Math 180, Final Exam, Spring 2008 Problem 1 Solution. 1. For each of the following limits, determine whether the limit exists and, if so, evaluate it.

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1 Math 80, Final Eam, Spring 008 Problem Solution. For each of the following limits, determine whether the limit eists and, if so, evaluate it. + (a) lim 0 (b) lim ( ) 3 (c) lim Solution: (a) Upon substituting 0 into the function f() + we find that which is indeterminate. We can resolve the indeterminacy by multiplying f() by the conjugate of the numerator divided by itself. lim lim ( + ) lim 0 ( + + ) lim 0 lim 0 ( + + ) (b) When substituting into the function f() ( )3 we find that ( ) 3 ( )3 0 0

2 which is indeterminate. We resolve the indeterminacy by computing the one-sided limits. ( ) 3 ( ) 3 lim lim lim ( ) ( ) 0 ( ) 3 ( ) 3 lim lim + + lim ( ) + ( ) 0 Thus, since the one-sided limits are equal to each other, we know that the limit eists and that: ( ) 3 lim 0 (c) When substituting into the function f() 0 0 we find that which is indeterminate. We resolve the indeterminacy by computing the one-sided limits. lim lim ( ) lim lim + lim + lim + Thus, since the one-sided limits are not equal to each other, we know that the limit does not eist.

3 Math 80, Final Eam, Spring 008 Problem Solution. Find the derivatives of the following functions. Show each step and do not simplify your answer. (a) f() cos (b) g(t) t e /t (c) h() + cos sin (d) k(t) ln( + t + ) Solution: (a) Use the Power Rule and basic derivative of cos. f () ( cos) sin (b) Use the Product and Chain Rules. g (t) ( t e /t) (t )(e /t ) + (t ) (e /t ) ( ) t e /t + te /t t t e /t ( t ) + te /t (c) Use the Quotient Rule. h () ( ) + cos sin ( sin )( + cos ) ( + cos)( sin ) ( sin ) ( sin )( sin ) ( + cos )( cos ) ( sin )

4 (d) Use the Chain Rule. [ k (t) ln( + t + ) ( + t + + ) t + ( ) + t + (t + ) / (t + ) ( ) + t + (t + ) / (t) ]

5 Math 80, Final Eam, Spring 008 Problem 3 Solution 3. The graph of f() is shown below. (a) For which values of does f () not eist? (b) For which values of is f () 0? (c) For which values of is f () > 0? (d) Sketch the graph of f () for 4 4. Solution: (a) f () does not eist at and. (b) f () 0 at 0 (c) f () > 0 on (, 0) (, ) (d) The graph of y f () is below: y

6 Math 80, Final Eam, Spring 008 Problem 4 Solution 4. Evaluate the following definite integrals: (a) (b) (c) 4 π/ dd + d sin() d Solution: (a) We use the Fundamental Theorem of Calculus, Part I. 3 3 d ( ) ( ) ( ) (4 8) (b) We use the substitution u +, du d. The limits of integration become u + and u Making the substitutions and using the Fundamental Theorem of Calculus, Part I the integral is then: d udu [ ] 7 3 u3/ [ 3 73/ ] 3 3/ ( 7 7 ) 3

7 (c) We use the Fundamental Theorem of Calculus, Part I. π/ 0 sin() d cos() π/ 0 ( π cos ( cos 0) ) 0 ( )

8 Math 80, Final Eam, Spring 008 Problem 5 Solution 5. Find an equation for the tangent line to the curve given by the equation at the point ( 4 3, 8 3 ). Solution: We must find dy d 3 +y 3 6y using implicit differentiation. 3 +y 3 6y d d 3 + d d y3 d d (6y) 3 +3y dy d 6dy d +6y 3y dy d 6dy d 3 +6y dy d (3y 6) 3 +6y dy d 3 +6y 3y 6 dy d y y + The value of dy d at (4, 8 ) is the slope of the tangent line. 3 3 dy d ( 4 3 ) ( 8 ) 3 ( 4 3,8 3 ) ( 8 3 ) +( 4) An equation for the tangent line at ( 4, 8 ) is then: 3 3 y ( 4 ) 3

9 Math 80, Eam, Spring 008 Problem 6 Solution 6. Find the maimum and minimum values for the function f() 3 3 on the interval [, ]. Solution: The minimum and maimum values of f() will occur at a critical point in the interval [, ] or at one of the endpoints. The critical points are the values of for which either f () 0 or f () does not eist. Since f() is a polynomial, f () eists for all R. Therefore, the only critical points are solutions to f () 0. f () 0 ( 3 3 ) ( ) 0 3( )( + ) 0 0, ± The critical points 0 lies in the interior of [, ] while ± are the endpoints of the interval. Therefore, we check the value of f() at, 0, and. f( ) ( ) 3 3( ) 4 f(0) 0 3 3(0) 0 f() 3 3() The minimum value of f() on [, ] is 4 because it is the smallest of the above values of f. The maimum is 0 because it is the largest.

10 Math 80, Final Eam, Spring 008 Problem 7 Solution 7. Sketch the graph of the function f(). Your graph must clearly indicate all local +3 maima, local minima, inflection points, and asymptotes. Include the - and y-coordinates of the local etrema and inflection points. Solution: The critical points of f() are the values of for which either f () does not eist or f () 0. f () 0 ( ) 0 +3 ( +3)( ) ( )( +3) 0 ( +3) ( +3)() ( )() ( +3) 0 6 ( +3) Thus, 0 is the only critical point of f. (Note: +3 > 0 for all.) The domain of f is (, ). We now split the domain into the intervals (,0) and (0, ). We then evaluate f () at a test point in each interval to determine the intervals of monotonicity. Interval Test Point, c f (c) Sign of f (c) (,0) f ( ) 3 8 (0, ) f () Using the table, we conclude that f is increasing on (0, ) because f () > 0 for all (0, ) and f is decreasing on (,0) because f () < 0 for all (,0). Also, since f changes sign from to + at 0 the First Derivative Test implies that f(0) 0 is a local minimum. Todeterminetheintervalsofconcavitywestartbyfindingsolutionstotheequationf () 0

11 and where f () does not eist. f () 0 ( ) 6 0 ( +3) ( +3) (6) (6)[( +3) ] 0 ( +3) 4 ( +3) (6) (6)(( +3)()) 0 ( +3) 4 6( +3) 4 ( +3) 3 0 6( +3) ± We now split the domain into the intervals (, ), (,), and (, ). We then evaluate f () at a test point in each interval to determine the intervals of concavity. Interval Test Point, c f (c) Sign of f (c) (, ) f ( ) (,) 0 f (0) 3 + (, ) f () Using the table, we conclude that f is concave down on(, ) (, ) because f () < 0 for all (, ) (, ) and f is concave up on (,) because f () > 0 for all (,). The inflection points of f() are the points where f () changes sign. We can see in the above table that f () changes sign at ±. Therefore, ± are inflection points. The values of f() at the inflection points are f( ) f() 4. f() has no vertical asymptotes. However, its horizontal asymptote is y because lim f() lim ± ± +3

12 y

13 Math 80, Final Eam, Spring 008 Problem 8 Solution 8. A sheep rancher plans to fence a rectangular pasture net to an irrigation canal. No fence will be needed along the canal, but the other three sides must be fenced. The pasture must have an area of 80,000 square yards to provide enough grass for the sheep. Find the dimensions of the pasture which require the least amount of fence. Solution: We begin by letting be the length of the side opposite the canal and y be the lengths of the remaining two sides. The function we seek to minimize is the length of the fence: Function : Length + y () The constraint in this problem is that the area of the pasture is 80,000 square meters. Constraint : y 80, 000 () Solving the constraint equation () for y we get: y 80, 000 Plugging this into the function () and simplifying we get: ( ) 80, 000 Length + 360, 000 f() + We want to find the absolute minimum of f() on the interval (0, ). We choose this interval because must be nonnegative ( represents a length) and non-zero (if were 0, then the area would be 0 but it must be 80, 000). The absolute minimum of f() will occur either at a critical point of f() in (0, ) or it will not eist because the interval is open. The critical points of f() are solutions to f () 0. f () 0 ( ) 360, , , 000 ±600 However, since 600 is outside (0, ), the only critical point is 600. Plugging this into f() we get: 360, 000 f(600) (3)

14 Taking the limits of f() as approaches the endpoints we get: ( ) 360, 000 lim f() lim ( ) 360, 000 lim f() lim both of which are larger than 00. We conclude that the length is an absolute minimum at 600 and that the resulting length is 00. The last step is to find the corresponding value for y by plugging 600 into equation (3). y 80, ,

15 Math 80, Final Eam, Spring 008 Problem 9 Solution 9. The graph below shows the velocity of a race car, in meters in second, during a time interval of 8 seconds. Estimate the distance that the car traveled during this time interval. You must eplain how you arrived at your estimate to receive credit. Solution: We will use L 8 to estimate the distance traveled. Since N 8 and the interval is [0, 8], we know that. The value of L 8 is then: L 8 [f(0) + f() + f() + f(3) + f(4) + f(5) + f(6) + f(7)] L 8 [ ] L 8 99 meters