 2 𝑥 −2𝑠𝑖𝑛 𝑥 4 has a zero in (0,2)
f(0) = 1 f(2) = Opposite signs, therefore a zero

9. Example cont. Now use bisection method to find zero more accurately
Find value halfway in between
f(1) =
Use interval that changes signs (0,1)
f(1/2) = (1/2,1)
f(3/4) = (3/4,1)
f(7/8) = (3/4, 7/8)
Function has zero in this interval

10. Assignment
Pg 106: #1-7 odd, 11, odd