MAT Calculus for Engineers I EXAM #1
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1 MAT 65 - Calculus for Engineers I EXAM # Instructor: Liu, Hao Honor Statement By signing below you conrm tat you ave neiter given nor received any unautorized assistance on tis eam. Tis includes any use of a graping calculator beyond tose uses specically autorized by te Scool of Matematical and Statistical Sciences and my instructor. Furtermore, I agree not to discuss tis eam wit anyone until te eam testing period is over. In addition, my calculator's memory and menus may be cecked at any time and cleared by any testing center proctor or Scool of Matematical and Statistical Sciences instructor. Signature Date Instructions:. Te eam consists of two parts: multiple coice, wort 60%, and free response sow your work, wort 40%. Please read eac problem carefully.. Tere are 0 multiple coice questions wort 6 points eac. Please circle your answer coice on te eam AND ll in te answer grid on te last page. 3. Provide complete and well-organized answers in te free response section. 4. Answers in te free response section witout supporting work will be given zero credit. Partial credit is granted only if work is sown. 5. No calculators wit Qwerty keyboards or ones like te Casio FX-, TI-89, TI-9, or TI-nspire tat do symbolic algebra may be used. 6. Proctors reserve te rigt to ceck calculators. 7. Please request scratc paper from your instructor. 8. Te use of cell pones is proibited. TURN YOUR CELL PHONE OFF! Do not allow your cell pone to ring wile you are taking te eam. Do not use te calculator on your cell pone. If a proctor sees you using a cell pone, tey will take your eam and you will be reported to te Dean of Students for ceating. 9. If you read te instructions, please place a ceck mark ere for one point!
2 MAT 65 - Calculus for Engineers I EXAM Hao Liu PART I - Multiple coice.. Te function epression tells us tat 0 5 0, and + 0 Hence te domain is { R, }. Or written in interval notation Answer: C. Since + = +, we ave Answer: D,, ] Notice tat plugging in = leads to a 0 0 radical conjugate case, so we need to simplify te function, by Answer: B Recall tat sin π =, so π g = sin Hence g DNE. Answer: C dierence of squares = = =, and + g = 3 = Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University
3 MAT 65 - Calculus for Engineers I EXAM Hao Liu 5. Note tat since wen f is a polynomial, f + = 4 f =. Hence f eists, but f is not continuous at. Furtermore te discontinuity is a removable type. Answer: D 6. On eiter branc, or, te function f is continuous, since it's polynomial. However, for it to be continuous over te entire real line, we need its continuity at =, namely f = f We already know f, and for f to eist we need equality from one-sided its f + f c 3 4c + c 3 = 4c 8c = 4c 4 4c c Also, since f is dened as c 3 for, it is left continuous, ie automatically we ave f = f. Terefore we c, we ave Answer: A f f = f. 7. Dividing by in bot numerator and denominator gives since Answer: C 0 as approaces We need to divide by in bot top and bottom / 5/ Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University
4 MAT 65 - Calculus for Engineers I EXAM Hao Liu In te above equation, we use te fact tat as approaces it is negative and. Closeup in te numerator Hence Answer: A since 4 0, 5 0 as 9. We again use radical conjugate Closeup in te denominator: / / divide by di. of squares 6 + = 6 + since we ave Hence since 0 wen Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University 3
5 MAT 65 - Calculus for Engineers I EXAM Hao Liu Answer: D 0. Recall tat sinπ = 0, wit direct substitution, we ave 0 = 6. Answer: E PART II - Free response. Sow all your work.. f g = fg = 4 3 From te epression we ave 4 3, wic gives us 5. Also from te domain of g, we ave 4. Hence te domain for f g is, 5 5, 4 ] Now we consider g f From te epression we seek g f = Now if 3 > 0 we ave However if 3 < 0 we ave Hence we obtain tat, 3 [3 4, Combine tis domain wit te domain for f we ave te domain for g f as, 3 3,, 3 [3 4, Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University 4
6 MAT 65 - Calculus for Engineers I EXAM Hao Liu. Te derivative of f = is f f + f + = Now at = 9 te slope of tangent line derivative is f 9 = 9 = 6 di. of squares radical conjugate Also 0 = 9 wit f 0 = 9 = 3. Terefore te equation for tangent line is or y f 0 = f 0 0 y 3 = 9 6 y = Using te it denition of derivative, we nd f Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University 5
7 MAT 65 - Calculus for Engineers I EXAM Hao Liu 4. By denition orizontal asymptote corresponds to eiter or Note tat for our function f f f f divide by Closeup in te denominator = 9 + = = 9 + since > 0, = Now we obtain tat f = 3. Similarly if we let, we use te fact tat since < 0, to obtain f divide by Summer 03 c Scool of Matematical & Statistical Sciences - Arizona State University 6
MAT Calculus for Engineers I EXAM #3
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