1. (a) Sketch the graph of a function that has two local maxima, one local minimum, and no absolute minimum. Solution: Such a graph is shown below.

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1 MATH 9 Eam (Version ) Solutions November 7, S. F. Ellermeer Name Instructions. Your work on this eam will be graded according to two criteria: mathematical correctness and clarit of presentation. In other words, ou must know what ou are doing (mathematicall) and ou must also epress ourself clearl. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, ou must suppl su cient detail in our solutions (relevant calculations, written eplanations of wh ou are doing these calculations, etc.). It is not su cient to just write down an answer with no eplanation of how ou arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what ou are tring to sa, the less credit ou will get. You ma use our calculator but ou ma not use an books or notes.. (a) Sketch the graph of a function that has two local maima, one local minimum, and no absolute minimum. Solution: Such a graph is shown below (b) Sketch the graph of a function that has three local minima, two local maima, and seven critical numbers. Solution: Such a graph is shown below. Note that two of the critical numbers correspond to neither local maima nor local minima. All of the critical points of this function are points where line segments with di erent slopes come together and hence the function is not di erentiable at these points.

2 Find the absolute maimum and absolute minimum values (and where the occur) of the function f () = 4 + on the interval [ Write our conclusions here: ; ]. You must include all details of our work. f has an absolute maimum value of occurring at. f has an absolute minimum value of occurring at. Show the details of our work here: The fact that f () = 4 4 = 4 = 4 ( + ) ( ) tells us that f has critical numbers = ; = ; and =. All of these critical numbers lie in the interval [ ; ]. Now we compute f ( ) = ( ) 4 ( ) + = f ( ) = ( ) 4 ( ) + = f () = () 4 () + = f () = () 4 () + = f () = () 4 () + = 66 and conclude that f has an absolute maimum value of 66 that occurs at the endpoint = and that f has an absolute minimum value of that occurs at both of the critical points = and =. The graph of f is shown below. 4 +

3 Sketch the graph of an eample of a single function, f, that satis es all of the following conditions. f () =! +! f () = f () =!! Solution: Such a graph is shown below. f () = ! + + =?

4 (a). (b). (c). (d). (e) =. (f) None of the above Circle the correct answer above AND eplain the reason for ou answer (including an necessar calculations). If the answer is choice f (none of the above), then give the correct value of the it in question. Note: Circling the correct choice is not worth an credit unless ou support our choice with correct reasoning. Reasoning: If is a number just slightl greater than, then + is ver close to and + is a positive number but ver close to. Therefore + + is a number ver close to divided b a positive number that is ver close to.. For the function f () =, (a) Find an vertical and horizontal asmptotes of f. (b) Find the intervals on which f is increasing and the intervals on which f is decreasing. (c) Find the local maimum values and local minimum values of f and where these local maima and minima occur. (d) Find the in ection points of f, the intervals on which f is concave up, and the intervals on which f is concave down. (e) Use the information ou found in parts a through d to sketch the graph of f. Solution a. Note that f is not de ned at = or at =. Furthermore! f () =! = f () =! +! + =! f () =! = f () =! +! + = which tells us that f has vertical asmptotes both at = and at =. 4

5 b. and c. Since Also note that f () =!! =! = = =! which tells us that f has a horizontal asmptote at = and that this asmptote is approached both as! and as!. f () = ( ) () () ( ) = ( ) we see that = is the onl critical number of f. Furthermore, since f () > for all < and f () < for all >, we see that f has a local maimum value that occurs at the critical number = (b the First Derivative Test). The value of f at is f () =. f is increasing on the intervals ( ; ) and ( ; ) and decreasing on the intervals (; ) and (; ). d. Since f () = ( ) ( ) ( ) ( ) () ( ) 4 = ( ) ( ) ( ) () ( ) = 6 + ( ) which is positive when > or < and negative when < <, we see that the graph of f is concave up on the intervals ( ; ) and (; ) and concave down on the interval ( ; ). The graph of f has no in ection points. (Note that the graph of f does change concavit but the change occurs at points ( and ) that are not in the domain of f.) e. A graph of f is shown below