Circle your answer choice on the exam AND fill in the answer sheet below with the letter of the answer that you believe is the correct answer.

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1 ircle your answer choice on the eam AND fill in the answer sheet below with the letter of the answer that you believe is the correct answer. Problem Number Letter of Answer Problem Number Letter of Answer. D. D 8.. A 9. B. D.. B. D. For Part I Free Response. Show all work (as described on page ).. [8 pts] Use linear approimation to approimate the value of to four decimal places. F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

2 . [9 pts] A beverage company works out a demand function for its sale of soda and find it to be q p where q is the quantity of sodas sold when the price per can, in cents, is p. At what price is the revenue a maimum?. [9 pts] Find the -values of all points where the function f ( ) has any relative etrema. Find the value(s) of any relative etrema. Write your final answer in the appropriate bo on the right. Answers must be eact. f () Set equal to and solve for. 8 ± 8 We also need to consider the places where the derivative does not eist. That will occur where Relative Maima -value value 8 8 Relative Minima -value value F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

3 Now calculate the values f() f() 8 8 f() 8 8. [8 pts] Find dy for y,, and.. [8 pts] Evaluate the indefinite integral ( e ) d F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

4 . [9 pts] A farmer decides to make three identical pens with feet of fencing. The pens will be net to each other sharing a fence and will be up against a barn. The barn side needs no fence (see picture to the right). What dimensions for the total enclosure (rectangle including all pens) will make the area as large as possible? F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

5 For Part II Multiple hoice: ircle your answer choice on the eam AND fill in the answer sheet on the front of page with the letter of the answer that you believe is the correct answer.. [ pts] The concentration of a certain drug in the bloodstream hours after being administered is approimately ( ). Use the differential to approimate the change in concentration as changes from to.. A.. B.... D..8 E. None of these d() ( ) () ( ) d d() d ( ) d() ( ) d () d() (. ) ( () ) d().8 8. [ pts] Find the elasticity of the demand function q p at the price p $8 and state whether the demand is elastic, inelastic or whether it has unit elasticity. 9 9 A. ; inelastic B. ; unit elasticity. ; elastic 8 D. ; elastic E. None of these p dq E q dp dq dp E 9 ( p ) p p p E ( p) 8 E ( 8) E ( ) > thus elastic p F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

6 9. [ pts] Find the open interval(s) where the function f ( ) is increasing. A. (, ) B. (, ). (, ) D. (, ) E. None of these f ( ) ( ) f ( ) ( ) ( ) ( ) f check - check ( ) ) f ) (( ) ) (() ( f ( ) positive () ) ( f () negative. [ pts] Suppose that the graph below is the graph of the derivative of the function f() (i.e. this is the graph of f () ). Find the locations of all etrema and tell whether each etrema is a relative maimum or relative minimum. f () A. relative minimum at - and ; relative maimum at - B. relative maimum at and relative minimum at -. relative maima at - and and relative minimum at - D. no relative etrema E. None of these The relative etrema occur when the derivative is equal to. This will be at -, -, and. If the derivative is positive to the left and negative to the right of the zero of the derivative, then the root represents a relative maimum of the function. If the derivative is negative to the left and negative to the right of the zero of the derivative, then the root represents a relative minimum of the function. F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

7 . [ pts] Find the absolute maimum and absolute minimum values of the function ( ) 8, f on the interval [ ] A. absolute maimum -, no absolute minimum B. absolute maimum, absolute minimum -. absolute maimum, absolute minimum D. absolute maimum, absolute minimum - E. None of these f ( ) 8. [ pts] Evaluate the indefinite integral 8 8 Find function values at,, and. f () 8() d f () f () 8() 8() A. B. ln. ln D. ln E. None of these We have to start by breaking up the integral. d d d d d d ln ln F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

8 F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University. [ pts] Evaluate the indefinite integral d A. ln B.. (ln ) D. E. None of these ln ln. [ pts] Evaluate the indefinite integral ( ) d A. B.. D. E. None of the above ( ) d d d d

9 . [ pts] The function g() is the inverse of the function f ( ) where >. Find g (). A. g ( ) B. g ( ). g ( ) D. g ( ) E. None of the above g ( y ) where, y is a point on f ( f ( ) ( ) ) We need to find that goes with the y-coordinate. ( )( ) or or The one that we use is since according to the problem is greater than. f ( ) g ( ) (). [ pts] Suppose that the graph to the right is the graph of the derivative of the function f() (i.e. this is the f () graph of f () ). Find all open intervals where the function f() is concave downward. A. (, ) B. (, ). (, ) (, ) D. f() is never concave downward E. None of these In general, concavity changes (an inflection point eists) where the second derivative is equal to (or does not eist). The second derivative is equal to where the first derivative has a maimum or a minimum. Thus the inflection points for f() are at - and at. The net thing that we need to do is check the sign of the second derivative to the left of -, between - and, and to the right of. When the second derivative is negative, this indicates that the slope of the tangent lines of the first derivative are negative. Thus the original function is concave down (occurs when the second derivative is negative) when the slope of the tangent lines of the first derivative are negative. Thus the function is concave downward on the interval (, ) (, ) F9B opyright 9 School of Mathematical and Statistical Sciences, Arizona State University

f'(x) = x 4 (2)(x - 6)(1) + (x - 6) 2 (4x 3 ) f'(x) = (x - 2) -1/3 = x 2 ; domain of f: (-, ) f'(x) = (x2 + 1)4x! 2x 2 (2x) 4x f'(x) =

f'(x) = x 4 (2)(x - 6)(1) + (x - 6) 2 (4x 3 ) f'(x) = (x - 2) -1/3 = x 2 ; domain of f: (-, ) f'(x) = (x2 + 1)4x! 2x 2 (2x) 4x f'(x) = 85. f() = 4 ( - 6) 2 f'() = 4 (2)( - 6)(1) + ( - 6) 2 (4 3 ) = 2 3 ( - 6)[ + 2( - 6)] = 2 3 ( - 6)(3-12) = 6 3 ( - 4)( - 6) Thus, the critical values are = 0, = 4, and = 6. Now we construct the sign chart

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