Table of Contents 8. Section 2.7 Intermediate Value Theorem

Essential Question What is the IVT used for?

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

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

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

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

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

Example Show that 𝑐𝑜𝑠 2 𝑥 −2𝑠𝑖𝑛 𝑥 4 has a zero in (0,2) f(0) = 1 f(2) = -0.786 Opposite signs, therefore a zero

Example cont. Now use bisection method to find zero more accurately Find value halfway in between f(1) = -0.203 Use interval that changes signs (0,1) f(1/2) = 0.521 ---- (1/2,1) f(3/4) = 0.163 ---- (3/4,1) f(7/8) = -0.231 ----- (3/4, 7/8) Function has zero in this interval

Assignment Pg 106: #1-7 odd, 15-21 odd