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Chapter 4. More Slope Sign Analysis

Volume 3, Why Slopes and More Math.

See the calculus previews 1 and 2 for quick or alternative view of the material in chapters 2 to 5.

Identifying intervals where a slope is positive or negative locates the uphill and downhill portions of a trail y = h(x). Several examples follow. Examples like these require and improve algebraic reasoning skills.

A Linear Function

The low-point or least value of the function occurs at the left end x = a. The high point or greatest value occurs at the right end x = b.

Another Linear Function

The low-point or least value of the function occurs at the right end x = b. The high point or greatest value occurs at the left end x = a.

[Play Video] 2¼ minutes: Slope Sign Analysis. Example of how to describe where a 2D hill has increasing height and decreasing height from sign analysis of a linear expression for the slope (derivative) of a function.

A Quadratic Function

Why this is so is an intellectual debt which you may owe yourself. From a sign analysis of the slope or derivative, what can be said about the behavior of the original function y = h(x)?

Solution. The sign analysis follows.

• For x > 3, the factor (x-3) and the slope m = 2(x-3) are both positive.

• For x < 3, the factor (x-3) and the slope m = 2(x-3) are both negative

• For x = 3, the factor (x-3) and the slope m = 2(x-3) are both zero.

The sign analysis leads to the following conclusion. The lowest point on the graph of the quadratic height function y = h(x) = x²-6x+2 is at x = 3.

Note that this conclusion also comes more from a previous knowledge of quadratics. For instance, by completing the square, y = h(x) = x²-6x+2 = (x-3)²-3²+2 = (x-3)²-7 ³ -7 with equality only at x = 3. The foregoing sign analysis gives the same information that could have been obtained by another method. In the case of quadratics, sign-analysis of slopes does not give much new information. The calculation of slopes and their sign analysis is of greater interest for more complicated height and slope formulas.

Exercises

For each of the following cases where the slope function m is given by a simple formula, find the x coordinate of the high and low points for the corresponding height function y = h(x).

1. m = 2 for 10 < x < 15

2. m = -8 for 2 < x < 4

3. m = 0 for 1 < x < 2.5

4. m = x-4 for 0< x < 8

5. m = (-1)(x-4) for 0 < x < 8

A Cubic

• the high and lowest points for x in the closed interval [-4,6].

• the greatest and least value of the height h(x) for x in the same interval [-4,6]

^Footnote: Factorizations of quadratics can also be done with the help of the quadratic formula. The case where there is no real roots can occur.

Observe

The first two subdiagrams 1 and 2 show the signs of the two factors (x+3) and (x+1) of the slope m = (x+3)(x+1). Subdiagram 3 shows or counts the number of negative signs in the computation of the slope m. This number depends on the factors. Subdiagram 4 shows where the slope m is positive and where it is negative. Based on subdiagram 4, the bottom diagram 5 employ arrows to show where the height y = h(x) is increasing and where it is decreasing. This information then gives or determines the locations of the low and high points in the interval [-4,6] where -4 < x < 6.

In particular, observe there are two high points in the interval [-4,6]. One is at x = -1 and the other is at x = 6. It is not possible to say which high point gives the greatest value of h(x) without computing h(-1) and h(6) or otherwise finding the sign of the difference h(6)-h(-1). Now a simple calculation gives h (-1) = 3.67 and h(6) = 20.0. Thus the highest point or peak occurs at x = 6 in this case. There the height is y = 20.0 = h(6). The lowest point in the interval can be found similarly.

[Play Video] 4¼ minutes: Sign Analysis for slope given by product of two linear terms, terms that appear here after the factorization of a quadratic.

Exercises

For each of the following cases where the slope function m is given by a simple formula, find the x coordinate of the high and low points for the corresponding height function y = h(x).

1. m = (-1)(x-4) for 0 < x < 8

2. m = (x-1)(x-2) for 0 < x < 4

3. m = x²-3x-2 for 0 < x < 4

4. m = x²+2x+4 for 0 < x < 10.

   Note that x²+2x+4 = (x+1)²-1+4 = (x+1)²+3 ³ 3 > 0. This quadratic is positive everywhere.
   

5. m = -10(x-1)(x-2) for 0 < x < 4

Sign Analysis Using More Factors

Problem: Find the x coordinates of the low and high points in the interval where 0 < x < 5.

Solution. The factors of m change sign at +1, +2, +4 and -3, respectively. We can almost ignore -3 as it is not in the interval of interest. A sign analysis of the factors and then of the slope m follows.

The first line subdiagrams 1 to 4 show the signs of the factors of the slope m. The line subdiagram 5 with expressions of the form (-1)^p = (-ve)^p indicates the number p of negative signs in the slope product m = (x-1)(x-2)(x-4)(x+3). This number depends on the factors and the location of x. The line in subdiagram 6 shows where the slope m is positive and where it is negative. The arrows below it indicate the behavior of the height y = h(x). That is, they show where the height y = h(x) is increasing and where it is decreasing. This information locates the low and high points in the interval [0,5] of interest.

In particular, from the above analysis, there are low points at x = 1 and at x = 4. There are also high points at x = 0, at x = 2 and x = 6. Note no conclusion can be drawn from the above analysis about which high point is highest or which low point is lowest. The ability to compute the height h(x) at these points would help in locating the highest and lowest points.

[Play Video] 6¾ minutes: Sign Analysis for slope given by product of three linear terms

Exercises

For each of the following cases where the slope function m is given by a simple formula, find the x coordinate of the high and low points for the corresponding height function y = h(x).

1. m = (x+2)(x-5)(x+1) for -5 < x < 4

2. m = (x-2)) (x-3)(x+1) for -5 < x < 4

Sign Analysis with Divisors (Rational Functions)

[Play Video] 5 minutes: (coming soon) Sign Analysis for slope given by quotient of linear terms

Postscript: (October 2008): Except at x = -3 and when x= +4, the value of m₁ = g₁(x) is undefined. That is small difference. The sign analysis for m = g(x) = (x-1)(x-2)(x-4)(x+3). and hence m₁ = g₁(x) is reproduced below with a minor change or two to avoid or identify division by zero.

Exercises

For each of the following cases where the slope function m is given by a simple formula, find the x coordinate of the high and low points for the corresponding height function y = h(x).

1. $m = \frac{(x+2)(x+5)}{x+1}$ for -5 < x < 4

2. $m = \frac{x+2}{(x+5)(x+1)}$ for -5 < x < 4

Pay attention to the end points of each interval. Each end point of an interval may be a low or a high point, that is a minimum or maximum.

When \[m =\frac{q(x)}{r(x)}\] is a ratio of two polynomials, sign analysis may be done by factoring of both the numerator (top) polynomial q(x) and the denominator (bottom) polynomial r(x) into linear and quadratic factors. Here the quadratic factors which have real roots should be replaced by a product of linear factors.

The Fundamental Theorem of Algebra (proven by Gauss), says in principle that any polynomial can be expressed as a product of linear and quadratic factors. But Galois theory, a specialized topic in advanced algebra, implies no single exact formula or method, involving roots of real and complex numbers for the factors, will suffice to factor a polynomial of degree n > 5. In contrast, formulas involving such roots are known for polynomials of degree n < 5. With the advent of the computer, approximate methods can also be used to approximately find the roots and factors of some polynomials. Moreover, for some special kinds of polynomials of a fixed degree n > 5, exact formulas involving roots of complex numbers can also be obtained - Galois theory just implies that no one formula will work for all polynomials of degree n > 5.

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
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5. Arith with units - science