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:
Let 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
This is done on the next page. Thereafter we give an application of the IVT to root finding.
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All rights reserved---1996
William A. Bogley