1 / 10

AP Calc AB

AP Calc AB. IVT. Introduction. What do we know GENERALLY about the solutions for the functions below? f(x)= cos x -2 g(x) = h(x) =. Because the function is continuous, it must take on every y value between and. Intermediate Value Theorem.

donagh
Download Presentation

AP Calc AB

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. AP Calc AB IVT

  2. Introduction What do we know GENERALLY about the solutions for the functions below? f(x)=cos x -2 g(x) = h(x) =

  3. Because the function is continuous, it must take on every y value between and . Intermediate Value Theorem If a function is continuous between a and b, then it takes on every value between and .

  4. The application of the IVT. If f(x) is continuous on a closed interval [a,b] and a<c<b, then there exists a value f(c) such that f(c) is between f(a) and f(b)

  5. Example 1 Prove that there exists a value c such that f(c) = 0 for f(x) = for the interval [3,5]

  6. Solution f(x) is continuous and f(3) = -6 and f(5) = 8 Therefore according to the IVT there exists a value c where -6 < f(c)< 8. Zero lies within that interval.

  7. Example 2 Prove that there exists a value c such that f(c)=2 for f(x) = for the interval [-10,-3]

  8. Solution f(x) is continuous and f(-10) = 8 and f(-3) = 1, therefore according to the IVT there exists a value c where 1<f(c)<8. 2 lies within this interval.

  9. Roots and the IVT

  10. Root solution p(-1)=8 and p(2) = -19 p(x) is continuous and p(-1) = 8 and p(2) = -19, therefore according to the IVT there exists a value c between [-1, 2] where -19<p(c)<8. (or, you can simply say that there is sign change in the y values guaranteeing a value c in this interval where p(c)=0.) Note: we have not found c, just shown that there is at least ONE VALUE c.

More Related