MATH 12002 - CALCULUS I 3.3 Preview: Sign Charts My Way Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.3 Preview: Sign Charts — My Way

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 5

Signs and Sign Charts

To determine the sign of a function g(x), we use the following basic facts:

1 If a function is continuous everywhere it is defined,

then it can change sign only where it is either 0 or undefined. Thus, on the interval between two such points, the sign of the function is constant.

2 A quotient of two functions (for example, a rational function)

is 0 where the numerator is 0 (and the denominator is not) and is undefined where the denominator is 0. Therefore, given a function g(x), we need to determine where it is 0 or undefined, and then determine the sign of g(x) on the intervals with these points as endpoints. There are two standard ways to determine the sign of a function on the interval between two zeros or undefined points: One way is to evaluate the function at some number in the interval. All of the function values on the interval will have the same sign as that y value. This method is a lot of work, even with a calculator, and I rarely use it.

D.L. White (Kent State University) 2 / 5

Signs and Sign Charts

The other method is to use a sign chart with the signs of the factors. This method is based on the following:

1 A linear factor, ax + b, will be zero at one point (x = −b

a)

and will be positive on one side of the zero and negative on the other.

2 Signs “multiply” and “divide” as follows:

+ · + = + + + = + + · − = − + − = − − · + = − − + = − − · − = + − − = +

3 A product/quotient of an even number of negative factors is positive.

A product/quotient of an odd number of negative factors is negative.

D.L. White (Kent State University) 3 / 5

Example

Example

Determine the intervals where the function g(x) =

x3+2x2 −x2+3x−2 is positive

and the intervals where it is negative.

Solution

First notice that g(x) = x2(x + 2) (x − 2)(1 − x). Therefore, g(x) = 0 at x = −2 and at x = 0, and g(x) is undefined at x = 1 and x = 2. The sign of g is constant on each of the intervals (−∞, −2), (−2, 0), (0, 1), (1, 2), (2, ∞). We begin our sign chart with a number line with these x values in order: −2 1 2 [Continued →]

D.L. White (Kent State University) 4 / 5

Example

Solution [continued]

g(x) = x2(x + 2) (x − 2)(1 − x): −2 1 2 x2 x + 2 x − 2 1 − x g(x) + + + + + − + + + + − − − − + + + + − −

X X

+ − − + − Therefore, g(x) is positive on (−∞, −2) ∪ (1, 2) and g(x) is negative on (−2, 0) ∪ (0, 1) ∪ (2, ∞).

D.L. White (Kent State University) 5 / 5