Math 1241 Calculus Test 1

Transcription

1 February 4, 2004 Name Te first nine problems count 6 points eac and te final seven count as marked. Tere are 120 points available on tis test. Multiple coice section. Circle te correct coice(s). You do not need to sow your work on tese problems. 1. Wic of te following is a factor of x 4 x? Circle all tose tat apply. (A) x (B) x 1 (C) x + 1 (D) x 2 + x + 1 (E) x 2 x + 1 Solution: Te expression factors as follows: x 4 x = x(x 3 1) = x(x 1)(x 2 + x + 1), so A, B, and D are all factors. 2. How many roots does te equation below ave? x 2 (x 2 3) 4(x 2 3) = 0 (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 Solution: E. Te expressions factors, so te equation can be written x 2 (x 2 3) 4(x 2 3) = (x 2)(x + 2)(x 3)(x + 3) = 0, so tere are four roots x 1 1 x = (A) x + 1 (B) x 1 (C) x 1 (D) 1 x (E) x x 1 x + 1 Solution: A. Find a common denominator for numerator and denominator to get (x + 1)/x (x 1)/x = (x + 1)/(x 1). 4. Wat is te radius of te circle wose equation is x 2 8x + y 2 + 6y = 24? (A) 4 (B) 24 (C) 5 (D) 6 (E) 7 Solution: E. Complete te two squares to get x 2 8x+16+y 2 +6y+9 = wic is te same as (x 4) 2 + (y + 3) 2 = 7 2, so te radius of te circle is Wic of te following is a solution to 2(5 3x) 2 5 3x = 108? Circle all tat apply. (A) 12 (B) 9 (C) 2 (D) 0 (E) none of tese Solution: A. Te equation reduces to 9x = 108 so te only solution is x = 12. 1

2 6. Wic of te following is a solution to 3(x 2) 3 (x+1) 2 2(x 2) 2 (x+1) 3 = 0? Circle all tat apply. (A) 2 (B) 1 (C) 0 (D) 2 (E) 8 Solution: B.D.E. Again factoring is te key. Te equation becomes (x 2) 2 (x + 1) 2 [3(x 2) 2(x + 1)] = 0, so we get te tree roots, x = 2, x = 1, and x = Consider te function y = a sin(bx), were a and b are constants, sown below. Wat is a + b? (Tick marks are located at unit positions.) 2π. (A) 2 (B) 4 (C) 5 (D) 6 (E) 7 Solution: E. Te amplitude is 6 and te period is 2π, so a = 6 and b = 1. 2

3 8. Suppose te functions f and g are given completely by te table of values sown. x f(x) x g(x) Wat is g 1 (f(3))? (A) 1 (B) 3 (C) 4 (D) 5 (E) 6 Solution: E. Note tat f(3) = 1, so we want to find g 1 (1) wic is seen from te table to be Referring again to te two functions in te previous question, solve te equation g(f(g(x))) = 5 for x. (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Solution: D. Note tat g(0) = 5, so we need to solve f(g(x)) = 0. Note f(6) = 0, so we need to solve g(x) = 6, wic we can see from te table is 4. 3

4 On all te following questions, sow your work. 10. (7 points) Find te (implied) domain of te function g(x) = x+1. Write your x 2 9 answer using interval notation. Solution: Te number tat must be excluded are tose for wic make te number in te square root negative and tose wic make te denominator zero. Tus, x and x We can write tis in interval notation as follows [ 1, 3) (3, ). 11. (7 points) Let f(x) = x 2 x. Compute in simplify f(4), f(x + 1), f(x + ), and f(x+) f(x), were 0. Solution: f(4) = = 12, f(x+1) = (x+1) 2 (x+1), and f(x+) = (x+ ) 2 (x+). Finally, f(x+) f(x) 2x+ 2 = (2x+ 1) = (x+)2 (x+) (x 2 x) = x2 +2x+ 2 x x 2 +x = 2x + 1. = 12. (7 points) Te slope of te line tangent to te grap of f(x) = 2x 2 x at te point (1, 1) is 3. Find an equation for tis tangent line. Solution: An equation is given in point-slope form by y 1 = 3(x 1), wic in slope-intercept form is y = 3x (10 points) Sketc te curve below represented parametrically by x = t sin(2t), y = t + cos(t) for 2 t 2 on te grid provided. Solution: Use your graping calculator to get.. 4

5 14. (7 points) Tis of te computation of J(x) = (x 2) as a sequence of four simple computations. Find four functions, f, g,, and k suc tat J(x) = f g k(x). Solution: Te operations in order are: subtract 2, square, add 3, take square root, so te functions are k(x) = x 2, (x) = x 2, g(x) = x + 3, f(x) = x. 15. (15 points) A. Does te function f(x) = 2x 5 ave an inverse. If it does, find it. If not, state wy it does not. Solution: Intercange te x and te y and solve for y to get 2y 5 = x or y = (x + 5)/2. B. Does te function f(x) = ln(2x 5) ave an inverse. If it does, find it. If not, state wy it does not. Solution: Again, intercange te x and te y to get x = ln(2y 5). Ten solve for y by treating eac side as an exponent. Tus e x = e ln(2y 5) = 2y 5. Terefore, y = (e x + 5)/2. 5

6 Tear tis page off your test booklet and take it ome. Return it Friday at class time (or before) wit a complete solution. 16. (20 points) A. First, complete te table below. 2 if x 2 f(x) = x if 2 < x x + 3 if x 1 g(x) = 2x if x > 1 Solution: x g(x) f g(x) π 2π 2π B. Find te composition f g of te two functions defined above. Remember tat f g(x) is, by definition f(g(x)). Your final answer sould not ave te absolute value symbol in it. Solution: First, compose all te part of g wit tose of f to get 2 if x 1 and x x + 3 if x 1 and x + 3 > 2 f g(x) = 2 if x > 1 and 2x 2 2x if x > 1 and 2x > 2 Next solve eac of te pairs of inequalities to eliminate some clauses and simplify oters. Te first clause disappears because te two inequalities x 1 and x 5 are incompatible. Te second inequality becomes x 1, te tird also disappears, and te fourt becomes x > 1. Tus te function is x + 3 if x 1 f g(x) = 2x if x > 1 6

7 Actually, te absolute value symbols are not needed ere because for all te values of x in te given ranges, te corresponding values in te absolute values is positive. So we can write te function as follows: f g(x) = x + 3 if x 1 2x if x > 1 Please sign te following pledge: On my onor I declare tat I ave neiter given nor received any elp from anyone else on tis test. 7