Practice Exam 2sol

Practice Exam 2sol - Practice Exam 1 This print-out should...

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Unformatted text preview: - Practice Exam 1 This print-out should have 10 questions. Multiple—choice questions may continue on the next column or page — find all choices before answering. 001 10.0 points Find the absolute maximum value of 2 1 y = §$3_§x2_3$+3 on the interval [0, 2]. 1. max value = 3 correct 2. max value 3. max value 4. max value coloo “[5 “I: wlxl 5. max value Explanation: Differentiating 2 1 y = §$3_§x2_3$+3’ we see that y’ = 2x2 — a: — 3. Thus the critical points occur at the solutions of the equation 2322—95—3 =(2m—3)(m+1)= 0, i.6., at cc = g, —1, only the first of which lies in the interval [0, 2]. But Consequently, on [0, 2] the abs max value = 3 . 002 10.0 points Let f be the function defined by 0<ac<7r. f(m) = a: — 231nm, On which interval(s) is f increasing? 1. [%7T, 3%] 2. [%7r, 7r] correct 3. [0, %7r] U [%W, 7r] 4. [0, éw] 5. [0, 7T] Explanation: After differentiation, f'(a:) = 1 — 2cosx. Now f will be increasing on an interval [a, b] if (x) > 0 on (a, b). Determining where f’(a:) > 0 can be done graphically or alge- braically. But the graph of 1 — 2cossc on [0, 7r] shows that f'(:c) > 0 only on the interval (g, 7r). Thus f will be increasing everywhere on gm]- [oh—I On the other hand, we know that cossc = on [0, 7r] only when x = g, and that < 1 7r < < cosa: — — at 7T. 2’ 3 _ Thus 1 — Zeossc > 0 holds everywhere on (g, 7r), showing algebraically that f will be increasing everywhere on g, 7r] 003 10.0 points Which of the following is the graph of {1:3 for) = —? 392—9 3. 4. cor- rect I! Explanation: Since 393 _ 310(302 — 9) + 93: x2 — 9 _ m2 — 9 _ as + 93: _ x2 — 9 ’ the graph will have an oblique (aka slant) asymptote y = :c; it will also have vertical asymptotes as = :|:3. This is enough to elimi— nate all but two of the graphs. On the other hand, , _ 302(302—27) J60”) — W, so f is decreasing on (—3, 3). Consequently, the graph of f must be If f and g are continuous functions such that f (3:) 2 0 for all x, which of the following 004 10.0 points must be true? III. 4. 5 I l 0" k5 A 3 9.. H + o\ 0" to A 3 9.. H . II only correct . III only . I and II only I only . II and III only Explanation: Property II is a special case of the linearity property of integrals, 216. “integral of the sum sum of the integrals”. 3 But there is no product rule or square root rule for integrals. For example, we know that /: f(ac)dx = 2 when f (as) = 1, because the value of the inte— gral is the area of a rectangle of height 1 and base length 2. So when f(:r) 2 9(39) 2 1 in I, /02 f(ac)g(:r)dac 2/02 1dr 2 2, while (/02 f($)d$)(/02 g(ac)d:r) 2 2X2 : 4_ On the other hand, in III, while 2 2 / f(a:)d:c= / 1d$=\/§. 0 0 Consequently, only II is true 005 10.0 points Find the derivative of F when 21: mm) = / (t2 — 3t) dt. E 1. F'(a:) = 5x2+6a§ 2. F'(a:) = 5x2+9x 3. F'(ac) 2 7952—69: 4. F'(x) = 7x2 — 9.7: correct 5. F'(a:) = 5:02—6:13 6. F'(ac) = 7302-1—93: Explanation: One version of the FTC tells us that g (f: N) dt) = f(x) for each fixed a. More generally, 9(96) % </ f(t)dt) = g'(a:)f(g<a:)> for each fixed a and differentiable function g. To apply this to F write F(ac) = /2m (1:2 — 3t) dt 0 2m 2/ (t2—3t)dt+/ (152—315) dt :6 0 :6 21: =—/ (t2—3t)dt+/ (t2—3t)dt. 0 0 By the FTC, therefore, F'(a:) = — (1:2 — 3t) ‘ +2 (1:2 — 3t) ‘ t=$ t=2$ = 2 (4332—650) — (:02 —3a:). Consequently, F'(a:) = 7:32—93: . 006 10.0 points Evaluate the integral 7r/4 I = / (3sec20—7sin0) d0. 0 7 1. I 2 10+— fl 7 2 I = 10—— fl 3 I — 4+1 7 4. I = —4+— correct fl 3 5. I = —4— — fl Explanation: By the Fundamental Theorem of Calculus, I = [3tan0—i— 70080]:/4 = (3+%)—7 = THE“ keywords: integral, FTC, trig function 007 10.0 points Find the area of the shaded region in y P a: bounded by the graph of f (96) = 4962, the tangent line at P(1, f(1)), and the m—axis. 1 1. area = 3 correct 2 2. = — area 3 1 3. = — area 4 1 4. = — area 2 5. area = 1 Explanation: To find the area of the shaded region it is simplest to compute separately the areas below the parabola and below the tangent line, then subtract. The area below the parabola is given by the definite integral A1 = /01 4$2d$ = [3x3]:= 3. On the other hand, the triangular region be- low the tangent line has area 1 A2 = 2 x base length X height = §-<1—xo)-f(1), where 300 is the m—intercept of the tangent line. Now f(1) = 4, f'(~’0) = 856, f'(1) = 8, so by the point slope formula an equation for the tangent line at P(1, f(1)) is y—4 = 8(39—1), 2'.e., y = Sac—4. Thus :00 = % , and so Consequently, the shaded region has 1 area = A1—A2 = 3 008 10.0 points Find all asymptotes of the graph of 3x2—16x+16 x2—6$+8 ' y : 1. vertical: 3: = 4, horizontal: y = 3 2. vertical: 3: = 2, 4, horizontal: y = 3 3. vertical: as = 2, horizontal: y = 3 cor- rect 4. vertical: 3: = 4, horizontal: y = —3 5. vertical: 3: = —2, horizontal: y = 3 6. vertical: 3: = 2, 4, horizontal: y = —3 7. vertical: 3: = 2, horizontal: y = —3 Explanation: After factorization _ (3,. — 4><x — 4) y _ (x—2)(x—4)' Thus y is not defined at a: = 4, but for :0 7E 4 339—4 9: $_2; notice, however, that _ 3x — 4 11m m—)4 :1:—2 : 4 exists, so the graph does not have a vertical asymptote at a: = 4. Since ‘339—4 39—2 ‘—>oo as a: —> 2 from the left and the right, the line cc = 2 will, however, be a vertical asymptote. On the other hand, _ 3a: — 4 11m m—>::oo .7; — 2 :3, so y = 3 will be a horizontal aymptote. Hence the graph has only the following asymptotes vertical: as = 2, horizontal: y = 3. 009 10.0 points Evaluate the definite integral 2 I 2/ $+6 dac. _1 :L'-I—2 Correct answer: 12.6667. Explanation: Set u = a: + 2. Then du = dac and 4 4 _ “+4 _ 1/2 —1/2 I—/1 Vida—A01, +411. )du. Consequently, _23/2 1/2“_38 I—[gu +8u]1— 3 010 10.0 points Find the area of the largest rectangle that can be inscribed in a semi-circle of radius 3. Explanation: Let’s take the semi-circle to be the upper half of the circle 502+y2 = 9 having radius 3 and center at the origin. For the rectangle we take one side on the x-axis and one corner at a point P(:c, y) on the semi- circle as shown in Then the area of the rectangle is given by 14(1)) = 2393; = 2mV9—x2. We have to maximize A(ac) on the interval [0, 3]. Now 2 2 2(9 — 2x2) 9 — :32 ' Thus the critical points of A(:I:) occur at m _ _i 1 \/§’ \/§’ only one of which lies in [0, 3]. But 3 A(0) _ 0, 4%) _ 9, A(3) _ 0. Consequently, max. area = 9 sq. units...
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