### The Intermediate Value Theorem

The Intermediate Value Theorem (often abbreviated as IVT) says that if a continuous function takes on two values y1 and y2 at points a and b, it also takes on every value between y1 and y2 at some point between a and b. A typical argument using the IVT is:
 Know: e0 = 1 and e1= e,1 < 2 < e,ex is continuous for all real x. Conclude: ex = 2 for some x with 0 x 1.

Graphically The IVT says that if y1 = f(a) and y2 = f(b) for a function f(x), and if we draw the horizontal line y = y0 for any y0 between y1 and y2, 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 y0 be a real number lying between f(a) and f(b), i.e. with f(a) y0 f(b) or f(b) y0 f(a). Then there is at least one c with a c b such that y0 = f(c).

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 u2 + cos(u) = 4.

To do this we apply the IVT to the function h(u) = u2 + 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) = 02 + cos(0*) = 1 h(2) = 22 + 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