to increase, passing once more through the value zero when x=about 4.1.

90. Relation of Graph of f(x) to Solution of f(x)=0.—The graph of a function f(x) is chiefly used, in algebra, as an aid to the solution of the equation f(x)=0. From the graph of

f(x)=(4x3 — 6x2 −45x+12)

just considered, it is evident that the values of x for which f(x)=0 are roughly 2.8, +0.2, +4.1; that is, these are the approximate roots of f(x)=0.

If the absolute term of f(x) is decreased by a units, the graph of the resulting function,

f1(x)=√(4x3 — 6x2-45x+12—10a),

may be got by moving the axis upward a units. This process will change the roots, but can not change their number, since f(x) and f(x) are of the same degree.

1

2

If a=4, we clearly have two negative roots in the vicinity of the maximum point, as the new x-axis A,B1 cuts across the curve just below that point; if a 5.5, the new x-axis A,B2 passes above the maximum point, and only one real root exists, the lefthand roots having disappeared. If a 5.25, the axis will be tangent to the curve at the maximum point, and there the points indicating the two roots will merge into one. The graph indicates, in these three cases, two distinct, two imaginary, and two equal roots, respectively, by its appearance in the vicinity of this maximum point.

If, however, the y-axis (i. e., the origin) be shifted to the right or left, the magnitude of the roots will be diminished or increased, but the character of the roots, whether real or imaginary, remains the same. If the origin be shifted one unit to the right, the values of x for the points where the curve crosses the x-axis are changed by -1.0, from -2.8 to 3.8, from +0.2 to 0.8, from +4.1 to +3.1. With the y-axis in this position, the curve is the graph of f(x+1)=√(4x3+6x2 −45x−35).

91. Multiple Roots.-When the axis of abscissas is tangent to the graph, i. e., when f(x) and f'(x) are zero for the same value of x, this value is a double root of f(x)=0.

In fact, if f(r) =f'(r) =0, r is at least a double root of f(x)=0; it may occur more than twice as a root.

It can be shown that if the common root r occurs k times as a root of f'(x)=0, it is a root (k+1) times of f(x)=0.

We may thus determine the equal roots of an equation by finding all the factors common to the functions f(x) and f'(x), which is most conveniently done by the process of finding the greatest common divisor.

For example, find the equal roots of the equation

f(x) = x2-6x3 +12x2-10x+3.

The derivative of f(x) is

f'(x)=4x3-18x2+24x-10.

Their G. C. D., by detached coefficients, is found below:

f'(x) = (x−1)2 (4x—10),

f(x)=(x-1) ( x— 3),

and the equal roots of f(x)=0 are 1, 1, 1.

Examples.

1. f(x)=x3 — 5x2+3x+9; determine the maximum and minimum points, and equal roots, if any.

2. Construct the graph for 3-2x2-5x+5=0, first finding the points of maximum and minimum values.

3. Show that the equation - 10x3 +36x2 — 54x+27=0 has equal roots, and find them.

4. Find the minimum point of the quadratic polynomial (x2+2x-4) and draw the graph.

5. Construct the graphs for the following, and estimate the roots of each equation to the nearest tenth (draw only one curve) :

6. What constant term would give equal roots to the equation 3x36x2-5x+a=0?

7. If f(x) = ax2+bx+c, find the minimum value of f(x), a being positive, and the conditions that this minimum shall be negative, zero, or positive.

8. Construct the graph for a3+2x2+3x-4=0 and locate the roots approximately.

9. Find the maximum and minimum values of x+px+q. What is the condition that two of the roots of x3+px+q=0 be equal?

10. How may the two graphs of x3 and (−px−q) be used to show the roots of x=-px-q? Apply to Example 8.

CHAPTER VI.

HORNER'S SOLUTION OF EQUATIONS.

92. Horner's Method.-Suppose that we wish to find a root of the equation

f(x)=2x3-15x2+24x+6=0.

x'

X

FIG. 7.

Any constant multiple of f(x) is zero whenever f(x)=0, and only then; so we first draw the graph of k f(x), giving k any convenient value. In the figure, k. From the graph, it appears that there is a root of f(x)=0 between -1 and 0, one between 2 and 3, and one between 4 and 5. We will compute the root between 2 and 3.

We first deduce the equation with roots each 2 less than those of f(x)=0; its graph will be the curve already drawn if the

axis of ordinates is moved 2 units to the right from 0 to 0'; the new equation has a root between 0 and 1, corresponding to the root of f(x) between 2 and 3. If this root is s, we have, substituting it in the new equation,

f1(s)=283-382-12s+10=0.

2) 2-15 +24 + 6

4 -22 + 4

-11+2+10

4

-14

7;-12

and

Now as s is less than 1, its square cube are much smaller, and we shall come pretty near to finding its true value by neglecting the first terms and solving 12s+10=0, whence s=.8, about. Trying .8, we find that f1(.8) is negative, showing, by the graph, that .8 is greater than the root of f1(s), because the corresponding point of the graph is below the x-axis. It is our object to take as large a value as we can, and still keep to the left of the root. fi(.7) is positive, hence the root of f1(s) =0 is between 7 and .8, and the corresponding root of f(x)=0 is between 2.7 and 2.8.

4

2;- 3

We next derive the equation having a root .7 less than the root of f1(s) =0, or 2.7 less than the root of f(x)=0, and so proceed progressively, getting finally as close an approximation to the actual root as we desire. We shall no longer need to try two values at each step, as with the decrease of the value substituted, the neglect of the terms of higher degree will cease to affect the single figure that we desire to obtain.

We shall use the principle of contraction, retaining only as many decimal places in the last column as we desire in the root, and using only as many in the earlier columns as are needed to determine those in the last. The coefficients of the successive reduced equations may be marked, in order to keep track of the work.