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4 Pages

### calculus test

Course: CALCULUS 1750, Spring 2002
School: Toledo
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Word Count: 772

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PracticeTest exams Calculus Solutions Name: 1. (3 points) Compute lim t2 t2 t2 - 4 . - 2t + 1 Solution Since the limit of the numerator is 0 and the limit of the denominator is not zero, we actually can use the plug-in theorem t2 - 4 0 (aka substitution theorem). Thus we have lim 2 = = 0. t0 t - 2t + 1 1 2. (6 points) Let f (x) = 3 2x - 1. Find the linearization L(x) to f (x) at x = 14. Then use this to...

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PracticeTest exams Calculus Solutions Name: 1. (3 points) Compute lim t2 t2 t2 - 4 . - 2t + 1 Solution Since the limit of the numerator is 0 and the limit of the denominator is not zero, we actually can use the plug-in theorem t2 - 4 0 (aka substitution theorem). Thus we have lim 2 = = 0. t0 t - 2t + 1 1 2. (6 points) Let f (x) = 3 2x - 1. Find the linearization L(x) to f (x) at x = 14. Then use this to approximate f (14.01). Solution The linearization is given by L(x) = f (14) + f (14)(x - 14), and since f (14) = 3, we have L(x) = 3 + f (14)(x - 14). It -2 2 1 2 2 = 27 . Thus L(x) = 3 + 27 (x - 14). If we remains to find f (14). We have that f (x) = 3 (2x - 1) 3 2, so f (14) = 2 3(27 3 ) use L to approximate f when x = 14.01, we obtain f (14.01) L(14.01) = 3 + 2 1 1 =3+ . 27 100 1350 3. (5 points) Ship A leaves Bridgetown harbor and sails due west at 10 miles per hour. Ship B leaves from the same harbor one hour later, and sails due south at 12 miles per hour. If t represents elasped time since ship B left the harbor (measured in hours), find an expression for z, the straight-line distance between the ships at time t. Solution After t hours ship A has sailed 10t + 10 miles, while ship B has sailed 12t miles. Using the Pythagorean theorem, we see that the distance between them is given by (12t)2 + (10t + 10)2 . http://math.furman.edu/PracticeTest/ c 2002 Mark R. Woodard 4. (5 points) Compute lim x (3x - 2)(2x + 10) . (2x + 1)(x + 2) Solution (3x - 2)(2x + 10) 6x2 + 26x - 20 = lim . If we divide the numerator and denominator by x2 , we obtain x (2x + 1)(x + 2) x 2x2 + 5x + 2 lim 6 + (26/x) - (20/x2 ) 6 = = 3. x 2 + (5/x) + (2/x2 ) 2 lim 5. (5 points) Does the function y = x 3 + 2 have any inflection points? If so, what are they? Solution 1 We must analyze y . Computing, we see that y = 3 x(-2/3) and y = -2 x(-5/3) . Clearly y is never zero, but it fails to 9 exist at x = 0. Note however that 0 is in the domain of the original function. Checking the sign of y on either side of zero, we see that y > 0 for x < 0 while y < 0 for x > 0. This, coupled with the fact that zero is in the domain of y shows that there is an inflection point at (0, 2). 1 6. (5 points) True/False (a) Every is polynomial a rational function. (b) Every rational function has domain R. (c) y = x-4 + x3 is a rational function. (d) The absolute value function has domain (-, 0) (0, ). (e) The function f (x) = x 3x-4 is a rational function. Solution (a) True. A polynomial f (x) can be written as (b) False. For example, f (x) = (c) True. It can be written as 1 x-1 x7 +1 x4 . f (x) 1 , which is the ratio of polynomials. doesn't have domain R. (d) False. The absolute value function has domain R. (e) False. It is algebraic but not rational because of the factor of x. http://math.furman.edu/PracticeTest/ c 2002 Mark R. Woodard 7. (4 points) What are the hypotheses of Rolle's Theorem? What is the conclusion? Solution They hypotheses are that f is continuous on [a, b], differentiable on (a, b), and that f (a) = f (b). The conclusion is that there exists a number c between a and b with f (c) = 0. 8. (a) (5 points) Suppose that f (x) = 2x + 5 and g(x) = (b) Find the domain of (f g)(x). x - 3 + 8. Find (f g)(x). Solution (a) (5 points) (f g)(x) = f (g(x)) = f ( x - 3 + 8) = 2( x - 3 + 8) + 5 = 2( x - 3) + 21. (b) The domain of (f g)(x) is {x : x 3}. 9. (5 points) If a ball is thrown up in the air so that its position about the earth at time t is s(t) = 80t - 16t2 (a) Find the average velocity of the ball between time t = 1 and t = 2. (b) Find the instantaneous velocity of the ball at time t = 1. Solution Between time t = 1 and time t = 2 the ball moves s(2) - s(1) feet, which is equal to (160 - 64) - (80 - 16) = 32 feet. Thus the ball is averaging 32 feet per second in this one elapsed second. The instantaneous velocity is given by s (1). Since s (t) = 80 - 32t, s (1) = 48 feet per second. http://math.furman.edu/PracticeTest/ c 2002 Mark R. Woodard 10. (5 points) Suppose that x and y are differentiable functions of t, and that the functions x and y are related by the equation y = 2x2 + x. What would dy/dt be at x = 3, if dx/dt = 5 when x = 3? Solution By differentiating both sides of the given equation with respect to t, we obtain the relationship dy/dt = 4x(dx/dt) + dx/dt. When x = 3 and dx/dt = 5, we obtain dy/dt = 4(3)(5) + 5 = 65. http://math.furman.edu/PracticeTest/ c 2002 Mark R. Woodard
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