Intermediate Value Theorem

1. Intermediate Value Theorem Mrs. King OCS Calculus Curriculum

2. Intermediate Value Theorem If f is a continuous function on a closed interval [a, b] and L is any number between f (a) and f (b), then there is at least one number c in [a, b] such that f(c) = L. f (b) f (c) = L f (a) a c b www.rowan.edu/open/depts/math/.../Limits%20and%20Continuity.pp

3. So what does this all mean? • The Intermediate Value Theorem is what is known as an existence theorem. • It tells us under what conditions c exists, but nothing about how to actually determine the value of c.

4. Examples • If between 7am and 2pm the temperature went from 55 to 70. • At some time it reached 62. • Time is continuous • If between his 14th and 15th birthday, a boy went from 150 to 165 lbs. • At some point he weighed 155lbs. • It may have occurred more than once. www.cdschools.org/.../lib/.../Intermediate_Value_Theorem.ppt

5. Example • f (x) is continuous (polynomial) • f (1) < 0 < f (2) • by the Intermediate Value Theorem there exists a c on [1, 2] such that f (c) = 0. www.rowan.edu/open/depts/math/.../Limits%20and%20Continuity.pp

6. Show that a “c” exists such that f(c)=2 for f(x)=x2 +2x-3 in the interval [0, 2] • f(x) is continuous on the interval f(0)= -3 f(2)= 5 www.cdschools.org/.../lib/.../Intermediate_Value_Theorem.ppt

7. Sample Problem • Use the Intermediate Value Theorem to show that the polynomial functionhas a zero in the interval [0,1].

8. Solution • First, we must comment on the continuity of the function. • Polynomial functions are continuous and so far the Intermediate Value Theorem applies. • Next, we must find f(a) and f(b).

9. Solution • We want to show that the polynomial function “has a zero in the interval [0,1]” so our k value is 0, and since –1  0  2, the Intermediate Value Theorem concludes that there is indeed a zero in the closed interval [0,1].

10. Why does the IVT fail to hold for f(x) on [-1, 1]? Not Continuous in interval! Point of discontinuity at x = 0 www.cdschools.org/.../lib/.../Intermediate_Value_Theorem.ppt

11. Show why a root exists in the given interval Continuous in interval f(-2)= -10 f(-1)= 1 www.cdschools.org/.../lib/.../Intermediate_Value_Theorem.ppt

12. For f(x) = x2 – 4, find, approximately, a value of x = c between 3 and 5 where f (c ) is exactly 12. Prove that there is a value c for which f ( c ) = 12.

13. For f(x) = x2 – 4, find, approximately, a value of x = c between 3 and 5 where f (c ) is exactly 12. F ( x) is a polynomial, hence it is continuous for all x  [3, 5]. f (a) < y < f (b)  5 < 12 < 19  There exists a c between 3 and 5 for which f (c) = 12.

14. Homework Page 80; #83-85, (87-88), 91-94