Practice 2nd Midterm Solutions

Math 102. Practice 2nd Midterm Solutions (i) What are the values f (−2), f (−1), and f (3)? (ii) Determine the domain and the range of f . (Use interval notation.) 1 For the function f (x) = 10x2 − 7x + 3 calculate the following values (simplify as much as possible): (iii) Determine the intervals on which f is increasing and the intervals on which f is decreasing. (i) f (a) (iv) Find the local maxima and the local minima of f . (ii) f (a + h) (v) Find the average rate of change of f between 3 and −2. f (a + h) − f (a) (iii) h Solution. (i) f (a) = 10a2 − 7a + 3 (ii) Plug in a+h into f thus f (a+h) = 10(a+h)2 −7(a+h)+3 = 10a2 + 10h2 + 20ah − 7a − 7h + 3 (iii) Using the results of (i) and (ii) we obtain f (a + h) − f (a) = 10a2 + 10h2 + 20ah − 7a − 7h + 3 − (10a2 − 7a + 3), and simplifying f (a + h) − f (a) = 10h2 + 20ah − 7h. Therefore f (a + h) − f (a) 10h2 + 20ah − 7h = = 10h + 20a − 7 h h √ x+1 2 Let f (x) = . x (i) Evaluate f (3), f (5), and f (a − 1). (ii) Find the domain of f . √ √ 2 6 a , f (5) = , and f (a − 1) = Solution. (i) f (3) = 3 5 a−1 (ii) The domain of f is all numbers x such that x , 0 (can’t divide by 0) and x + 1 ≥ 0 (can’t take the square root of a negative number); that is, the domain is {x | x ≥ −1 and x , 0}. Using interval notation, [−1, 0) ∪ (0, ∞) Solution. (i) The values are obtained directly from the graph: f (−2) = −2 , f (−1) = 2 , and f (3) = −3 (ii) The domain of f is [−4, 7] and the range of f is [−3.4] (iii) f is increasing on [−2, 0], on [3, 4] and on [6, 7] and f is decreasing on [−4, −2], on [0, 2] and on [4, 6] (iv) The local extrema are the point in the graph with coordinates (−4, 1), (−2, −2), (0, 3), (4, 4), (6, 1), and (7, 2). From these, the local maxima are (−4, 1), (0, 3), (4, 4) and (7, 2) and the local minima are (−2, −2), and (6, 1) . (v) Average rate of change is − 1 5 5 For the piecewise function f defined by ( 1 − x2 x ≤ 2 f (x) = 2x − 1 x > 2. (a) Find the values f (0), f (−2) and f (2). (b) Sketch the graph of f . Solution. (i) f (0) = 1 − 02 = 1 , f (−2) = 1 − (−2)2 = −3 , and f (2) = 1 − (2)2 = 1 − 4 = −3 (ii) The graph is 3 Determine the average rate of change of the function f (x) = x + x4 between x = −1 and x = 3. 6 Solution. The average rate of change is 4 f (3) − f (−1) 3 + 34 − ((−1) + (−1)4) 84 = = = 21 3 − (−1) 4 4 4 The function f has the following graph. 2 -2 -1 1 2 3    -2

1 1 -4 6 (i) Sketch the graph of the function f (x) = (x + 1)5 − 1. (ii) Is the function f one-to-one? If yes, find an equation for the inverse function f −1 ; if no, justify your answer. Solution. (i) To sketch the graph of f we use transformations of graphs: start with the graph of y = x5 , translate it 1 unit to the left, then 1 unit downward. The vertex is (4, 36). The maximum of P(x) is then 36, and so the maximum monthly profit is $36,000 . (i) The x-coordinate of the vertex gives us the point where the maximum of P(x) is attained. Thus, in order to maximize the profit, the monthly advertisement should be $4,000 8 Given f (x) = 2 -1 -2 1 + x and g(x) = √ 1 − x, find: (i) the functions f + g, f − g, f · g, and f /g; (ii) the domains of the functions in Part (i). √ √ √ Solution. (i) ( f + g)(x) = 1 + x + 1 − x, ( f − g)(x)r = 1 + x− √ √ 1+x 1 − x, ( f · g)(x) = (1 + x)(1 − x), and ( f /g)(x) = . 1−x (ii) The domain of f is {x | x ≥ −1} = [−1, ∞) and the domain of g is {x | 1 ≥ x} = (−∞, 1]. Therefore, the domain of f + g, of f − g and of f · g is the interval [−1, 1]. Since g(1) = 0, the domain of f /g is the interval [−1, 1). 1 -3 √ 1 2 3 -1 -2 9 For the functions f (x) = (ii) It is one-to-one because it passes the horizontal line test (the graph intersects any horizontal line in at most one point). To find the inverse function f −1 , set up the equation y = f (x) and solve for x, like this y = (x + 1)5 − 1 y + 1 = (x + 1)5 1 (y + 1) 5 = x + 1 1 (y + 1) 5 − 1 = x (Fifth root) (Add −1) 1 7 Odyssey Travel Agency’s monthly profit P (in thousands of dollars) depends on the amount of money x (in thousands of dollars) spent on advertising per month according to the rule P(x) = −x2 + 8x + 20. (i) What would Odyssey’s monthly advertising be in order to maximize its monthly profit? (ii) What is this maximum monthly profit? Solution. (Note the order) (ii) The graph of y = P(x) is a parabola that opens downward, so the maximum value of P(x) is the y-coordinate of the vertex of that parabola. The find the vertex we write P(x) = −x2 + 8x + 20 in standard form 2 −(x − 4)2 + 36 (ii) Find the domains of f ◦ g, g ◦ f , f ◦ f , and g ◦ g. (Split out part with x’s) (Add and subtract square half the coeff. of x) (Perfect square) (Dist. outermost “–” sign) (Simplify) 1 1 = , x , −1 (x − 2) + 3 x + 1 −2x − 5 1 −2= , x , −3 g ◦ f (x) = g( f (x)) = x+3 x+3 1 x+3 10 f ◦ f (x) = f ( f (x)) = = , x , −3, − 1 3x + 10 3 +3 x+3 g ◦ g(x) = g(g(x)) = (x − 2) − 2 = x − 4, all numbers f ◦ g(x) = f (g(x)) = 1
− (x − 4)2 − 16 + 20 −(x − 4)2 + 16 + 20 (i) Find ( f ◦ g)(x), (g ◦ f )(x), ( f ◦ f )(x) and (g ◦ g)(x) Solution (i) and (ii) and write the inverse f −1 (x) = (x + 1) 5 − 1 −(x2 − 8x + 16 − 16) + 20 and g(x) = x − 2 (Add 1) Now interchange x and y in this last expression: y = (x + 1) 5 − 1, −x + 8x + 20 −(x2 − 8x) + 20 1 x+3 10 (i) If f (x) = 2 − √ 3 − x, find the inverse function f −1 . (ii) Sketch the graphs of f and f −1 on the same coordinate axes. √ Solution. Set the equation y = 2 − 3 − x and solve for x: √ y−2=− 3−x 2 (y − 2) = 3 − x (y − 2)2 − 3 = −x 3 − (y − 2)2 = x (Subtr. 2) (Square both sides) (Add 3) (Mult. by −1) Now interchange x and y in this last expression: y = 3 − (x − 2)2 , and write the inverse f −1 (x) = 3 − (x − 2)2 The graph of f −1 is the graph of f reflected in the main diagonal. To sketch the√graph of f we use transformations: √ Start with the graph of y = x, then reflect in y axis√ to get y = −x, then shift right 3 units to√get the graph of y = −x + 3, then reflect in x-axis to get y = − 3 − x, then shift up 2 units. 12 For the polynomial P(x) = x3 + 3x2 − 4x − 12 (i) Find x and y-intercepts. 2 (ii) Determine end behavior. -4 -2 2 (iii) Sketch its graph. (Make sure that your graph shows all intercepts and exhibits the proper end behavior.) Solution. (i) The x-intercepts are the zeros of P(x). To find them we use factoring by grouping: -2 (x3 + 3x2 ) − (4x + 12) x2 (x + 3) − 4(x + 3) (Group terms) (Factor x2 and 4) -4 (x2 − 4)(x + 3) (x + 3 common factor) (x − 2)(X + 2)(x + 3) (Difference of squares) -6 The x-intercepts are 2, −2, −2 . The y-intercept is P(0) = −12 . (ii) The asymptotic behavior is P(x) → ∞ as x → ∞ and P(x) → −∞ as x → −∞. (iii) The graph of P(x) is 11 Here is the graph of a polynomial P. 10 5 -3 -2 -1 1 2 -5 -3 -2 2 -10 -12 (i) How many real zeros does P have? Mark them in the graph. (ii) How many local maxima and how many local minima does P have? Mark them in the graph. (iii) What is the smallest possible degree that P can have? 13 For the polynomial P(x) = −2x7 + 3x5 − 4x3 − 11x2 + 7x − 1 (i) Find the quotient and reminder when P is divided by (x − 2)(x + 3) (iv) Is the degree of P even or odd? (v) Is the sign of the leading coefficient of P positive or negative? Explain your answer. Solution. (i) The polynomial P has 5 different real zeros because the graph of P intersects the x-axis in 5 different points. (ii) There are two local maxima and two local minima (iii) P must have at least degree 5 because a polynomial of degree n can have at most n real zeros. degree of P is odd and the (iv) and (v) The leading coefficient is negative because the end behavior of P is like the end behavior of −x3 . (ii) Find the quotient and remainder when is divided by x − 2. (iii) Use synthetic division to find P(−3). Solution. (i) The divisor is D(x) = (x−2)(x+3) = x2 +x−6. Using long division we obtain that the quotient is Q(x) = −2x5 + 2x4 − 11x3 +23x2 −93x+220 and the remainder is R(x) = −771x+1319. (ii) We use synthetic division. 2 −2 0 3 0 −4 −11 7 −4 −8 −10 −20 −48 −118 −2 −4 −5 −10 −24 −59 −111 −1 −222 −223 The quotient is −2x6 − 4x5 − 5x4 − 10x3 − 24x2 − 59x − 111 and the remainder is −223 (b) The value P(−3) is the remainder of dividing P(x) by x − (−3). We divide using synthetic division: −3 −2 0 3 0 −4 −11 7 6 −18 45 −135 417 −1218 −2 6 −15 45 −139 406 −1211 −1 3633 3632 Therefore P(−3) = 3632 14 The following curve is the graph of a polynomial of degree 3. Find the polynomial. The remainder is 0, so −1 is a zero of P. Factor P(x) √ = (x + 7 ± 49 − 24 2 , 1)(2x − 7x + 3). The quadratic formula gives x = 4 or x = 3, 1/2. The zeros of P are −1.3.1/2 (iii) The complete factorization of P is P(x) = 2(x + 1)(x − 3)(x − 1/2) 16 Find a fourth degree polynomial with real coefficients which has zeros 1 + 3i and −1, with −1 a zero of multiplicity 2, and such that the constant coefficient is 20. Solution. The general form for P is P(x) = a(x − c1 )(x − c2 )(x − c3 )(x − c4 ) The zeros of P are 1 + 3i, 1 − 3i, −1, −1, so we can write P(x) = a(x − (1 + 3i))(x − (1 − 3i))(x − (−1))(x − (−1)) 1 -1 or P(x) = ax4 + 7ax2 + 18ax + 10a. This polynomial has constant coefficient 10a, so we must have 10a = 20, or a = 2. Thus P(x) = ax4 + 7ax2 + 18ax + 10a 1 €€€€€ 2 17 Find a polynomial P with real coefficients that has degree 4, zeros i and 2i, and whose graph passes through the point (1, 5). Solution. The general form for P is Solution. From the graph we see that the polynomial, P(x), has zeros at −1 and 1/2. Furthermore, the graph does not cross at −1 and this means that the polynomial has a zero of even multiplicity at −1. Because P(x) has degree is 3, it must be of the form P(x) = a(x − 1/2)(x + 1)2 , where a is a real number to be found. To find it, we observe that the y-intercept is 1, which means that P(0) = 1. But P(0) = a(0 − 1/2)(0 + 1)2 = −a/2, so we must have −a/2 = 1, or a = −2. Therefore the polynomial is P(x) = −2x3 − 3x2 + 1 3 P(x) = a(x − c1 )(x − c2 )(x − c3 )(x − c4 ) The zeros of P are i, −1, 2i, −2i, so P(x) = a(x−i)(x+i)(x−2i)(x+ 2i), or P(x) = ax4 + 5ax2 + 4a. If we want that the graph of P goes through (1, 5), then we must have P(1) = 5, or 4a + 5a + a = 5, or 1 5 a = 1/2. Therefore P(x) = x4 + x2 + 2 2 2 18 Find all the zeros and the complete factorization of the polynomial P(x) = 2x4 − 8x3 + 28x2 − 72x + 90 given that 2 + i is a zero. 2 15 Let P(x) = 2x − 5x − 4x + 3. (i) List all the possible rational zeros of P. (ii) Find all the zeros of P. (iii) Find the complete factorization of P. Solution. (i) The possible rational zeros are ±1, ±1/2, ±3, ±3/2 (ii) Divide by x + 1 −1 2 −5 −4 −2 7 2 −7 3 3 −3 0 Solution. Since P(x) has real coefficients and 2 + i is a zero, the conjugate 2 − i is also a zero of P, by the Conjugate Zero Theorem. We also know that to say that 2 ± i are zeros of P is the same as saying that x − (2 ± i) are factors of P. That is, (x − (2 + i))(x − (2 − i)) = x2 − 4x + 5 is a factor of P. After long 2 division we find that P(x) = (x2 − 4x + 5)(2x √ + 18). We obtain 2 that the solutions to 2x + 18 = 0 are x = ± −9, or x = ±3i. The zeros of P are 2 + i, 2 − i, 3i, −3i and the complete factorization of P is P(x) = 2(x − (2 + i))(x − (2 − i))(x − 3i)(x + 3i)