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Additional problems address continuity by determining a parameter, tangent line and second-derivative computations for an implicit curve, matching derivative values from a graph, and an applications section involving exponential decay and inverse functions with domains.",{"@graph":14,"@context":72},[15,34,55],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/calculus-221-first-exam-50-minutes-answers/466346/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/calculus-221-first-exam-50-minutes-answers/466346.png","ImageObject",300,407,{"name":42,"@type":43},"Jasmine","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-28","2026-09-30",true,{"@type":52,"interactionType":53,"userInteractionCount":26},"InteractionCounter",{"@type":54},"ViewAction",{"@type":56,"mainEntity":57},"FAQPage",[58,64,68],{"name":59,"@type":60,"acceptedAnswer":61},"How are trigonometric limits evaluated in these solutions?","Question",{"text":62,"@type":63},"Trigonometric limits are handled using bounding and substitution, then evaluated using the Squeeze theorem or continuity where appropriate.","Answer",{"name":65,"@type":60,"acceptedAnswer":66},"What method is used to prove the product rule (fg)'?",{"text":67,"@type":63},"The proof starts from the definition of the derivative for w(x)=f(x)g(x), applies algebraic expansion, and uses limit laws plus the fact that differentiability implies continuity.",{"name":69,"@type":60,"acceptedAnswer":70},"How is the constant c determined for continuity in the piecewise function?",{"text":71,"@type":63},"The solutions equate the left-hand and right-hand limits at the discontinuity point, using limit laws, and solve the resulting equation for c.","https://schema.org",{"og:url":32,"og:type":74,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":76,"canonical":32},"index,follow",{"doc_id":78,"site_id":7},466346,1793029533,{"code":4,"msg":81,"data":82},"success",[83,87,91,94,99,104,109,114,119,122,126],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":84,"show_sort_weight":85,"slug":86},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":88,"show_sort_weight":89,"slug":90},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":92,"slug":93},70,"exam",{"id":95,"doc_module":4,"doc_module_name":25,"category_name":96,"show_sort_weight":97,"slug":98},5,"Comic",60,"comic",{"id":100,"doc_module":4,"doc_module_name":25,"category_name":101,"show_sort_weight":102,"slug":103},6,"Technology",50,"technology",{"id":105,"doc_module":4,"doc_module_name":25,"category_name":106,"show_sort_weight":107,"slug":108},7,"Healthcare",40,"healthcare",{"id":110,"doc_module":4,"doc_module_name":25,"category_name":111,"show_sort_weight":112,"slug":113},8,"Research & Report",30,"research-report",{"id":115,"doc_module":4,"doc_module_name":25,"category_name":116,"show_sort_weight":117,"slug":118},9,"Religion & Spirituality",20,"religion-spirituality",{"id":117,"doc_module":4,"doc_module_name":25,"category_name":120,"show_sort_weight":117,"slug":121},"World Cup","world-cup",{"id":123,"doc_module":4,"doc_module_name":25,"category_name":124,"show_sort_weight":123,"slug":125},10,"Lifestyle","lifestyle",{"id":127,"doc_module":4,"doc_module_name":25,"category_name":128,"show_sort_weight":95,"slug":129},19,"General","general",{"code":4,"msg":81,"data":131},{"doc_id":78,"user_id":132,"nickname":42,"user_avatar":133,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":134,"file_id":135,"file_url":136,"file_type":137,"file_size":138,"view_count":26,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":139,"language":140,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":141,"faqs":142,"seo_title":143,"seo_description":12,"update_tm":144,"read_time":145},2336478487870,"https://ap-avatar.wpscdn.com/davatar_085a072bc5b1113ac321206ff7593b45","Calculus 221-First Exam (50 Minutes)  \nFriday October 4 1996  \nI Find the limit or show that it does not exist. Justify your answer.  \nt3 􀀀 1  \n(1) limt!􀀀1 t2 􀀀 1  \nAnswer:  \nlim t3 􀀀 1 = lim t 􀀀 1t2  = 􀀀1  \nt!􀀀1 t2 􀀀 1 t!􀀀1 1 􀀀 1t2  \nt3 􀀀 1  \n(2) lim ~~ ~~t!3 t2 􀀀 1  \nAnswer:  \nltim!3 t~~ ~~3t2~~ ~~􀀀~~ ~~11 = 3332~~ ~~􀀀~~ ~~11 = 268  \nt3 􀀀 1  \n(3) lim ~~ ~~t! 1 t2 􀀀 1  \nAnswer:  \nlti1 t~~ ~~3t2~~ ~~􀀀~~ ~~11 = lti1 (t~~ ~~􀀀(t~~ ~~)~~ ~~(t12)(+t~~ ~~~~ ~~1) = lti1 (t2(+t~~ ~~~~ ~~1) = 32  \nII Find the limit or show that it does not exist. Justify your answer.  \nsin(3x)(1) lx x  \nAnswer: Since 􀀀1 􀀔 sin(3x) 􀀔 1 for all x we have  \n1   sin(3x)  1  \n􀀀 􀀔 􀀔  \nx x x  \nsin(3x)  \nfor x > 0. Hence lim = 0 by the Squeeze theorem.  \nx!1 x  \nsin(x2 )  \n(2) lim ~~  ~~Ansxwe!r:0 x  \nsin(x2 ) sin(x2 ) sin(x2 ) sin(u)  \nlim ~~ ~~ = lim ~~ ~~ 􀀁 x = lim ~~ ~~ 􀀁 lim x = lim ~~ ~~ 􀀁 lim x = 1 􀀁 0 = 0 : x!0 x x!0 x2 x!0 x2 x!0 u!0 u x!0  \n(3) lim sin(x) 􀀀 sin(􀀙=3)) x!􀀙=3 x 􀀀 􀀙=3  \nAnswer: Let f (x) = sin(x) and a = 􀀙=3 . Then f 0 (x) = cos(x) and  \nsin(x) 􀀀 sin(􀀙=3)  f (x) 􀀀 f(a)) 0  1  \nlim = lim = f (a) = cos(􀀙=3) = :  \nx!􀀙=3 x 􀀀 􀀙=3 x!a x 􀀀 a 2  \n\n| III (1) Find f0 (x) and f 00 (x) if f(x) = sin(x3 􀀀 2) .\u003Cbr>Answer:\u003Cbr>f 0 (x) = 􀀀cos(x3 􀀀 2) 􀀁 3x2 ; f 00 (x) = 􀀀􀀀sin(x3 􀀀 2) 􀀁 􀀀3x2 􀀁 2 +􀀀cos(x3 􀀀 2) 􀀁 6x:\u003Cbr>(2) Find g0 (3) if h0 (9) = 17 and g(x) = h (x2 ) .\u003Cbr>Answer:\u003Cbr>g0 (x) = h0 (x2 )2x; g0 (3) = h0 (9)6 = 102 : |\n| --- |\n| IV Find the constant c which makes g continuous on ( 􀀀1 ; 1 ), g (x) = ( x2cx + 20 if xif x  4,\u003Cbr>4.\u003Cbr>Answer: By the limit laws, g (x) is continuous at any x  4.\u003Cbr>lim g (x) = lim x2 = 16 ; lim g (x) = lim cx + 20 = 4c + 20:\u003Cbr>x!4 􀀀 x!4 􀀀 x!4+ x!4+\u003Cbr>The function g(x) is continuous when these are equal, i.e. when c = 􀀀1.\u003Cbr>Answer: |\n\nV Consider the curve  \ny 2 + xy 􀀀 x2 = 11 :  \n(1) Find the equation of the tangent line to the curve at the point P (2; 3) .  \nAnswer: Di􀀋erentiate:  \n2y dydx + y + x dydx 􀀀 2x = 0 :  \nEvaluate at the point P (2; 3):  \n6 dy 􀀌 + 3 + 2 dy 􀀌 􀀀 4 = 0 :  \ndx􀀌 (x;y)=(2 ;3) dx􀀌 (x;y)=(2 ;3)  \nSolve:  \ndy 􀀌 1  \n~~ ~~ 􀀌 = ~~ ~~ :  \ndx􀀌 (x;y)=(2 ;3) 8  \nThis is the slope of the tangent line. The point P (2; 3) lies on the tangent line so the equation of the tangent line is  \n(x 􀀀 2)  \n(y 􀀀 3) = :  \n8  \nd2 y  \n(2) Find ~~ ~~ at the point P (2 3) .  \ndx2 ;  \nAnswer: Di􀀋erentiate again:  \n2 􀀔 􀀐 dydx 􀀑 2 + y d2~~ ~~ydx2 􀀕 + dydx + 􀀔 dydx + x d2~~ ~~ydx2 􀀕 􀀀 2 = 0 : Evaluate at (x; y) = (2; 3):  \n2 \" 􀀐 18 􀀑 2 + 3 d2~~ ~~ydx2 􀀌 (x;y)=(2 ;3)\\# + 18 + \" 18 + 2 d2~~ ~~ydx2 􀀌 (x;y)=(2 ;3)\\# 􀀀 2 = 0 :  \nSolve:  \nd2 y 􀀌 = 􀀀2 􀀀 ~~1~~8 􀀁 2 􀀀 ~~1~~8 􀀀 ~~1~~8 + 2 :  \ndx2 􀀌 (x;y)=(2 ;3) 6 + 2  \nAnswer:  \nVI Consider the function y = f (x) whose graph is shown below. Match the expression in the left column with the correct corresponding value in the right column.  \nf 0 (0) 􀀀6  \nf 0 (0:9) 0  \nf 0 (1) 3  \nf 0 (1:732) 0:6  \n-2 -1.5 -1 -0.5 0 0.5 1 1.5 2  \nAnswer: For positive x, the slope of the tangent line is decreasing as x increases, sof 0 (0) > f 0 (0:9) > f 0 (1) > f 0 (1:732) . The only possibility is f 0 (0) = 3, f 0 (0:9) = 0:6, f 0 (1) = 0, f 0 (1:732) = 􀀀6.  \nAnswer:  \nVII Prove the product rule (fg)0 = f 0g + fg0 using the de􀀌nition of the derivative, high school algebra, and the appropriate limit laws. Justify each step.  \nAnswer: Let w (x) = f (x)g (x) and assume that f and g are di􀀋erentiable.  \nw (x + h) 􀀀 w(x)  \nw0 (x) = lim (1)  \nh!0 h  \n= lim f (x + h)g (x + h) 􀀀 f(x)g (x) (2) h!0 h  \n= lim f (x + h) 􀀀 f(x)g (x + h) + f(x)g (x + h) 􀀀 g(x) (3)  \nh!0 h h  \n= lim f (x + h) 􀀀 f(x) lim g(x + h) + f(x) lim g (x + h) 􀀀 g(x) (4) h!0 h h!0 h!0 h  \n= f 0 (x)g (x) + f(x)g0 (x): (5)  \nReasons:  \n(1) De􀀌nition of w0 (x) .  \n(2) De􀀌nition of w(x) .  \n(3) High school algebra.  \n(4) Limit laws.  \n(5) De􀀌nition of f0 (x), continuity of g(x), and de􀀌nition of g0 (x) .  \nIn step (5) we used the fact that a di􀀋erentiable function is continuous. Here is the proof:  \n􀀐 lhi0 g (x + h)􀀑 􀀀 g(x) = lhi0 (g(x + h) 􀀀 g(x))  \n","cbCainTvzClVMVMb","https://ap.wps.com/l/cbCainTvzClVMVMb","pdf",254197,39,"English","# I Find the limit or show that it does not exist\n## (1) Limit with justification\n## (2) Limit with justification\n## (3) Limit with justification\n# II Find the limit or show that it does not exist\n## Trigonometric limits and squeeze arguments\n# III Differentiation: first/second derivatives and chain rule\n## (1) Derivatives for sin(x^3-2)\n## (2) Derivative via related functions\n# IV Continuity: find c for a piecewise function\n# V Implicit curve: tangent line and second derivative\n## (1) Tangent at P(2,3)\n## (2) Find d2y/dx2 at P(2,3)\n# VI Match f'(x) values from a graph\n# VII Prove the product rule\n# Calculus 221-Second Exam - Two Hours\n## Exponential derivative, logarithms\n## Radon-222 half-life and decay to 10%\n## Inverse functions and domains","[{\"question\":\"How are trigonometric limits evaluated in these solutions?\",\"answer\":\"Trigonometric limits are handled using bounding and substitution, then evaluated using the Squeeze theorem or continuity where appropriate.\"},{\"question\":\"What method is used to prove the product rule (fg)'?\",\"answer\":\"The proof starts from the definition of the derivative for w(x)=f(x)g(x), applies algebraic expansion, and uses limit laws plus the fact that differentiability implies continuity.\"},{\"question\":\"How is the constant c determined for continuity in the piecewise function?\",\"answer\":\"The solutions equate the left-hand and right-hand limits at the discontinuity point, using limit laws, and solve the resulting equation for c.\"}]","Calculus 221-First Exam - 50 Minutes - Answers | PDF",1790771099,98]