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Math 115 Spring 2011 Written Homework 8 Due Friday, March 18 1. For each function, determine whether the function is a polynomial function or not. It it is not a polynomial function, explain why not. If it is a polynomial function, determine the leading coefficient, the leading term, the degree, and the constant term. √ (a) f (x) := πx3 + 17x − 3.175 Solution: The definition of a polynomial is that the coefficients are real numbers and the powers on the variable terms xn are always positive integers (n ∈ N). The function in (a) is a polynomial. The leading coefficient is π, the leading term is πx3 , the degree is 3, and the constant term is −3.175. (b) f (x) := x−3 + x2 + 1 Solution: This is not a polynomial because of the x−3 term. The power is not a positive integer. (c) f (x) := 3xπ + 7 Solution: This is not a polynomial because of the xπ term. The power is not an integer. (d) f (x) := (x + 3)(2x − 5)(2x2 − 1)( 12 x + 5) Solution: This is a polynomial. The leading term is (x)(2x)(2x2 )( 21 x) = 2x5 . Thus the leading coefficient is 2 and the degree is 5. The constant term is (3)(−5)(−1)(5) = 75. 2. Calculate lim f (x) and lim f (x) for each function. x→−∞ 3 x→∞ 2 (a) f (x) := 2x − 11x + 4x + 5 Solution: lim 2x3 h i = 2 lim x3 x→∞ h i3 = 2 lim x lim f (x) = x→∞ x→∞ x→∞ = ∞ lim 2x3 3 = 2 lim x x→−∞ 3 = 2 lim x lim f (x) = x→−∞ x→−∞ x→−∞ = −∞ since (−‘ve number)3 < 0 (b) h(x) := x4 − 4x3 + 3x2 + 4x − 4 Solution: lim x4 h i4 = lim x lim h(x) = x→∞ x→∞ x→∞ = ∞ lim x4 4 = lim x lim h(x) = x→−∞ x→−∞ x→−∞ = ∞ since (−‘ve number)4 > 0 (c) g(x) := −3x4 + 5x3 + 6x5 + x2 + 3x − 9 Solution: lim 6x5 since this is the highest degree term h i = 6 lim x5 x→∞ h i5 = 6 lim x lim g(x) = x→∞ x→∞ x→∞ = ∞ lim 6x5 5 = 6 lim x x→−∞ 5 = 6 lim x lim g(x) = x→−∞ x→−∞ x→−∞ = −∞ since (−‘ve number)5 < 0 (d) p(x) := (πx2 − 4)(3x + 2)(4x2 − 5)(2 − x) Solution: Here, we need the leading term for p(x). Note that the highest degree term for the polynomial is (πx2 )(3x)(4x2 )(−x) = −12πx6 . lim −12πx6 x→∞ h i 6 = −12π lim x x→∞ h i6 = −12π lim x lim p(x) = x→∞ x→∞ = −∞ since the coefficient − 12π is negative −12πx6 x→−∞ 6 = −12π lim x x→−∞ 6 = −12π lim x lim p(x) = x→−∞ lim x→−∞ = −∞ since − 12π < 0 and (−‘ve number)6 > 0 Theorem: The Rational Root Theorem p Let be a rational number written in fully reduced form. Consider the polynomial equation q cn xn + cn−1 xn−1 + . . . + c1 x + c0 = 0 p is a zero (or q a solution) of the equation then p must divide the constant coefficient c0 and q must divide where all of the coefficients ci are integers and cn 6= 0. If the rational number the leading coefficient cn . Example: Let f (x) := 6x4 + 7x3 − 12x2 − 3x + 2. For a rational number p q to be a zero, p must be a factor of c0 = 2 and q must be a factor of c4 = 6. Thus, p can be ±1 or ±2, and q can be ±1, ±2, ±3, or ±6. The possible rational zeros, pq , are 1 1 1 2 ±1, ±2, ± , ± , ± , ± 2 3 6 3 3. Find all the rational roots of f (x) := 6x4 + 7x3 − 12x2 − 3x + 2. Solution: This is just plug-and-chug. Evaluate f at each of the possible rational roots. f (1) = 6(1)4 + 7(1)3 − 12(1)2 − 3(1) + 2 = 6 + 7 − 12 − 3 + 2 = 0 hence x = 1 is a root of f f (−1) = 6(−1)4 + 7(−1)3 − 12(−1)2 − 3(−1) + 2 = 6 − 7 − 12 + 3 + 2 = −8 hence x = −1 is not a root of f f (2) = 6(2)4 + 7(2)3 − 12(2)2 − 3(2) + 2 = 100 hence x = 2 is not a root of f f (−2) = 6(−2)4 + 7(−2)3 − 12(−2)2 − 3(−2) + 2 = 0 hence x = −2 is a root of f f (1/2) 6= 0 hence x = 1/2 is not a root of f f (−1/2) = 0 hence x = −1/2 is a root of f f (1/3) = 0 hence x = 1/3 is a root of f f (−1/3) 6= 0 hence x = −1/3 is not a root of f f (1/6) 6= 0 hence x = 1/6 is not a root of f f (−1/6) 6= 0 hence x = −1/6 is not a root of f f (2/3) 6= 0 hence x = 2/3 is not a root of f f (−2/3) 6= 0 hence x = −2/3 is not a root of f 4. (Some Algebra Review ) Let f (x) := 3x3 + x2 − 38x + 24. (a) For a rational number p q to be a zero, p must be a factor of c0 = 24 and q must be a factor of c3 = 3. Thus, p can be ±1, ±2, ±3, ±4, ±6, ±8, ±12 or ±24, and q can be ±1 or ±3. The possible rational zeros, pq , are 1 2 3 4 6 8 12 24 ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± , ± , ± , ± , ± , ± , ± , ± . 3 3 3 3 3 3 3 3 Removing the rational numbers that repeat when simplified, we get 1 2 4 8 ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ± , ± , ± , ± . 3 3 3 3 (b) Show that x = −4 is a zero (or root) of the function f (x) := 3x3 + x2 − 38x + 24. What does this tell you about a factor of this polynomial function? Solution: To show x = −4 is a zero, we just need to show that f (−4) = 0. f (−4) = 3(−4)3 + (−4)2 − 38(−4) + 24 = 3(−64) + 16 + 152 + 24 = −192 + 192 = 0 The linear term x − (−4) is a factor of the polynomial function. (c) Determine the quotient and remainder when 3x3 + x2 − 38x + 24 is divided by the factor you found in part (a). 3x2 − 11x + 6 Solution: x+4 3x3 + x2 − 38x + 24 − 3x3 − 12x2 − 11x2 − 38x 11x2 + 44x 6x + 24 − 6x − 24 0 Hence, f (x) := 3x3 + x2 − 38x + 24 = (x + 4)(3x2 − 11x + 6). (d) Use what you found in parts (b) and (c) to completely factor the polynomial function f (x) := 3x3 + x2 − 38x + 24. Solution: Continuing the factoring by factoring the quadratic term; f (x) := (x + 4)(3x − 2)(x − 3) 5. (More Algebra Review ) Determine the zeros of the polynomial function g(x) := x4 + 4x3 − 10x2 − 28x − 15 and write it in factored form. (Hint: Use the Rational Root Theorem to determine all possible rational roots and then use the division technique of problem 4 to fully factor g(x).) Solution: For a rational number p q to be a zero, p must be a factor of c0 = 15 and q must be a factor of c4 = 1. Thus, p can be ±1, ±3, ±5 or ±15, and q can be ±1 only. The possible rational zeros, pq , are ±1, ±3, ±5, ±15. Plugging these potential roots into g(x), the first one we see is that x = −1 is a root. Hence, x − (−1) is a factor of g. x3 + 3x2 − 13x − 15 x+1 x4 + 4x3 − 10x2 − 28x − 15 − x4 − x3 3x3 − 10x2 − 3x3 − 3x2 − 13x2 − 28x 13x2 + 13x − 15x − 15 15x + 15 0 Hence, g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x3 + 3x2 − 13x − 15). Continuing our use of the potential rational roots, we see that x = 3 is a root (since 32 + 3(32 ) − 13(3) − 15 = 0). Hence x − 3 is a factor of both g(x) (and of x3 + 3x2 − 13x − 15). x2 + 6x + 5 x−3 x3 + 3x2 − 13x − 15 − x3 + 3x2 6x2 − 13x − 6x2 + 18x 5x − 15 − 5x + 15 0 Hence, g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x − 3)(x2 + 6x + 5). We can continue using our list of potential rational roots, but the quadratic term is easy to factor. g(x) := x4 + 4x3 − 10x2 − 28x − 15 = (x + 1)(x − 3)(x + 5)(x + 1) The roots/zeros of g(x) are x = −5, x = −1, and x = 3. 6. Use the end behavior, the intercepts, and continuity to draw a rough sketch of each polynomial function. (Note: you may use information you found in previous problems as part of these solutions.) (a) h(x) := (x − 6)(x2 − 3)(7 − 2x) Solution: 1. end behavior Need to examine the limits to infinity. Recall that long-run behavior of a polynomial is dictated by the leading term. Here, lim h(x) = lim (x)(x2 )(−2x) = lim −2x4 = −∞. x→∞ x→∞ x→∞ Since h(x) is an even degree polynomial (degree is 4), lim h(x) = lim h(x) = −∞. x→−∞ x→∞ 2. intercepts (a) y-intercept Since h(0) = (−6)(−3)(7) = 126. The graph of h(x) intersects the y-axis at (0, 126). (b) x-intercept(s) Fortunately, h(x) is already factored. For the x-intercepts, we need to identify all of the roots of the polynomial. Here, we see that the roots are √ 7 x = 6, x = ± 3, and x = 2 These correspond to the x-intercepts, √ √ (6, 0), ( 3, 0), (− 3, 0) and 7 ,0 . 2 3. graph We use continuity to connect the end behavior and the intercepts. We need to test a √ point in the interval ( 3, 72 ). I choose x = 2: h(2) = (2 − 6)(22 − 3)(7 − 2(2)) < 0. We also need to test a point in the interval ( 72 , 6). I choose x = 5: h(5) = (5 − 6)(52 − 3)(7 − 2(5)) > 0. Putting all of this information together gives the following rough sketch: (b) f (x) := 3x3 + x2 − 38x + 24 Solution: 1. end behavior Need to examine the limits to infinity. lim f (x) = lim 3x3 = ∞. x→∞ x→∞ Since f (x) is an odd degree polynomial (degree is 3), lim f (x) = − lim f (x) = −∞. x→−∞ x→∞ 2. intercepts (a) y-intercept Since f (0) = 24. The graph of f (x) intersects the y-axis at (0, 24). (b) x-intercept(s) To determine the roots, we need to factor f (x). Fortunately, we factored f (x) in problem 1; f (x) := (x + 4)(3x − 2)(x − 3). Here, we see that the roots are x = −4, x = 3, and x = 2 3 These correspond to the x-intercepts, (−4, 0), (3, 0), and 2 ,0 . 3 3. graph We use continuity to connect the end behavior and the intercepts. We need to test a point in the interval ( 72 , 3). I choose x = 2: f (2) = (2 + 4)(3(2) − 2)(2 − 3) < 0. Putting all of this information together gives the following rough sketch: (c) g(x) := x4 + 4x3 − 10x2 − 28x − 15 Solution: 1. end behavior Need to examine the limits to infinity. lim g(x) = lim x4 = ∞. x→∞ x→∞ Since g(x) is an even degree polynomial (degree is 4), lim g(x) = lim g(x) = ∞. x→−∞ x→∞ 2. intercepts (a) y-intercept Since g(0) = −15. The graph of g(x) intersects the y-axis at (0, −15). (b) x-intercept(s) To determine the roots, we need to factor g(x). Fortunately, we factored g(x) in problem 2; g(x) := (x + 1)2 (x − 3)(x + 5). Here, we see that the roots are x = −1, x = 3, and x = −5 These correspond to the x-intercepts, (−1, 0), (3, 0), and (−5, 0). 3. graph We use continuity to connect the end behavior and the intercepts. We need to test a point in the interval (−5, −1). I choose x = −3: g(−3) = (−3+1)2 (−3−3)(−3+5) < 0. Putting all of this information together gives the following rough sketch: 7. In class we used a proof by contradiction to show that √ 2 was an irrational number. Armed with the Rational Root Theorem, we have now have an alternative proof. √ (a) Consider the function f (x) := x2 − 2. Algebraically show that ± 2 are the roots of this polynomial function. Solution: f (x) = 0 x2 − 2 = 0 x2 = 2 √ x = ± 2 (b) Now, use the Rational Root Theorem to determine all of the the possible rational roots of f . Solution: For a rational number p q to be a zero of f , p must be a factor of c0 = 2 and q must be a factor of c2 = 1. Thus, p can be ±1 or ±2, and q can be ±1 only. The possible rational zeros, pq , are ±1 or ± 2. (c) Explain why your answers in (b) prove that √ 2 must be an irrational number. √ 2 is a root of f but is not contained in the list of possible rational √ roots (±1 or ±2), x = 2 can not be a rational number. Hence, by the definition of irrational √ numbers, 2 is an irrational number. Solution: Since x =