Intermediate Value Theorem

Suppose a < b and f(a) ¹ f(b), where f is a continuous function. Then the Intermediate Value Theorem says that for any value k where k is between f(a) and f(b), there exists a value cÎ(a, b) such that f(c) = k.

This is easy to understand if you recall that cÎ (a, b) means a < c < b, and you draw a picture: In this picture there were three c’s that worked for the chosen k. There are other values of k that would have resulted in only one c.

This theorem is often used to show that a function has a zero between two given x values, but that is just a special case of the theorem. When showing that the function has a zero between a and b, you need to show that f(a) and f(b) have different signs, i.e. one is positive and the other is negative, where f is a continuous function.

Notice the theorem only applies if f is a continuous function. An easy way to understand continuous, is by looking at the drawing. You can draw a continuous function without any jumps or holes. You don’t need to lift your pencil. All polynomial functions are continuous. Rational functions are not continuous in general.

Practice

1.) Let f(x) be a polynomial function where f(-3) =7 and f(4) = 12.

a.) Does f(x) = 2 for some xÎ (7, 12)? yes, no, possibly
b.) Does f(x) = 8 for some xÎ (-5, 5)? yes, no, possibly
c.) Does f(x) = 9 for some xÎ (-2, 3)? yes, no, possibly

2.) Let g(x) = 4x³ – 200x² + 2400x. Does g(x) = 5000 for some xÎ (10, 15)? Use the Intermediate Value Theorem to justify your answer.

Answer:

1.) a.) possibly, b.) yes, since we know it equals 8 for some xÎ (-3, 4) and (-3, 4) Ì (-5, 5) c.) possibly, we know x = 9 for some xÎ (-3, 4)

2.) g(10) = 8000 and g(15) = 4500, so g(15) < 5000< g(10). g(x) is a continuous function, since it is a polynomial function. Thus by the Intermediate Value Theorem g(x) must equal 5000 for some x between 10 and 15.