Math 180, Exam 2, Study Guide Problem 1 Solution

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1 . Let f() = +. Math 80, Eam, Study Guide Problem Solution Determine the intervals on which f is increasing and those on which it is decreasing. Determine the intervals on which f is concave up and those on which it is concave down. Find the critical points of f and determine if they correspond to local etrema. Find the asymptotes of f. Determine the global etrema of f. Sketch the graph of f. Solution: First, we etract as much information as we can from f (). We ll start by computing f () using the Quotient Rule. f () = ( + )() ()( + ) ( + ) f () = ( + )() ()() ( + ) f () = ( + ) The critical points of f() are the values of for which either f () does not eist or f () = 0. f () is a rational function but the denominator is never 0 so f () eists for all R. Therefore, the only critical points are solutions to f () = 0. f () = 0 ( + ) = 0 = 0 = ± Thus, = and = are the critical points of f. The domain of f is (, ). We now split the domain into the three intervals (, ), (, ), and (, ). We then evaluate f () at a test point in each interval to determine the intervals of monotonicity.

2 Interval Test Point, c f (c) Sign of f (c) (, ) f ( ) = 3 5 (, ) 0 f (0) = + (, ) f () = 3 5 Using the table, we conclude that f is increasing on (, ) because f () > 0 for all (, ) and f is decreasing on (, ) (, ) because f () < 0 for all (, ) (, ). Furthermore, since f changes sign from to + at = the First Derivative Test implies that f( ) = is a local minimum and since f changes sign from + to at = the First Derivative Test implies that f() = is a local maimum. We then etract as much information as we can from f (). We ll start by computing f () using the Quotient and Chain Rules. f () = ( + ) ( ) ( )[( + ) ] ( + ) 4 f () = ( + ) ( ) ( )[( + )( + ) ] ( + ) 4 f () = ( + ) ( )( + )() ( + ) 4 f () = ( + ) 4( ) ( + ) 3 f () = 3 6 ( + ) 3 The possible inflection points of f() are the values of for which either f () does not eist or f () = 0. Since f () eists for all R, the only possible inflection points are solutions to f () = 0. f () = ( + ) 3 = = 0 ( 3) = 0 = 0, = ± 3 We now split the domain into the four intervals (, 3), ( 3, 0), (0, 3), and ( 3, ). We then evaluate f () at a test point in each interval to determine the intervals of concavity.

3 Interval Test Point, c f (c) Sign of f (c) (, 3) f ( ) = 4 5 ( 3, 0) f ( ) = + (0, 3) f () = ( 3, ) f () = Using the table, we conclude that f is concave up on ( 3, 0) ( 3, ) because f () > 0 for all ( 3, 0) ( 3, ) and f is concave down on (, 3) (0, 3) because f () < 0 for all (, 3) (0, 3). Furthermore, since f changes sign at = 3, = 0, and = 3, all three points are inflection points. f() does not have a vertical asymptote because it is continuous for all R. The horizontal asymptote is y = 0 because lim f() = 0 + The absolute minimum of f() is at = and the absolute maimum is at =. y

4 Math 80, Eam, Study Guide Problem Solution. Let f() = e. i) Find and classify the critical points of f. ii) Find the global minimum of f over the entire real line. Solution: i) The critical points of f() are the values of for which either f () = 0 or f () does not eist. Since f() is a product of two infinitely differentiable functions, we know that f () eists for all R. Therefore, the only critical points are solutions to f () = 0. f () = 0 (e ) = 0 ()(e ) + (e )() = 0 e + e = 0 e ( + ) = 0 = = is the only critical point because e > 0 for all R. We use the First Derivative Test to classify the critical point =. The domain of f is (, ). Therefore, we divide the domain into the two intervals (, ) and (, ). We then evaluate f () at a test point in each interval to determine where f () is positive and negative. Interval Test Number, c f (c) Sign of f (c) (, ) e (, ) 0 + Since f changes sign from to + at = the First Derivative Test implies that f( ) = e is a local minimum. ii) From the table in part (a), we conclude that f is decreasing on the interval (, ) and increasing on the interval (, ). Therefore, f( ) = e is the global minimum of f over the entire real line.

5 Math 80, Eam, Study Guide Problem 3 Solution 3. Find the minimum and maimum of the function f() = 6 3 over the interval [0,]. Solution: The minimum and maimum values of f() will occur at a critical point in the interval [0,] or at one of the endpoints. The critical points are the values of for which either f () = 0 or f () does not eist. The derivative f () is found using the Chain Rule. [ (6 f 3 () = ) ] / [ (6 ) ] (6 3 / 3) f () = f () = f () = [ (6 ) 3 / ] (6 3 ) f () does not eist when the denominator is 0. This will happen when 6 3 = 0. The solutions to this equation are obtained as follows: 6 3 = 0 (6 ) = 0 = 0, = ± 6 The critical point = 0 is an endpoint of [0,]. The critical points = ± 6 both lie outside [0,]. Therefore, there are no critical points in [0,] where f () does not eist. The only critical points are points where f () = 0. f () = = = 0 = = ± The critical point = lies outside [0,]. Therefore, = is the only critical point in [0,] where f () = 0. We now evaluate f() at = 0,, and. f(0) = 6(0) 0 3 = 0 ( ) ( ) ) 3 f = 6 ( 4 = f() = 6() 3 =

6 The minimum value of f() on [0,] is 0 because it is the smallest of the above values of f. The maimum is 4 because it is the largest. y 3 4 0

7 Math 80, Eam, Study Guide Problem 4 Solution 4. Let f() = 3 3. i) On what interval(s) is f increasing? ii) On what interval(s) is f decreasing? iii) On what interval(s) is f concave up? iv) On what interval(s) is f concave down? v) Sketch the graph of f. Solution: i) Webeginbyfindingthecriticalpointsoff(). Thecriticalpointsoff()arethevalues of for which either f () does not eist or f () = 0. Since f() is a polynomial, f () eists for all R so the only critical points are solutions to f () = 0. f () = 0 ( 3 3 ) = = 0 = = ± The domain of f is (, ). We now split the domain into the three intervals (, ), (,), and (, ). 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) (, ) f ( ) = 9 (,) 0 f (0) = 3 + (, ) f () = 9 Using the table we conclude that f is increasing on (,) because f () > 0 for all (,) ii) Fromthetableabove we conclude that f is decreasing on (, ) (, ) because f () < 0 for all (, ) (, ).

8 iii) To determine the intervals of concavity we start by finding solutions to the equation f () = 0 and where f () does not eist. However, since f() is a polynomial we know that f () will eist for all R. The solutions to f () = 0 are: f () = 0 6 = 0 = 0 We now split the domain into the two intervals (,0) and (0, ). 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) (,0) f ( ) = 6 + (0, ) f (0) = 6 Using the table we conclude that f is concave up on (,0) because f () > 0 for all (,0). iv) Fromtheabovetableweconcludethatf isconcavedownon (0, ) becausef () < 0 for all (0, ). y - - v) -

9 Math 80, Eam, Study Guide Problem 5 Solution 5. For a function f() we know that f(3) = and that f (3) = 3. Give an estimate for f(.9). Solution: We will estimate f(.9) using L(.9), the linearization L() of the function f() at a = 3 evaluated at =.9. The function L() is defined as: Using f(3) = and f (3) = 3 we have: Plugging =.9 into L() we get: L() = f(3) + f (3)( 3) L() = 3( 3) L(.9) = 3(.9 3) L(.9) =.7 Therefore, f(.9) L(.9) =.7.

10 Math 80, Eam, Study Guide Problem 6 Solution 6. Let f() = +. Find the best linear approimation of f around the point = 0 and + use it in order to estimate f(0.). Would this be an underestimate or an overestimate? Solution: The linearization L() of f() at = 0 is defined as: L() = f(0) + f (0)( 0) The derivative f () is found using the Quotient Rule: f () = At = 0, the values of f and f are: ( ) + + = ( + )( + ) ( + )( + ) ( + ) = ( + )() ( + )() ( + ) = + ( + ) f (0) = 0 + (0) (0 + ) = f(0) = = The linearization L() is then: L() = Since f(0.) L(0.) we find that: f(0.) L(0.) The actual value of f(0.) is: So L(0.) = 0.8 is an underestimate. f(0.) = =.04. = 3 5 > 5 = 0.8

11 Math 80, Eam, Study Guide Problem 7 Solution 7. A rectangular farm of total area 0,000 sq. feet is to be fenced on three sides. Find the dimensions that are going to give the minimum cost. Solution: We begin by letting be the length of one side, y be the lengths of the remaining two fenced sides, and C > 0 be the cost of the fence per unit length. The function we seek to minimize is the cost of the fencing: Function : Cost = C( + y) () The constraint in this problem is that the area of the garden is 0,000 square meters. Constraint : y = 0, 000 () Solving the constraint equation () for y we get: y = 0, 000 Plugging this into the function () and simplifying we get: [ ( )] 0, 000 Cost = C + ( ) 40, 000 f() = C + (3) 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 0, 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 ( ) 40, 000 C + = 0 ( ) 40, 000 C = 0 = 40, 000 = ±00 However, since = 00 is outside (0, ), the only critical point is = 00. Plugging this into f() we get: ( ) 40, 000 f(00) = C 00 + = 400C 00

12 Taking the limits of f() as approaches the endpoints we get: ( ) lim f() C 40, = C (0 + ) = ( ) lim f() C 40, = C ( + 0) = both of which are larger than 400C. We conclude that the cost is an absolute minimum at = 00 and that the resulting cost is 400C. The last step is to find the corresponding value for y by plugging = 00 into equation (3). y = 0, 000 = 0, = 00

13 Math 80, Eam, Study Guide Problem 8 Solution 8. Let f() = Find the critical points of f. Determine the intervals on which f is increasing and the ones on which it is decreasing. Determine the intervals on which f is concave up and the ones on which it is concave down. Determine the inflection points of f. Sketch the graph of f. Solution: The critical points of f() are the values of for which either f () does not eist or f () = 0. Since f() is a polynomial, f () eists for all R so the only critical points are solutions to f () = 0. f () = 0 ( 3 5 3) = = 0 3 (5 ) = 0 = 0, = ± 5 Therefore, the critical points of f are = 0, ± 5. The domain of f is (, ). We now split the domain into the four intervals (, 5 ), ( 5, 0), (0, 5 ), and ( 5, ). 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) (, 5 ) f ( ) = + ( 5, 0) 5 f ( 5 ) = 5 (0, 5 ) f ( ) = ( 5, ) f () = +

14 Using the table we conclude that f is increasing on (, 5 ) ( 5, ) because f () > 0 for all (, 5 ) ( 5, ) and f is decreasing on ( 5, 0) (0, because f () < 0 for all ( 5, 0) (0, To determine the intervals of concavity we start by finding solutions to the equation f () = 0 and where f () does not eist. However, since f() is a polynomial we know that f () will eist for all R. The solutions to f () = 0 are: 5 ). f () = 0 ( ) = = 0 6(0 ) = 0 = 0, = ± 0 5 ) We now split the domain into the four intervals (, 0 ), ( 0, 0), (0, 0 ), and ( 0, ). 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) (, 0 ) f ( ) = 54 ( 0, 0) 0 f ( 0 ) = (0, 0 ) 0 f ( 0 ) = 7 50 ( 0, ) f () = 54 + Using the table we conclude that f is concave up on ( 0, 0) ( 0, ) because f () > 0 for all ( 0, 0) ( 0, ) and that f is concave down on (, (0, 0 ) because f () < 0 for all (, 0 ) (0, 0 ). 0 ) An inflection point of f() is a point where f () changes sign. From the above table we conclude that = 0, ± 0 are inflection points.

15 y

16 Math 80, Eam, Study Guide Problem 9 Solution 9. A rectangle has its left lower corner at (0, 0) and its upper right corner on the graph of i) Epress its area as a function of. f() = + ii) Determine for which the area is a minimum. iii) Can the area of such a rectangle be as large as we please? Solution: i) The dimensions of the rectangle are and y. Therefore, the area of the rectangle has the equation: Area = y () We are asked to write the area as a function of alone. Therefore, we must find an equation that relates to y so that we can eliminate y from the area equation. This equation is y = + () because (, y) must lie on this curve. Plugging this into the area equation we get: Area = ( + ) g() = 3 + ii) We seek the value of that minimizes g(). The interval in the problem is (0, ) because the domain of f() is (, 0) (0, ) but (, y) must be in the first quadrant. 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 ( 3 + ) = 0 3 = = 0 = ± 4 3

17 However, since = 4 3 is outside (0, ), the only critical point is = 4 3. Plugging this into g() we get: ( ) ( ) 3 f = = Taking the limits of f() as approaches the endpoints we get: ( lim f() 3 + ) = 0 + = ( lim f() 3 + ) = + 0 = both of which are larger than We conclude that the area is an absolute minimum at = 4 3 and that the resulting area is iii) We can make the rectangle as large as we please by taking 0 + or.

18 Math 80, Eam, Study Guide Problem 0 Solution 0. A bo has square base of side and constant surface area equal to m. i) Epress its volume as a function of. ii) Find the maimum volume of such a bo. Solution: i) We begin by letting be the length of one side of the base and y be the height of the bo. The volume then has the equation: Volume = y () We are asked to write the volume as a function of width,. Therefore, we must find an equation that relates to y so that we can eliminate y from the volume equation. The constraint in the problem is that the total surface area is. This gives us the equation + 4y = () Solving this equation for y we get + 4y = + y = 6 y = 6 We then plug this into the volume equation () to write the volume in terms of only. Volume = y ( ) 6 Volume = (3) f() = 3 3 (4) ii) We seek the value of that maimizes f(). The interval in the problem is (0, 6]. We know that > 0 because must be positive and nonzero (otherwise, the surface area would be 0 and it must be ). It is possible that y = 0 in which case the surface area constraint would give us + 4(0) = = 6 = 6.

19 The absolute maimum of f() will occur either at a critical point of f() in (0, 6], at = 6, or it will not eist. The critical points of f() are solutions to f () = 0. f () = 0 ( 3 ) 3 = = 0 = = ± However, since = is outside (0, 6], the only critical point is =. Plugging this into f() we get: ( ) f ( ) = 3 ( ) 3 = Evaluating f() at = 6 and taking the limit of f() as approaches = 0 we get: lim f() (3 ) 3 = ( ) ( ) f 6 = 3 6 ( ) 3 6 = 0 both of which are smaller than. We conclude that the volume is an absolute maimum at = and that the resulting volume is m 3.

20 Math 80, Eam, Study Guide Problem Solution. Use the Newton approimation method in order to find as an estimate for the positive root of the equation 5 = 0 when 0 = 5. Solution: The Newton s method formula to compute is = 0 f( 0) f ( 0 ) where f() = 5. The derivative f () is f () =. Plugging 0 = 5 into the formula we get: = = (5) = = 3 The Newton s method formula to compute is Plugging = 3 into the formula we get: = f( ) f ( ) = 5 = (3) = = 7 3

21 Math 80, Eam, Study Guide Problem Solution. Use L Hôpital s Rule in order to compute the following limits: lim 0 lim 4 ln(3 + ) ln(5 + ) ( 4 4 lim ln 0 ) + lim + lim 0 Solution: Upon substituting = 0 into the function ln(3+) ln(5+) ln(3(0) + ) ln(5(0) + ) = 0 0 e 3 tan e + ln we get which is indeterminate. We resolve the indeterminacy using L Hôpital s Rule. lim 0 ln(3 + ) ln(5 + ) L H 0 (ln(3 + )) (ln(5 + )) = 3 5 5(0) + 3(0) + = 3 5 As 0 + we find that ln 0 ( ) which is indeterminate. However, it is not of the form 0 or which is required to use L Hôpital s Rule. To get the limit into one of the two 0 required forms, we rewrite ln as follows: ln = ln As 0 +, we find that ln. We can now use L Hôpital s Rule. / ln lim ln L = H (ln ) lim 0 + ( ) = 0

22 Upon substituting = 0 into the function e3 tan e 3(0) tan0 we get which is indeterminate. We resolve the indeterminacy using L Hôpital s Rule. lim 0 e 3 tan = 0 0 L H 0 (e 3 ) (tan ) 0 3e 3 sec 0 3e 3 cos = 3e 3(0) cos 0 = 3 Upon substituting = 4 into the function 4 4 we get = which is indeterminate. In order to use L Hôpital s Rule we need the limit to be of the form. To get the limit into one of these forms, we rewrite the function as follows: 0 0 or 4 4 = 4 4( ) ( )( 4) = ( )( 4) = ( )( ) ( )( 4) = 4 Upon substituting = 4 into the 4 we get = 0 0

23 which is now of the indeterminate form 0. We can now use L Hôpital s Rule. 0 ( lim 4 ) L = H ( ) lim 4 ( 4) 4 = 4 = 4 As +, we find that using L Hôpital s Rule. e +ln lim + which is indeterminate. We resolve the indeterminacy e + ln L H + + = = + (e ) ( + ln ) e + 3

24 Math 80, Eam, Study Guide Problem 3 Solution 3. Compute the following indefinite integrals: ( ) d 3 ( ) d e 3 d Solution: Using the linearity and power rules, the first integral is: ( ) d = d 5 d + 6 d = ( ) () + C = C Using some algebra and the linearity and power rules, the second integral is: 3 ( ) d = /3( /) d ( = 7/3 5/6) d = 3 0 0/3 6 /6 + C Using the rule e k d = k ek + C, the third integral is: e 3 d = 3 e3 + C

25 Math 80, Eam, Study Guide Problem 4 Solution 4. Consider the function f() = on [0, ]. Compute L 4 and R 4. Solution: For each estimate, the value of is: = b a N = 0 = 4 The L 4 estimate is: [ ( ) ( )] 3 L 4 = f(0) + f + f() + f [ = (0 0 ) ( ( ) ) + + ( ) ( (3 ) )] + 3 = [ ] 4 = 4 The R 4 estimate is: [ ( ) ( ) ] 3 R 4 = f + f() + f + f() [( ( = ) ) + ( ) ( (3 ) ) ( )] = [ ] + = 5 4

26 Math 80, Eam, Study Guide Problem 5 Solution 5. Use the Fundamental Theorem of Calculus in order to compute the following integrals: 0 ( + + ) d 4 d π 0 sin() d Solution: The first integral has the value: 0 ( + + ) [ d = ] + 0 [ = ] [ ] [ ] 8 = [ ] = 0 3 The second integral has the value: 4 4 d = / d = [ 3 3/ ] 4 = 3 43/ 3 3/ = = 4 3 The third integral has the value: π [ sin() d = ] π 0 cos() 0 = [ ] cos(π) [ ] cos((0)) [ = ] [ ] = 0