Solutions Part 2

x ( x – 1 ) – ( x – 1 ) / x > 0

[ x²( x – 1 ) - ( x – 1 ) ] / x > 0

[ ( x² – 1 ) ( x – 1 ) ] / x > 0

[ ( x + 1 ) ( x – 1 )²] / x > 0

So the critical numbers are –1, 1 and 0. –1 and 0 are transitive and 1 is intransitive. The ratio of the leading coefficients is positive, so the expression is positive on the interval ( 1, ∞ ). The sign doesn’t change at 1, so the expression is also positive on the interval ( 0, 1 ). The sign changes at 0, so the expression is negative on the interval ( -1, 0 ). The sign changes at –1, so the expression is positive on the interval ( -∞, -1 ). Thus the solution is ( -∞, -1 ) U ( 0, 1 ) U ( 1, ∞ )

The critical numbers are 1/2, -1, and 1. 1/2 and –1 are transitive and 1 is intransitive. The ratio of leading coefficients is positive, so the expression is positive on the interval ( 1, ∞ ). The sign doesn’t change at 1, so the expression is still positive on the interval ( 1/2, 1 ). The sign changes at 1/2, so the expression is negative on the interval ( -1, 1/2 ). The sign changes at –1, so the expression is positive on the interval ( -∞, 1 ). Thus, the solution is the interval ( -1, 1 / 2 ).

The solution is ( 2, ∞ )

3 x > 12

x > 4

The solution is ( 4, ∞ )

f ( x ) = [ ( x - √3 ) ( x + √3 ) ] / [ ( x – 2 ) ( x + 1 ) ]

So the x intercepts are at ( ±√3, 0 ) and the vertical asymptotes are at 2 and –1. All the intercepts and asymptotes are transitive, so the function changes sign at each intercept and asymptote.

The quotient is 1 and the remainder is x – 1. So y = 1 is the horizontal asymptote and there is a transitive quotient intercept when x – 1 = 0, so the graph crosses its horizontal asymptote when x = 1. The ratio of the leading coefficients is positive, so the graph is above the x-axis on the interval ( 2, ∞ ).

Since the denominator has no zeros, there are no vertical asymptotes. The quotient is 2 and the remainder is

x – 2, so the graph has a horizontal asymptote y = 2, and the graph crosses the horizontal asymptote when

x = 2. The graph has two x intercepts ( 0, 0 ) and ( - 1/2, 0 ), both of which are transitive. The ratio of the leading coefficients is positive, so the graph is above the x-axis on the interval ( 0, ∞ ). The graph crosses the horizontal asymptote at ( 2, 2 ) and then approaches the horizontal asymptote from above. The graph dips below the x-axis on the interval ( - 1 / 2, 0 ), but lies above the x-axis and below the horizontal asymptote on the interval ( -∞, - 1/2 ).

The quotient is 2, so y = 2 is the horizontal asymptote. The zeros of the denominators are ±1, so the vertical asymptotes are x = 1 and x = -1.

The zeros of the denominator are ±1, so the equations of the vertical asymptotes are x = 1 and x = -1.

g ( - x ) = x³ – 5 x – 2. If we find the least integral upper bound c on the set of zeros of g ( - x ), then – c will be the greatest integral lower bound on the set of negative zeros of g ( x ).

The first entry to give a row of all non-negative numbers is 3, thus –3 is the greatest integral lower bound on the set of zeros of g ( x ).

f ( - x ) = – x³ + 5 x + 2. If we find the least integral upper bound c on the set of zeros of f ( - x ), then – c will be the greatest integral lower bound on the set of negative zeros of f ( x ). But we will never get a row of all zeros, because the first entry will always be –1. So we look for the bounds on the zeros of – f ( x ), since – f ( x ) has the same zeros as f ( x ). – f ( - x ) = x³ - 5 x - 2

The first entry to give a row of all non-negative numbers is 3, thus –3 is the greatest integral lower bound on the set of zeros of f ( x ).

Since the leading coefficient is negative, we’ll never get a row of all positive numbers. Since the set of zeros of g ( x ) is the same as the set of zeros of – g ( x ), we use – g ( x ) instead of g ( x ).

Thus 4 is the smallest positive integer which is an upper bound on the set of zeros of the function.

Thus 4 is the smallest positive integer which is an upper bound on the set of zeros of the function.

There are three changes of sign in the coefficients of p ( x ) and no changes of sign in the coefficients of p ( -x ). Thus there are either three or one positive zeros and no negative zeros. The chart showing the possible distribution of zeros follows:

There are five changes of sign in the coefficients of g ( x ), thus there are at most five positive zeros. Since there can be fewer zeros than the maximum possible, and since the zeros decrease only by an even amount, there can be five, three or one positive zeros.

The possible rational zeros are ±1, ±5, ±1/2, ±5 / 2. Applying synthetic division, we find

So 1/2 is a zero of the function. We also see that the function factors as

Thus, the complex zeros are the zeros of .

Using the quadratic formula, the other two zeros are

f ( x ) = x² + 2 x + 1 = ( x + 1 )², thus –1 is a zero twice.

Applying the quadratic formula, we get

Using synthetic division

Which also shows us that

Use the abbreviated form of synthetic division where the middle row is omitted and the carry is done mentally.

q ( x ) = 3 x³ 4 x² + 2 x – 3, and the remainder r = – 2. So

f ( x ) = ( x – 1 ) (3 x³ 4 x² + 2 x – 3 ) – 2

So f ( 3 ) = 238.

( ( ( 3 x + 1 ) x – 2 ) x – 5 ) x + 1

( ( 2 x – 3 ) x – 11 ) x + 6 = ( 2 x² – 3 x – 11 ) x + 6

= 2 x³ – 3 x² – 11 x + 6

The zeros are – 2, – 1, 1 and 3. –2, 1 and 3 are transitive zeros, while –1 is intransitive. Since the leading coefficient is negative, the graph lies below the x-axis on the interval ( 3, ∞ ). The graph crosses the x-axis at 3, 1 and – 2, but at – 1, it touches the x-axis, but does not cross.

Detail in vicinity of –1 (notice that the vertical scale is different from first depiction):

The leading coefficient is 1, which is a positive number. This means that the graph of the function on the interval to the right of the largest x-intercept lies above the x-axis. Since the graph crosses the x-axis at 1 and 2, but does not cross at the x-intercept at –1, the graph will look like the following:

Detail in the vicinity of 2 (different scale on the vertical axis):

From the factor ( x + 1 )², we get the zero – 1. Since the zero is of multiplicity two and since 2 is an even number, the zero is of even multiplicity. Thus the zero is intransitive. This means that the sign of the factor

( x + 1 )² does not change at the zero – 1, thus the graph of the function does not cross the x-axis at – 1.

From the factor ( x – 1 ), we get the zero 1. Since the zero is of multiplicity 1 and since 1 is an odd number, 1 is a zero of odd multiplicity. Thus it is a transitive zero. The sign of the factor ( x – 1 ) changes at 1, so the graph of the functions crosses the x-axis at 1.

From the factor ( x – 2 )³, we get the zero 2. Since the zero is of multiplicity 3 and since 3 is an odd number, 2 is a zero of odd multiplicity. Thus the sign of the factor ( x – 2 )³ changes at 2. So the graph of the function crosses the x-axis at 2. There is a special feature of zeros of odd multiplicity greater than 1: The graph not only crosses the x-axis at the zero, it flattens out to horizontal at the cross-over point.

Detail in the vicinity of 1 and 2:

The possible rational zeros are ±1, ±2, ±3, ±6, ± 1/2 and ± 3 / 2.

f ( 1 ) = - 6, f ( -1 ) = 12, f ( 2 ) = - 12, f ( - 2 ) = 0 so ( x + 2 ) is a factor.

f ( 3 ) = 0 so ( x – 3 ) is a factor.

f ( 6 ) = 264, f ( - 6 ) = 468, f ( 1/2 ) = 0, so ( x - 1/2 ) is a factor.

A cubic polynomial can have only three first degree factors, so we have found all first degree factors. However, when we multiply these three factors together, we get

x³ – ( 3/2 ) x² – ( 11/2 ) x + 3 which is only one-half of f ( x ). Thus,

f ( x ) = 2 ( x + 2 ) ( x – 3 ) ( x - 1/2 )

We can multiply the 2 by the ( x - 1/2 ) to get

f ( x ) = ( x + 2 ) ( x – 3 ) ( 2 x - 1 )

We probably would have found the zeros of – 2 and 3 by trial and error, but not the zero 1/2. If it were not for the rational zero theorem, we would not have known to try 1/2.

f ( 1 ) = 4, f ( 2 ) = 0, so ( x – 2 ) is a factor. Knowing this factor, we can take a shortcut and divide this factor into f ( x ) to get ( x² – 2 x – 3 ) which we can factor by conventional means as ( x – 3 ) ( x + 1 ). Thus,

f ( x ) = ( x – 2 ) ( x – 3 ) ( x + 1 ).

f ( 1 ) = 1 + 2 – 1 – 2 = 0

f ( - 1 ) = - 1 + 2 + 1 – 2 = 0

f ( - 2 ) = - 8 + 8 + 2 – 2 = 0

In order to complete the square, the coefficient of the x² term must be one. If the coefficient is some other number besides one, that number must be factored out of both x terms.

f ( x ) = 2 ( x² – 2 x ) + 3

f ( x ) + 2 ( 1 ) = 2 ( x² – 2 x + 1 ) + 3

f ( x ) + 2 = 2 ( x – 1 )² + 3

f ( x ) = 2 ( x – 1 )² + 1

So the vertex is ( 1, 1 ). The graph is the same as the graph of y = 2 x² shifted one unit right and one unit up.

Since the zeros are – 1 and 5, the x-intercepts are ( - 1, 0 ) and ( 5, 0 ).

The x-coordinate of the vertex is – b / 2 a = 2, so the y-coordinate is f ( 2 ) = 9.

f ( x ) = - ( x² – 4 x – 5 ) = - ( x – 5 ) ( x + 1 ), so the zeros are 5 and – 1.

Given, f ( x ) = 2 x² + 8 x + 1, the zeros are the values of x which result in an output value of y = 0. Thus the zeros are the solutions of the equation

2 x² + 8 x + 1 = 0

Using the quadratic formula, we get

$\begin{displaymath}x=\frac{-8\pm\sqrt{8^2-4(2)(1)}}{2(2)}=\frac{-8\pm\sqrt{56}}{4}=-2\pm\frac{\sqrt{14}}{2} \end{displaymath}$

The quadratic formula:

If a x² + b x + c = 0, then

Given f ( x ) = 2 x² + 8 x + 1. The parabola opens upwards, since a = 2, which is a positive number. The vertex of the parabola has x-coordinate – b / ( 2a ) = - 2. The y-coordinate of the vertex is f ( -2 ) = -7. Thus, -7 is the smallest output of the function. The range of the function is [-7,∞).

If a = 0, then the polynomial will be a first degree polynomial, not a second degree polynomial. A quadratic function has a second degree polynomial equation.

The slope a = [ T ( 98.6 ) – T ( 32 ) ] / ( 98.6 – 32 ) = 37 / 66.6 = 370 / 666 = 5 / 9.

Thus, T ( x ) = ( 5 / 9 ) x + b, for some constant value of b. To find the value of b, use the fact that

T ( 32 ) = 0 to get T ( 32 ) = ( 5 / 9 ) ( 32 ) + b = 0. Thus, b = - 160 / 9. Therefore,

T ( x ) = ( 5x – 160 ) / 9