MATH 12002 - CALCULUS I 1.5: Intermediate Value Theorem Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §1.5: Intermediate Value Theorem

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Intermediate Value Theorem

The Intermediate Value Theorem says that if a continuous function has two different y-values, then it takes on every y-value between those two values. The precise statement of the theorem is the following.

Intermediate Value Theorem

If y = f (x) is continuous on the interval [a, b] and N is any number strictly between f (a) and f (b), then there is a number c, with a < c < b, such that f (c) = N.

✲ ✛ ✻ ❄

a f (a)

q

b f (b)

q

N c f (c) = N

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Intermediate Value Theorem

Alternatively, suppose we have two points in the xy-plane and we draw a horizontal line between them at y = N. In order to draw the graph of a continuous function y = f (x) between the points, we must cross the line y = N somewhere. The x-coordinate of the intersection point is the number c in the theorem.

✲ ✛ ✻ ❄

a

q

b

q

N

✲ ✛ ✻ ❄

a f (a)

q

b f (b)

q

N c

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Finding Zeros

Letting N = 0 in the IVT gives us the following special case.

Corollary

If y = f (x) is a continuous function and there are numbers a and b such that f (a) is positive and f (b) is negative, then there is a zero of f between a and b; that is, there is a c between a and b such that f (c) = 0.

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Finding Zeros

The corollary can be used to approximate zeros of functions, as follows.

Example

Show that f (x) = x3 + 3x − 5 has a root between x = 1 and x = 2 and estimate the value of the root to within 0.01. First observe that f is a polynomial, so is continuous everywhere, and that f (1) = 13 + 3(1) − 5 = −1 < 0 and f (2) = 23 + 3(2) − 5 = 9 > 0. By the Corollary, there is a number c, 1 < c < 2, such that f (c) = 0. Now compute the value of f for a number between 1 and 2; say f (1.5). If f (1.5) > 0, then f has a root between x = 1 and x = 1.5. If f (1.5) < 0, then f has a root between x = 1.5 and x = 2. Check that f (1.5) = 2.875 is positive. Since f (1) is negative, there is a root between 1 and 1.5. [Continued →]

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Finding Zeros

Now f (1.2) = 0.328 is also positive. Again, f (1) is negative, so there is a root between 1 and 1.2. Continue this way: f (1.1) = −0.369 is negative and f (1.2) is positive ⇒ root in (1.1, 1.2). f (1.15) ≈ −0.029 is negative and f (1.2) is positive ⇒ root in (1.15, 1.2). f (1.18) ≈ 0.183 is positive and f (1.15) is negative ⇒ root in (1.15, 1.18). f (1.16) ≈ 0.041 is positive and f (1.15) is negative ⇒ root in (1.15, 1.16). Therefore, we know that f has a root in the interval (1.15, 1.16), which has length 0.01. NOTE: Continuing this method, you can show that the root is between 1.1541714951 and 1.1541714952.

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