Chapter Five Notes N P U2C5

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1 Chapter Five Notes N P UC5 Name Period Section 5.: Linear and Quadratic Functions with Modeling In every math class you have had since algebra you have worked with equations. Most of those equations have probably involved the use of polynomial functions equations that have exponents on variables that are whole numbers. By definition a polynomial function of degree n is a function given by f(x) = a n x n + a n- x n- + + a x + a x + a 0 where n is a nonnegative integer and a 0 a a n- a n are real numbers where a n is not zero. For this definition the leading coefficient is a n and n is the degree of the polynomial (the biggest exponent on a variable). Realize that there are special types of polynomial functions. Polynomial Functions and Their Degrees Name of Degree Form (a 0) Example function Constant 0 f(x) = a f(x) = Linear f(x) = ax + b f(x) = 3x + Quadratic f(x) = ax + bx + c f(x) = x + 3x Cubic 3 f(x) = ax 3 + bx + cx + d f(x) = 5x 3 + x 7 Quartic 4 f(x) = ax 4 + bx 3 + cx + dx + e f(x) = x 4 3x 3 If a function is a polynomial function it must have the following properties: Exponents on variables must be zero or positive integers. A polynomial function can be simplified into polynomial form if under a radical. Example : Determine which are polynomial functions. For those that are state the degree and leading coefficient. For those that are not explain why not. #: f(x) = 3x 5 + 7: This is not a polynomial function since one of the variable exponents is a negative number. All exponents must be zero or positive integers. 5 #: f ( x) x x 9 : This is a polynomial function. The degree of the polynomial is 5 (the biggest exponent on a variable) and the leading coefficient is #3: f ( x) 7x 8x : This is not a polynomial function since the function cannot be simplified into a polynomial function outside the radical. A power function is a special type of function that takes some value x to a power. By definition a power function is as follows: f(x) = kx a where k and a are both not zero and are constants (numbers not variables) There are specific names for each part of a power function: a: power of the power function k: constant of variation (or proportion) page N P UC5

2 For this function we say that f(x) varies as the a th power of x or that f(x) is proportional to the a th power of x. We have already discussed a number of types of polynomial functions based on their degree or largest exponent on a variable. Below are the main types discussed: Type of function Degree of function Graph Constant 0 Straight horizontal line Linear Slanted line Quadratic Parabola Qubic Quartic Recall that a polynomial is written as follows: a n x n + a n- x n- + + a 0 Each monomial is called a term of the polynomial. Polynomials are written in standard form if terms are written left to right with descending (decreasing) degrees. a n a n- etc. are coefficients of the polynomial. a n x n is the leading term (the term with the highest power on the variable). a 0 is the constant term. n is the degree of the polynomial. In Chapters Three and Four you learned how to graphically transform functions based on adding/subtracting values directly to the x term adding/subtracting terms to the function and multiplying the x term by some number. The same graphical transformations can be applied to basic monomial functions. In addition you can find intercepts for those graphs by recalling the following from advanced algebra: The x-intercept is the point on the graph where the graph crosses the x-axis. An x-intercept will always be of the form (x 0) where the x is found by substituting 0 in for y (or f(x)). Also x- intercepts will often be referred to as zeros since the y-value for these points is zero. The y-intercept is the point on the graph where the graph crosses the y-axis. A y-intercept will always be of the form (0 y) where the y is found by substituting 0 in for x. Finally you might remember finding minimums and maximums during the previous chapter. These are also referred to as turning points because they turn increasing to decreasing or vice versa. Example : Describe how to transform the graph of an appropriate monomial function f(x) = x 3 into the graph of g(x) = (x 3) 3. Sketch the transformed graph by hand and support your answer with a grapher. Compute the location of the y-intercept as a check on the transformed graph. From the resulting graph it would be safe to say that the original function was f(x) = x 3. This is a monomial function and to get it to result into g(x) you multiplied the function by and subtracting 3 from the x term. Recalling from Chapter One what this means as far as translations and transformations the following occurred: The original function was vertically stretched by a factor of. The result was then translated horizontally 3 units to the right. Knowing this you can take the basic function f(x) = x 3 (one of the basic functions you learned in Chapter One) stretch it vertically by a factor of and move it 3 units to the right. To find the y-intercept for the graph simply substitute 0 for x and solve for y. page N P UC5

3 g(0) = (0 3) 3 = ( 3) 3 = 54 y-intercept: (0 54) Verify this by graphing it on your calculator. Also graph the function below. One can also graph combinations of monomial functions albeit the results are a bit unpredictable. Example 3: Graph the polynomial function f(x) = x 4 + x locate its extrema and zeros and explain how it is related to the monomials from which it is built. The basic graph of a function always comes from its leading term so you are in essence graphing y = x 4. As the graph approaches the origin though it behaves like y = x which you might expect considering you are adding graphs of monomial functions. To find the local extrema and zeros first graph the function. After doing so you can see quite visibly where the extrema are. To find specifically where the zeros are press nd and then TRACE on your calculator and select maximum minimum or zero and set appropriate left and right bounds to find the following extrema: Local maximum: (0.79.9) Zeros (areas where the graph crosses the x-axis): (0 0) and (.6 0) Note also that the function is neither even nor odd since it is not symmetric with respect to either the y- axis or origin. In advanced algebra you also learned how to predict the maximum number of local extrema and zeros for any function. RULE: A polynomial function of degree n will have at most n local extrema and at most n zeros. WHY??!! Recall that a quadratic equation has at most two factors meaning two zeros. The same goes for cubic functions quartic functions pentic (is that a word?) functions and so on. Finding the end behavior of any polynomial function is easy to do as well because it can be easily predicted. page 3 N P UC5

4 Recall that linear functions have odd degrees as do cubic functions. All odd-degree polynomial functions have the same end behavior: If the leading coefficient is positive If the leading coefficient is negative x x x x Notice that for odd-degree polynomials one end always goes up and one end always goes down. Also recall that quadratic functions have even degrees as do quartic functions. All even-degree polynomial functions have the same end behavior: If the leading coefficient is positive If the leading coefficient is negative x x x x Notice that for even-degree polynomials both ends always go up or down depending on the sign of the leading coefficient. Example 4: Describe the end behavior of the polynomial f(x) = 3x 4 5x + 3 using limit notation. Since you are dealing with an even-degree polynomial either both ends go up or both ends go down. Because you have a positive leading coefficient both ends go up or approach positive infinity. In other words x x Recall from earlier in this section that zeros of a polynomial are the same thing as x-intercepts or when y or f(x) is zero. You don t necessarily need to graph a function to find its zeros; factoring the polynomial when it is set equal to zero and then solving can be much easier. Example 5: Find the zeros of the function f(x) = 3x 3 x x algebraically. page 4 N P UC5

5 To solve this first let f(x) = 0 and then factor to solve. The zeros are 0 /3 and. 0 = 3x 3 x x 3x 3 x x = 0 x(3x x ) = 0 x(3x + )(x ) = 0 x = 0 3x + = 0 x = 0 x = 0 x = /3 x = Graphing this you will see that the graph crosses the axis 3 times meaning there are three distinct zeros. You can also work backwards to construct a function realize that if you have a zero c the factor that presents it is (x c). Example 6: Using only algebra find a cubic function with the zeros 3 4 and 6. Support by graphing your answer. Since your zeros are 3 4 and 6 the factors that make them up are (x 3) (x + 4) and (x 6). f(x) = (x 3)(x + 4)(x 6) f(x) = (x + x )(x 6) f(x) = x 3 5x 8x + 7 However have you ever noticed with some graphs that the graph may hit the x-axis but not cross it? This deals with multiplicity. Example 7: Graph both f(x) = (x ) 3 and g(x) = (x ). page 5 N P UC5

6 What you will see is the cubic functions crosses the x-axis at 0 while the quadratic function bounces back from the x-axis at zero. This relates to the concept of multiplicity. Relating to zeros of polynomial functions If f is a polynomial function and (x c) m is a factor of f but (x c) m + is not then c is a zero of multiplicity m of f (essentially a repeated zero or factor). If m is odd the graph of f crosses the x-axis at (c 0) and the value of f changes sign when x = c. If m is even the graph of f bounces back from the x-axis at (c 0) and the value of f doesn t change sign when x = c. One other consideration when dealing with graphs is what is known as the Intermediate Value Theorem: If a and b are real numbers with a < b and if f is continuous on the interval [a b] then f takes on every value between f(a) and f(b). (In easy terms if you have some value between a and b for your x the y you get out has to be between the values f(a) and f(b). If f(a) and f(b) have opposite signs then f(c) = 0 for some c in [a b]. One last thing: always look for hidden zeros. Although you might graph a function and appear to see only one zoom in to make sure only one is there; there might be more. Example 8: Form a polynomial functions whose real zeros and degree are listed with a leading coefficient of. Express the polynomial in standard form. Zeros: 3 (multiplicity of ) 4 (multiplicity of ) degree 3 Since your zeros are 3 and 4 the factors that make them up are (x + 3) and (x 4). Since the multiplicity of 3 is and 4 is the factors when multiplied are (x + 3) (x 4) or (x + 3) (x 4). Since the leading coefficient is this changes simply to (x + 3) (x 4). f(x) = (x + 3) (x 4) Function f(x) = (x + 3)(x + 3)(x 4) Definition of power f(x) = (x + 6x + 9)(x 4) Multiplication of (x + 3)(x + 3) f(x) = (x + x + 8)(x 4) Distribution of f(x) = x 3 8x + x 48x + 8x 7 Multiplication of (x + x + 8)(x 4) f(x) = x 3 + 4x 30x 7 Combining like terms HOMEWORK: Pages of your online textbook questions 5 60 (multiples of 3) (multiples of 3) page 6 N P UC5

7 Section 5.: The Real Zeros of a Polynomial Function Back in elementary school you learned how to do long division. For example This meant that you had a remainder of 3 when dividing the dividend 95 by the divisor to get a quotient of 6 with a remainder of 3. This can also be expressed as the following: or 7 7 The same process of long division can be done to divide polynomials. If you have some function f(x) divided by a divisor d(x) this results in a quotient q(x) and remainder r(x) so that the following is true: f ( x) r ( x) q( x). d( x) d( x) Example : Divide (x + 7x 4) by (x ) using long division. x 9 x x 7x 4 x x 9x 4 9x 8 4 This leaves a quotient of 4 x 9 x. As in long division with numbers you did not omit roots because they were placeholders. The same is true when dividing polynomials; if a term appears to be missing such as in x 3 + 5x + 0 you would need to fill in a zero for the missing x term (in other words write out x 3 + 0x + 5x + 0). THIS MUST BE DONE FOR BOTH THE DIVIDEND AND THE DIVISOR. Example : Divide (5x 4 + 4x 3 + 9x) by (x + 3x) using long division. Do not forget to fill in the missing terms. In other words be sure that you realize this is the same as dividing (5x 4 + 4x 3 + 0x + 9x + 0) by (x + 3x + 0). x 3x 0 5x 5x 4 4 4x 5x 3 3 x x 3 3 5x x 3 0x 9x 0 0x 0x 3x 3x 3x 9x 0x 9x 0 9x 0 0 This leaves a quotient of 5x x + 3. Realize that there is no remainder for this division. page 7 N P UC5

8 For each of the previous problems you can easily double-check your results by multiplying your answer back onto the original divisor and adding the remainder back. The last example relates to what you might notice if you get a remainder of zero. To reinforce this divide 050 by Notice that the remainder is zero. If you end up with a remainder of zero you know that the divisor evenly goes into the dividend and is said to be a factor of the number. The same is true of polynomials. In the previous example you could say that x + 3x is a factor of the polynomial 5x 4 + 4x 3 + 9x. Long-division will work to see if a remainder is zero or not but an easier way for a divisor of the form x h is to use synthetic division a type of division you learned about back in advanced algebra. If you use synthetic division and the remainder is zero you know the factor x h is a factor of the polynomial. Also the remainder is just as if you substituted the value of h for each x of the function. The Remainder Theorem If a polynomial f(x) is divided by x k then the remainder is r = f(k). Synthetic division is another way to divide. For example let s say you wanted to divide f(x) = 4x 3 + 3x 5 by x. In synthetic division you put the coefficients of all terms (including missing ones in between) on the top row put the zero by which you are dividing in a box in the left (in this case since that is the solution for when x = 0) drop the first coefficient multiply it by the divisor and add the top. As follows is a picture to demonstrate: In using synthetic substitution f(x) = 4x 3 + 3x 5 is the same as saying f(x) = 4x 3 + 0x + 3x 5. The top row is the coefficients of the polynomial in standard form. Also since the remainder (last number calculated) is 3 this means 3/(x ) the remainder from long division. Example 3: Divide (x 3 7x 6) by (x ) using synthetic division. Do not forget to fill in the missing terms. In other words be sure that you realize this is the same as dividing (x 3 + 0x 7x 6) by (x ). Therefore goes into the dividing box and the coefficients that go at the top are 0 7 and page 8 N P UC5

9 Based on the result dividing (x 3 7x 6) by (x ) results in x x 3. x Also notice that if you had substituted into the polynomial x 3 7x 6 This is the remainder you received. () 3 7() 6 = =. Example 4: Divide (0x 4 + 5x 3 + 4x 9) by (x + ) using synthetic division. Do not forget to fill in the missing terms. In other words be sure that you realize this is the same as dividing (0x 4 + 5x 3 + 4x + 0x 9) by (x + ). Therefore goes into the dividing box (since x ( ) is the same as x + ) and the coefficients that go at the top are and Based on the result dividing (x 3 7x 6) by (x ) results in 0x 3 5x + 9x 9. There is no remainder. You could also double-check the remainder by substituting into the x s of the dividend. 0( ) 4 + 5( ) 3 + 4( ) 9 = 0() + 5( ) + 4() 9 = = 0. This is the remainder you received. Also the remainder of zero tells you that x + is a factor of 0x 4 + 5x 3 + 4x 9. What you just saw from the previous example shows the meaning of what is called the Factor Theorem. The Factor Theorem A polynomial f(x) has a factor x h if and only if f(h) = 0. In other words if a divisor being divided into a dividend results in a remainder of zero the divisor is a factor of the dividend. Example 5: Factor the polynomial f(x) = x 3 5x x + 4 given that f( ) = 0. This means that x ( ) or x + is a factor of the polynomial x 3 5x x + 4. In other words synthetically dividing x 3 5x x + 4 by x + will give the quotient which can be factored to provide the other factors This leaves x 7x +. This can be factored into (x 3)(x 4). Therefore x 3 5x x + 4 = (x + )(x 3)(x 4). Example 6: Factor the polynomial f(x) = 4x 3 4x 9x + 9 given that f() = 0. This means that x is a factor of the polynomial 4x 3 4x 9x + 9. In other words synthetically dividing 4x 3 4x 9x + 9 by x will give the quotient which can be factored to provide the other factors page 9 N P UC5

10 This leaves 4x + 0x 9 or 4x 9. This can be factored into (x 3)(x + 3). Therefore 4x 3 4x 9x + 9 = (x )(x 3)(x + 3). For the sake of argument if you were asked to solve 4x 3 4x 9x + 9 = 0 you know that this is the same as saying (x )(x 3)(x + 3) = 0 since this is the factoring of the polynomial 4x 3 4x 9x + 9. If this is true any factor could equal zero meaning you could solve for x. This means the following solutions are present: x = 0 x = x 3 = 0 x = 3 3 x x + 3 = 0 x = 3 3 x 3 3 and are said to be roots of the polynomial 4x 3 4x 9x + 9 since any of these values of x will make the polynomial equal to zero. If you were to multiply the factors (3x + )(5x + 7)(x + ) you would find that the polynomial that results 7 is 5x x + 355x The roots as a result would be and. Notice that the 3 5 numerator of each zero is a factor of the constant term 54 and the denominator of each zero is a factor of the leading term 5. This pattern of finding roots is generalized by the Rational Roots Theorem. The Rational Roots Theorem If f(x) = a n x n + + a x + a 0 and all terms have integer coefficients then every rational zero of f(x) has the following form: p factor of constant term a0 q factor of leading coefficient The Rational Roots Theorem provides a list of suspects for what could be rational roots for a polynomial. Not all of the possible roots actually work but a few of the roots that actually are roots of the polynomial that are rational actually work. a n Example 7: List the possible rational roots of f(x) using the Rational Roots Theorem if f(x) = 6x 4 3x 3 + x + 0. The possible rational roots come from dividing the factors of 0 by the factors of 6. p q To test these suspects you can use synthetic division to see if it is a factor by getting a remainder of zero. page 0 N P UC5

11 Example 8: Use synthetic division to decide if and/or is a root of the function f(x) = x 4 + 3x 3 + 3x 3x Since the remainder is 0 is a root of the function Since the remainder is 0 is a root of the function Since the remainder is 4 is not a root of the function Since the remainder is 6 is not a root of the function. In order to find all the real root of a function you can use the Rational Roots theorem to give you a suspect list and then narrow it down to get the real roots. Example 9: Find all the real roots of the function f(x) = x 4 + x 3 x 9x + 8. The possible rational roots come from dividing the factors of 8 by the factors of. p q There appears to be numbers that could be rational roots. If you graph the function though you can see it appears that there are only 4 that work: 3 and 3. You can use synthetic division to see if each works Since the remainder is 0 is a root of the function. Now try and see if 3 works with what is left over from the synthetic substitution: x 3 + x 9x page N P UC5

12 Since the remainder is 0 3 is a root of the function. Once you have the leading term down to the second degree you can factor or use the quadratic formula to find the other roots as you did previously. In other words solve for what is remaining: x + 5x + 6 = 0. x + 5x + 6 = 0 (x + )(x + 3) = 0 x + = 0 x + 3 = 0 x = x = 3 HOMEWORK: Pages of your online textbook questions 5 30 (multiples of 3) (multiples of 3) 63 7 (multiples of 3) page N P UC5

13 Section 5.3: Complex Zeros; Fundamental Theorem of Algebra Before we get into the dreaded i-word (imaginary) let s use what we learned in the previous section to lead into this section. Example : Given one zero of the polynomial function is 3 find the other roots of the polynomial function f(x) = x 3 + x + x + 4. This means that x ( 3) or x + 3 is a factor of the polynomial x 3 + x + x + 4. In other words synthetically dividing x 3 + x + x + 4 by x + 3 will give the quotient which can be factored to provide the other factors and as a result the other roots This leaves x x + 8. Since this cannot be factored normally the quadratic formula must be used to find the other roots. In this case the values to substitute into the quadratic formula are a = b = and c = 8. b x b 4ac a ( ) ( ) () 4()(8) i 7 i 7 This means the other roots are i 7 and i 7. Notice that when you get imaginary numbers they always form conjugates meaning the imaginary component is added in one and subtracted in another. Anytime there is an imaginary number as a solution its conjugate is another solution. Irrational Root Theorem Let a and b be rational numbers and let b be an irrational number. If a b is a root of a polynomial equation with rational coefficients then the conjugate a b is also a root. Example : Find all the real roots of the function f(x) = x 5 + x 4 3x 6. The possible rational roots come from dividing the factors of 6 by the factors of. p q There appears to be numbers that could be rational roots. If you graph the function though you can see it appears that there are only 3 that work: and. You can use synthetic division to see if each works page 3 N P UC5

14 Since the remainder is 0 is a root of the function. Now try and see if works with what is left over from the synthetic substitution: x 4 3x 3 + 6x x Since the remainder is 0 is a root of the function. Now try and see if from the synthetic substitution: x 3 + x + 8x + 4. works with what is left over Since the remainder is 0 is a root of the function. Once you have the leading term down to the second degree you can factor or use the quadratic formula to find the other roots as you did previously. However what is left over is x + 8. Setting this equal to zero x + 8 = 0 x = 8 x = 4 x = i Since i and i are not real numbers the only real roots are and You can also do the process in reverse.. Example 3: Find all the real roots of the function f(x) = x 5 + x 4 3x 6. The possible rational roots come from dividing the factors of 6 by the factors of. p q There appears to be numbers that could be rational roots. If you graph the function though you can see it appears that there are only 3 that work: and. You can use synthetic division to see if each works page 4 N P UC5

15 Since the remainder is 0 is a root of the function. Now try and see if works with what is left over from the synthetic substitution: x 4 3x 3 + 6x x Since the remainder is 0 is a root of the function. Now try and see if from the synthetic substitution: x 3 + x + 8x + 4. works with what is left over Since the remainder is 0 is a root of the function. Once you have the leading term down to the second degree you can factor or use the quadratic formula to find the other roots as you did previously. However what is left over is x + 8. Setting this equal to zero x + 8 = 0 x = 8 x = 4 x = i Since i and i are not real numbers the only real roots are and. Imaginary Root Theorem If one zero of a function is of the form a + bi another function of the zero is a bi. As previously discussed if a zero of a function is h a factor of the function is x h. This means that if you know the roots of a function you can write a function by finding its factors and multiplying them together. Example 4: Find all the roots of the function f(x) = x 4 3x 3 x + x. The possible rational roots come from dividing the factors of by the factors of. p q. There appears to be 6 numbers that could be rational roots. If you graph the function though you can see it appears that there are only that work: and. You can use synthetic division to see if each works page 5 N P UC5

16 Since the remainder is 0 is a zero of the function. Now try and see if works with what is left over from the synthetic substitution: x 3 + x + 0x Since the remainder is 0 is a zero of the function. Once you have the leading term down to the second degree you can factor or use the quadratic formula to find the other roots as you did previously. In other words solve for what is remaining: x x + = 0. x b b 4ac a i 7 i Therefore the roots are i and i Example 5: A fifth-degree function has zeros of with a multiplicity of 3 and a zero of 4i. Write the function in standard form. Since this is a fifth-degree function there should be five zeros. Three of them are (because of the multiplicity). A fourth one is 4i. Because of zero is 4i the other one has to be its conjugate meaning the other zero is + 4i. To create the function let s turn these zeros into factors of x (to the third power because of its multiplicity) x ( 4i) and x ( + 4i). f(x) = (x )(x ( 4i))( x ( + 4i)) Function creation f(x) = (x )(x + 4i)(x 4i) Distributive property f(x) = (x )(x x 4ix x i + 4ix 8i 6i ) Distributive property f(x) = (x )(x 4x + 4 6i ) Combining like terms f(x) = (x )(x 4x + 4 6( )) i = f(x) = (x )(x 4x ) Multiplication f(x) = (x )(x 4x + 0) Addition f(x) = x 3 4x + 0x x + 8x 40 Addition f(x) = x 3 6x + 8x 40 Addition HOMEWORK: Pages of your online textbook questions 3 36 (multiples of 3) page 6 N P UC5

17 Section 5.6: Polynomial and Rational Inequalities Back in Section 4.5 you solved for quadratic inequalities. You can use those same methods to solve for polynomial inequalities. Example : Solve the inequality x 3 > x graphically and algebraically. Graphically: If you got everything compared to 0 you could more easily see where the function is compared to 0. x 3 > x x 3 x > 0 Original inequality Substraction At this point on a graphing calculator (or by graphing by hand) you could graph some points and then the function to see what is shown at right for the function f(x) = x 3 x. Notice that the graph of the function is above y = 0 when x is between and 0 and from out to infinity. This means that the solution is as follows: ( 0) ( ) Notice that parentheses are used instead of brackets since it is not greater than and equal to zero. Algebraically: Going back to the original function you could still treat it the same way but what if you not only compared the function to zero but factored it down as if you were solving to find zeros? x 3 > x x 3 x > 0 x(x ) > 0 x(x + )(x ) > 0 Original inequality Substraction Factoring Factoring You could know like you did in previous sections what the zeros of the functions would be based on setting each factor equal to zero. x = 0 x + = 0 x = 0 x = x = page 7 N P UC5

18 Even those these points will not work because you cannot equal zero for this inequality you can test points in between that might. 0 Try into the Try 0.5 into the inequality. Try 0.5 into the Try into the inequality. x(x + )(x ) > 0 ( + )( ) > 0 ( )( 3) > 0 6 > 0 Since this is not a true statement the inequality doesn t work from ( ). x(x + )(x ) > 0 0.5( )( 0.5 ) > 0 0.5(0.5)(.5) > > 0 Since this IS a true statement the inequality DOES work from ( 0). inequality. x(x + )(x ) > 0 0.5(0.5 + )(0.5 ) > 0 0.5(.5)( 0.5) > > 0 Since this is not a true statement the inequality doesn t work from (0 ). inequality. x(x + )(x ) > 0 ( + )( ) > 0 (3)() > 0 6 > 0 Since this IS a true statement the inequality DOES work from ( ). NO!!! YES!!! NO!!! YES!!! Combining the areas where the inequality works you have the following solution: ( 0) ( ) Example : Solve the inequality x(x ) 3 (x + ) 0 algebraically. You can use the same algebraic techniques you did in the previous example but this time focus only on whether it makes the function positive or negative instead of getting actual final numbers in the inequality (since the numbers are going to be bigger because of the powers). You could know like you did in previous sections and in the previous example what the zeros of the functions would be based on setting each factor equal to zero. x = 0 x = 0 x + = 0 x = x = These points will work for the inequality because you can equal zero for this inequality. Now you can test points in between that might also work. Try into the inequality. x(x ) 3 (x + ) 0 ( ) 3 ( + ) 0 ( ) 3 ( ) Since this is NOT a true statement the inequality does not work from ( ) but DOES work exactly at. 0 Try 0.5 into the inequality. Try into the inequality. Try 3 into the inequality. x(x ) 3 (x + ) 0 x(x ) 3 (x + ) 0 0.5( 0.5 ) 3 ( ) 0 ( ) 3 ( + ) 0 ( ) 3 (+) 0 +( ) 3 (+) Since this is NOT a true statement the inequality does not work from ( 0) but DOES work exactly at and at 0. Since this IS a true statement the inequality DOES work from [0 ]. x(x ) 3 (x + ) 0 3(3 ) 3 (3 + ) 0 +(+) 3 (+) Since this is NOT a true statement the inequality does not work from ( ) but DOES work exactly at. NO!!! NO!!! YES!!! NO!!! Combining the areas where the inequality works you have the following solution: [0 ] page 8 N P UC5

19 HOMEWORK: Pages of your online textbook questions 0 8 (even) page 9 N P UC5