The Intermediate Value Theorem (often abbreviated as IVT) says that if a continuous function takes on two values y_{1} and y_{2} at points a and b, it also takes on every value between y_{1} and y_{2} at some point between a and b. A typical argument using the IVT is:

Know: | e^{0} = 1 and e^{1}= e,1 < 2 < e, e ^{x} is continuous for all real x. |

Conclude: | e^{x} = 2 for some x with 0 x 1. |

Graphically The IVT says that if y_{1} = f(a) and y_{2} = f(b) for a function f(x), and if we draw the horizontal line y = y_{0} for any y_{0} between y_{1} and y_{2}, then the horizontal line intersects the graph of y = f(x) at (at least one) point whose x-coordinate is between a and b.

The formal statement of the IVT is:

The Intermediate Value TheoremLet f(x) be a function which is continuous on the closed interval [a,b] and let y Then there is at least one c with a c b such that y |

Note: "continuous on the closed interval [a.b]" means that f(x) is continuous at every point x with a < x < b and that f(x) is right-continuous at x = a and left-continuous at x=b.

**Example:** Show that there is some u with 0 < u< 2 such that u^{2} + cos(u) = 4.

To do this we apply the IVT to the function h(u) = u^{2} + cos( u). h(u) is the sum of two functions which we know are continuous (these are Field Guide functions), so h(u) is continuous. We can compute that

h(0) | = | 0^{2} + cos(0*) | = | 1. |

h(2) | = | 2^{2} + cos(2) | = | 5. |

Since h(0) = 1 < 4 < 5 = h(2), and since h is continuous, the IVT implies that h(u) = 4 for some u in the closed interval [0,2]. This u cannot be one of the endpoints, so 0 < u< 2.

In order to understand the IVT, we should take a closer look at

- What the IVT says and does not say.
- Why continuity is a necessary hypothesis.

This is done on the next page. Thereafter we give an application of the IVT to root finding.

/Stage4/Lesson/IVT.html

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William A. Bogley

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