MIDTERM SOLUTIONS (CHAPTERS AND 3: POLYNOMIAL, RATIONAL, EXP L, LOG FUNCTIONS) MATH 141 FALL 018 KUNIYUKI 150 POINTS TOTAL: 41 FOR PART 1, AND 109 FOR PART PART 1: USING SCIENTIFIC CALCULATORS (41 PTS.) 1) The Vertex Form for the equation of a parabola in the usual xy-plane is given by y = 3 x + 4 6. (4 points total) a) Which way does this parabola open? Box in one: Upward Downward Vertex Form is given by y = a x h + k. The leading coefficient a = 3 < 0, so the parabola opens downward. b) What is the vertex of this parabola? Here, h, k = ( 4, 6). ) The profit P (in dollars) for the Superdoom computer game company is given by P or P x = 0x + 400x 1500, where x is the number of game DVDs produced and sold. You may assume that the domain of P is 0, ). For parts a) and b), write units! (8 points total) a) Write and use a formula we used in class to find the number of DVDs (produced and sold) for which profit is maximized. (4 points) The graph of P versus x is a piece of a parabola opening downward. We want the x-coordinate of the vertex (the maximum point) of the parabola. x = b a = 400 0 = 10 DVDs b) What is the corresponding maximum profit? (4 points) The profit when 10 DVDs are produced and sold is given by: = 0( 10) + 400( 10) 1500 = 000 + 4000 1500 = $500 P 10 Note: Here is a graph of P versus x. We assume that the domain of P is 0, ), though that may be technically unrealistic.
3) Consider s( r) = r 3 3r 4r + 4 in parts a) and b) below. Hint: One of the zeros is 3. (16 points total) a) Write the two other complex zeros of s in simplest, standard form. Show all work, as in class. Box in your answers! (13 points) By the Factor Theorem, because 3 is a zero, r + 3 is a factor of s( r).. Use Synthetic Division to check this and to help us factor s r Therefore, s( r) = ( r + 3) ( r 6r +14). Find the zeros (roots) of the quadratic factor, r 6r + 14. Observe: a = 1, b = 6, c = 14. The discriminant is: b 4ac = ( 6) 4( 1) ( 14) = 36 56 = 0. 0 is not a perfect square, and GCF = 1, so we cannot factor r 6r + 14 over. Use the Quadratic Formula (QF): r = b ± b 4ac a = 6 ± i 0 ± 0 ( 1) = 6 = 6 ± i 4 5 = 6 ± i 5 We know the discriminant. = 6 ± i 5 = 3± i 5 b) Write the polynomial s( r) as a product of three linear factors over, the set of complex numbers. We basically want the Linear Factorization Theorem (LFT) Form of the factorization. (3 points) s( r) = ( r + 3) r 3+ i 5 r 3 i 5, or r + 3 r 3 i 5 r 3+ i 5
4) An exponential decay model for the number of fish remaining in a toxic lake is given by: P t = P 0 e 0.073t, where P( t) is the number of fish in the lake t years after January 1, 000. It was estimated that there were 5800 fish in the lake on January 1, 000. (9 points total) a) In how many years after January 1, 000 will there be only 1000 fish remaining in the lake? Give both an exact answer (which may look ugly; you don t have to simplify it) and an approximate answer rounded off to three significant digits. Write units! (8 points) Model: P t = P 0 e 0.073t Solve: 1000 = 5800e 0.073t for t. 1000 5800 = e 0.073t ln 5 9 t = 0.073, or 5 9 = e 0.073t ln 5 9 = ln e 0.073t t 4.3 years approximate ln 9 5 10,000 ln 9, or 0.073 73 ln 5 9 ln( 5) years = 0.073t ( exact) b) In what year will there be only 1000 fish remaining in the lake? (1 point) In the year 04, there will only be 1000 fish in the lake. 5) Approximate log 5 ( 134) to four decimal places. Show work by using a formula we have discussed in class. (4 points) Use the Change of Base Formula: log 5 ( 134) = ln( 5) ln 134 log 5 log 134 or 4.47 PART : NO CALCULATORS ALLOWED! (109 POINTS) 6) Fill in each blank below with or. (4 points total; points each) a) If f ( x) = x 3 + 5x 3 + 1, then lim x f ( x ) = x lim x 1 x = 0, so focus on the polynomial part. The leading term is x3, so the graph of y = f x without bound to the right. is a falling snake in the long run. The graph shoots down b) If g ( x) = x 4 + 3x +, then lim g ( x ) = x The leading term is x 4, so the graph of y = g x is an upward-opening bowl in the long run. The graph shoots up without bound to the left.
Graph of f in a) Graph of g in b) 7) How many turning points (TPs) can the graph of y = f ( x) have if f is a 4 th -degree polynomial function? (3 points) Answer: 3 or 1. Start with (degree 1) and count down by twos; stop before the negative numbers. The graph must be a bowl in the long run, so the number of TPs must be odd. 8) A polynomial (in x) of degree 4 has the following zeros: 0, 3, and. Its leading coefficient is 7, and 3 is a zero of multiplicity. Write this polynomial; you may leave it in factored form. (5 points) By the Factor Theorem and the Linear Factorization Theorem (LFT), use f ( x) = 7( x 0) ( x 3) x ( ) = 7x x 3 x +, or 7x 4 8x 3 1x +16x 9) Use Long Division to perform the division: 8x4 + 1x 3 6x 7x + 1. Write 4x 3 the answer in the form: ( polynomial) + ( proper rational expression). (11 points) The terms in red are deleted after their opposites are written in the subtraction process. x + 3x 4x + 0x 3 8x 4 +1x 3 6x 7x +1 8x 4 + 0x 3 6x 8x 4 0x 3 + 6x 1x 3 + 0x 7x +1 1x 3 +0x 9x 1x 3 0x + 9x x +1 x + 1 has degree 1, which is less than, the degree of 4x 3. Therefore, x + 1 is an appropriate remainder. Answer: x + 3x + x +1 4x 3 is a nonzero polynomial with real coefficients such that one of its 10) If f x zeros is 3+ 5i, what other complex number must also be a zero of f ( x)? (1 point). 3 5i, the complex conjugate of 3+ 5i, must also be a zero.
11) Write the list of the possible rational zeros of f, where f x = 7x 5 + 1x 3 4x +, based on the Rational Zero Test (Rational Roots Theorem). You do not have to determine which of these candidates are, in fact, zeros. (6 points) is a polynomial with integer coefficients and a nonzero constant term, so the f x theorem applies. p : ±1, ± q : ±1, ± 7 ( Factors of, the constant term) Think : ±1, ± ±1, ± 7 ( Factors of 7, the leading coefficient) ( Informal) p q : ±1, ± 1 7, ±, ± 7 ( The candidates) 1) Factor x 3 3x + 9x 7 completely over, the set of complex numbers. Hint: You may use Factoring by Grouping first to obtain the shortest solution. (7 points) x 3 3x + 9x 7 = x 3 3x x 3 = x + 9 + 9x 7 = x ( x 3) + 9( x 3) = ( x + 3i) ( x 3i) ( x 3) Note: Instead, we can use the Rational Zero Test (Rational Roots Theorem), identify 3 as a rational zero, and use Synthetic Division by x 3 13) Simplify i 447. ( points). to obtain x + 9 When we divide 447 by 4, we get a remainder of 3, so i 447 = i 3 = i. (Observe that 444 is a nice multiple of 4; 447 is 3 more than that.) = 3x 6 7x 4 + x 3 + 1. Using only Descartes s Rule of Signs, (8 points total) 14) Consider f x a) List the possible numbers of positive real zeros of f (accounting for multiplicity: double roots are counted twice, for example). (3 points) has variations in sign. Counting down by twos, there are either or 0 positive real zeros. Note: There are two: about 0.7875 and 1.3074, each with multiplicity 1. b) List the possible numbers of negative real zeros of f (accounting for multiplicity: double roots are counted twice, for example). (5 points) = 3( x) 6 7( x) 4 + ( x) 3 +1 Show work. f x has variations in sign. Counting down by twos, there are either or 0 negative real zeros. Note: There are two: about 1.641 and 0.5689, each with multiplicity 1.
15) Consider the graph of y = x +1 ( x + ) ( x +1) ( x + ) ( = x +1 ) ( x + ) = ( x +1) ( x + ) 4 4 in the usual xy-plane. (6 pts.; each) x +1 Let f x ( x 1). ( x + ) 3 a) Give the x-coordinate(s) of the hole(s), if any. (Holes correspond to removable discontinuities. ) as a common variable factor, factors are entirely canceled (divided) out in the denominator The numerator and denominator have x +1 and all x +1 after simplifying. There is a hole at x = 1. Note: The hole is at ( 1, 0), since lim f x x +1 = lim x 1 x 1 ( x + ) = 1+1 3 1+ b) Find the equation(s) of the vertical asymptote(s) (VAs), if any. 3 = 0. Find the real zeros of the denominator of the simplified expression x +1 ; observe that the x + x + 3 x 1 factors are not entirely canceled (divided) out in the denominator. ( x + ) 3 = 0 x + = 0 x = The one VA has equation: x =. c) Find the equation of the horizontal asymptote (HA), if any. x +1 In the simplified expression ( x 1), the degree of the numerator ( x + ) 3 (1) is less than the degree of the denominator (3), so the expression is bottom-heavy (proper) in degree. The HA has equation: y = 0. Note 1: There are no x-intercepts, since f x expression, the sole real zero of the numerator 1 The hole at 1, 0 does not count as an x-intercept. Note : The y-intercept is at 0, 1 8, since f 0 is never 0. In the simplified is not in the domain of f. = 1 8.
16) Consider the graph of y = 3x +1 in the usual xy-plane. If an answer to 6x 6 a part below is none, write NONE. Box in the answers! (14 points total) Let f ( x) = 3x +1, which is rational with polynomial numerator and denominator. 6x 6 a) Find the x-intercept(s), if any. (3 points) Set y or f x = 0 and solve for x. We want the real zeros of f. 0 = 3x + 1 6x 6 0 = 3x + 1 provided 6x 6 0 ( provided 6x 6 0) 1 = 3x provided 6x 6 0 x = 1 3 This has no real solutions, so there are no x-intercepts. Answer: NONE. b) Find the y-intercept, if any. (3 points) Set x = 0. f 0 = 3 ( 0 ) +1 6( 0) 6 = 1 6 = 1 6 0, 1 6 is the y-intercept. c) Find the equation(s) of the vertical asymptote(s) (VAs), if any. (5 points) Find the real zeros of the denominator, 6x 6 : 6x 6 = 0 6 x 1 Note: You may also divide both sides by 6. = 0 ( x 1) = 0 6 x +1 x = 1 or x = 1 These are not zeros of the numerator, 3x +1, so x +1 and ( x 1) are not factors of the numerator that could lead to holes. In fact, 3x +1 6x 6 The VAs have equations: x = 1 and x = 1. is simplified. d) Find the equation of the horizontal asymptote (HA), if any. (3 points) The HA is at y = 1, the ratio of the leading coefficients of the numerator and denominator; note that 3 6 = 1. This is because the degree of the numerator of 3x + 1 6x 6 is equal to the degree of the denominator ().
Here is the graph of y = f ( x) ; observe that f is even: 1, using interval form x 16 (the form using parentheses and/or brackets). (5 points) Due to the even-root radical (the index,, is even), the radicand must be nonnegative: x 16 0. 17) Write the domain of f, where f ( x) = We further require x 16 0 so as to avoid a zero denominator. The domain is the solution set of the inequality: x 16 > 0. See the Notes for Section.7 to see various methods of solution. Method 1: Sign Chart Method x 16 > 0 ( x + 4) ( x 4) > 0 Answer: (, 4) ( 4, ) Method : Parabola Method (for Quadratic Inequalities) The graph of y = x 16 is an upward-opening parabola with x-intercepts at 4, 0 and ( 4, 0). To solve x 16 > 0, we pick up the x-coordinates corresponding to the parts of the parabola lying strictly above the x-axis. Answer: (, 4) ( 4, )
Methods 3, 4: Test Value / Interval / Window Method; Evaluation or Factoring 4 4 Test x-values 5 0 5 f ( 5) = 9 f ( 0) = 16 f ( 5) = 9 0 0 ( x 4) ( ) ( ) ( + )( ) ( + )( + ) Sign of x 16 + 0 0 + Value of x 16 Sign Analysis: x + 4 Answer: (, 4) ( 4, ) 18) Write the domain of f, where f x parentheses and/or brackets). (1 point)., = 10 x, in interval form (the form using 19) Write the range of f from 18). (1 point). ( 0, ) 0) Write the domain of f, where f x = ln x parentheses and/or brackets). (1 point). 0,, in interval form (the form using 1) Write the range of f from 0). (1 point). (, ) ) Simplify the following: (4 points total; points each) a) log 5 ( 5) = log 5 5 1 b) log 4 16 = log 5 5 1/ = log 1 4 = log 4 4 4 = 1 3) Simplify: log 4 ( 3) + log 4 ( ). (3 points) By the Product Rule for Logarithms, Inverse Properties ; 5 = 5 = 5 = Inverse Properties log 4 ( 3) + log 4 ( ) = log 4 ( 3) ( ) ; 4 = 1 4 = 1 16 = log 64 4 = 3 Observe: 4 3 = 64. 1
4) Expand and evaluate where appropriate: ln e ( 3 x ) ln y z 5 e 3 = ln( e ) 3 + ln( x ) ln( y ) ln( z ) 5 ( x ). x, y, z > 0. (10 points) y z 5 by the Product and Quotient Rules for Logarithms = 3 + ln x 1/ ln( y ) ln( z ) 5 = 3, by the Inverse Properties Observe: ln e 3 = 3 + 1 ln ( x ) ln( y) 5ln( z) by the Power / "Smackdown" Rule for Logarithms Note: If y < 0 were allowed, then write ln y instead of ln y 5) Find all real solution(s) of the equation: log( 3x + 1) log( x 1) = log( x + ). Write the solution set. Show all work; do not use trial-and-error! (10 points) log 3x +1 log( x 1) = log( x + ) 3x +1 log x 1 = log x +. by the Quotient Rule for Logs 3x +1 = x + by the One-to-One Properties; see (*) x 1 3x +1= ( x + ) ( x 1), x 1 1 will fail the check, anyway. 3x +1= x + x 0 = x x 3 Isolate 0 on one side. ( x 3) 0 = x +1 x +1= 0 or x 3 = 0 x = 1 or x = 3 Checking tentative solutions is required when solving log equations. Only x = 3 checks out in the original equation; we only take logs of positive values. Check x = 1: Check x = 3: log 3 1 +1 log( 1 1) = log( 1+ ) log( ) log( ) = log( 1) not real not real (Check fails.) log 3 3 +1 log( 3 1) = log( 3+ ) log( 10) log( ) = log( 5) log 10 = log ( 5 ) [by the Quotient Rule for Logs] log 5 = log( 5) (Checks out.)
Solution set = { 3}. Note 1 (*): We could have performed base-10 exponentiation on both sides: 10 log 3x+1 10 x 1 log = 10 10 ( x+) 3x +1 x 1 = x + Note : Domain issues are considered at the end, when we checked our tentative solutions. 6) Find the real solution of the equation: 1 9 34x+1 = 3. The solution is a rational number, and you must write it in simplified form. Show all work, as in class; do not use trial-and-error! (6 points) Method 1 1 9 34x+1 = 3 3 4x+1 = 7 Isolate the exponential. 3 4x+1 = 3 3 4x +1= 3 4x = Use the One-to-One Property of the 3 x function. Method x = 4 x = 1 Solution set = 1. 1 9 34x+1 = 3 log 3 3 4x+1 3 4x+1 = 7 Isolate the exponential. = log 3 7 4x + 1 = 3 4x = Note: log 3 ( 7) = 3, since 3 3 = 7. Use the Inverse Properties. x = 4 x = 1 Solution set = 1.