[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"detail-sidebar-cat-0-en-105":3,"doc-seo-466336-105":59,"doc-detail-466336-en":131},{"code":4,"msg":5,"data":6},0,"success",[7,13,18,23,28,33,38,43,48,51,55],{"id":8,"doc_module":4,"doc_module_name":9,"category_name":10,"show_sort_weight":11,"slug":12},1,"Document","Story & Novel",90,"story-novel",{"id":14,"doc_module":4,"doc_module_name":9,"category_name":15,"show_sort_weight":16,"slug":17},2,"Literature",80,"literature",{"id":19,"doc_module":4,"doc_module_name":9,"category_name":20,"show_sort_weight":21,"slug":22},4,"Exam",70,"exam",{"id":24,"doc_module":4,"doc_module_name":9,"category_name":25,"show_sort_weight":26,"slug":27},5,"Comic",60,"comic",{"id":29,"doc_module":4,"doc_module_name":9,"category_name":30,"show_sort_weight":31,"slug":32},6,"Technology",50,"technology",{"id":34,"doc_module":4,"doc_module_name":9,"category_name":35,"show_sort_weight":36,"slug":37},7,"Healthcare",40,"healthcare",{"id":39,"doc_module":4,"doc_module_name":9,"category_name":40,"show_sort_weight":41,"slug":42},8,"Research & Report",30,"research-report",{"id":44,"doc_module":4,"doc_module_name":9,"category_name":45,"show_sort_weight":46,"slug":47},9,"Religion & Spirituality",20,"religion-spirituality",{"id":46,"doc_module":4,"doc_module_name":9,"category_name":49,"show_sort_weight":46,"slug":50},"World Cup","world-cup",{"id":52,"doc_module":4,"doc_module_name":9,"category_name":53,"show_sort_weight":52,"slug":54},10,"Lifestyle","lifestyle",{"id":56,"doc_module":4,"doc_module_name":9,"category_name":57,"show_sort_weight":24,"slug":58},19,"General","general",{"code":4,"msg":60,"data":61},"ok",{"site_id":62,"language":63,"slug":64,"title":65,"keywords":66,"description":67,"schema_data":68,"social_meta":124,"head_meta":126,"extra_data":128,"updated_unix":130},105,"en","calculus-final-exam-study-guide-solutions","Calculus Final Exam Study Guide - Solutions","","Calculus final exam study guide solutions covering seven-plus problem types: evaluating limits (including indeterminate forms via L’Hopital’s Rule), finding the tangent line to a given function at a specified point using derivative and point-slope form, analyzing when a particle speeds up by comparing velocity and acceleration sign changes on a time interval, and verifying which derivative-based limit statements are true/false for a sine expression. Additional solutions include absolute extrema on a closed interval, simplifying radical limits at infinity, implicit differentiation to compute a second derivative at a point, and piecewise differentiability via continuity and matching derivatives at the join.",{"@graph":69,"@context":123},[70,84,106],{"@type":71,"itemListElement":72},"BreadcrumbList",[73,77,79,82],{"item":74,"name":75,"@type":76,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":78,"name":9,"@type":76,"position":14},"https://docshare.wps.com/document/",{"item":80,"name":20,"@type":76,"position":81},"https://docshare.wps.com/document/exam/",3,{"item":83,"name":65,"@type":76,"position":19},"https://docshare.wps.com/document/calculus-final-exam-study-guide-solutions/466336/",{"url":83,"name":65,"@type":85,"image":86,"author":91,"headline":65,"publisher":94,"fileFormat":97,"inLanguage":63,"description":67,"dateModified":98,"datePublished":99,"encodingFormat":97,"isAccessibleForFree":100,"interactionStatistic":101},"DigitalDocument",{"url":87,"@type":88,"width":89,"height":90},"https://docshare.wps.com/thumbnails/calculus-final-exam-study-guide-solutions/466336.png","ImageObject",300,407,{"name":92,"@type":93},"Dipper","Person",{"url":74,"name":95,"@type":96},"DocShare","Organization","application/pdf","2026-10-31","2026-09-30",true,{"@type":102,"interactionType":103,"userInteractionCount":105},"InteractionCounter",{"@type":104},"ViewAction",12,{"@type":107,"mainEntity":108},"FAQPage",[109,115,119],{"name":110,"@type":111,"acceptedAnswer":112},"How do you evaluate the indeterminate limit in Problem 1?","Question",{"text":113,"@type":114},"Substitute the target value to confirm the indeterminate form, then apply L’Hopital’s Rule by differentiating the numerator and denominator separately before plugging in again.","Answer",{"name":116,"@type":111,"acceptedAnswer":117},"How do you find the tangent line in Problem 2?",{"text":118,"@type":114},"Compute the function value at x=0 for the point, find the derivative using the chain rule to get the slope at x=0, then use the line equation y = m(x-0) + y0.",{"name":120,"@type":111,"acceptedAnswer":121},"When is the particle speeding up in Problem 3?",{"text":122,"@type":114},"The particle speeds up when velocity and acceleration have the same sign. The solution finds where each changes sign by setting the velocity and acceleration expressions equal to zero and then checking intervals within 0 to 5.","https://schema.org",{"og:url":83,"og:type":125,"og:title":65,"og:site_name":95,"og:description":67},"article",{"robots":127,"canonical":83},"index,follow",{"doc_id":129,"site_id":62},466336,1790884550,{"code":4,"msg":5,"data":132},{"doc_id":129,"user_id":133,"nickname":92,"user_avatar":134,"doc_module":4,"category_id":19,"category_name":20,"doc_title":65,"doc_description":67,"doc_content":135,"file_id":136,"file_url":137,"file_type":138,"file_size":139,"view_count":105,"is_deleted":4,"is_public":8,"is_downloadable":8,"audit_status":8,"page_count":140,"language":141,"language_code":63,"site_id":62,"html_lang":63,"table_of_contents":142,"faqs":143,"seo_title":144,"seo_description":67,"update_tm":145,"read_time":146},1374404997633,"https://ap-avatar.wpscdn.com/davatar_a8503ba1806abce46bf441b54a3ca4cd","Calculus Final Exam Study Guide  \n1. (No Calc) What is the value of lim 􀯫 2 −3􀯫−4?  \n􀯫→−1 􀯫 2 −1  \nStart by plugging in:  \n(−1)2 − 3 (−1) − 4 1 + 3 − 4 0  \n=  =  =   \n(−1)2 − 1 1 − 1 0  \nWe have the indeterminate form, we need to use L’Hopital’s Rule. Take the derivative ofthe top and bottom (separately), then try plugging in again.  \n2􀝔 − 3 2 (−1) − 3 −2 − 3 −5 5 = lim  =  =  =   =   \n􀯫→−1 2􀝔 2 (−1) −2 −2 2  \n2. (No Calc) What is the equation of the line tangent to 􀝂(􀝔) = 3􀝔 − 5 cos 2􀝔 at 􀝔 = 0?  \nWe will use the form: 􀝕 = 􀝉 (􀝔 − 􀝔1) + 􀝕1. We have our 􀝔1, we only need 􀝉 and 􀝕1.  \nTo get 􀝕1, plug into the original equation:  \n􀝂 (0) = 3 (0) − 5 cos(2 ∙ 0) = 0 − 5 cos 0 = −5  \nTo get 􀝉, plug into the derivative (remember to use the chain rule!):  \n􀝂 ′(􀝔) = 3 + 10 sin 2􀝔  \n􀝂 ′(0) = 3 + 10 sin(2 ∙ 0)  \n= 3  \nAnd now we just plug into the equation ofthe line:  \n􀝕 = 3 (􀝔 − 0) − 5  \n3. (No Calc) A particle moves along the 􀝕 −axis so that its position at any time 􀝐, for 0 ≤ 􀝐 ≤ 5 , is given by 􀝕(􀝐) = 􀝐 4 − 18􀝐2. In which interval(s) is the particle speeding up? (Hint: Remember, speed and velocity don’t mean the same for derivatives. )  \nThe particle will be speeding up when the velocity and the acceleration have the same sign. We need to see where they each change sign.  \nLet’s get the velocity first and find the intervals we’ll be looking at:  \n􀝒 (􀝐) = 4􀝐3 − 36􀝐  \nSetting this equal to zero, we get:  \n0 = 4􀝐(􀝐2 − 9)  \n􀝐 = 0, ±3  \nNow, let’s look at the acceleration:  \n􀜽 (􀝐) = 12􀝐2 − 36  \n0 = 12 (􀝐2 − 3)  \n􀝐 = ±√3  \nRemember, they told us to use the interval from 0 to 5, so the only ones that matter here are 3, and √3. The rest ofthe numbers we got are not in the interval.  \nSo, our intervals are: From 0 to √3, from √3 to 3, and from 3 to 5. I’m going to pick numbers from each of these intervals to see what’s going on with velocity and acceleration. I’ll make a table so it’s easier to see:  \n\n| Interval | My Test Point | Velocity | Acceleration | Matching? |\n| --- | --- | --- | --- | --- |\n| 0 \u003C 􀝔 \u003C √3 | 􀝔 = 1 | 4 (1)3 − 36 (1)= −32 | 12 (1)2 − 36\u003Cbr>= −24 | Yes, both\u003Cbr>negative. |\n| √3 \u003C 􀝔 \u003C 3 | 􀝔 = 2 | 4 (2)3 − 36 (2)= −40 | 12 (2)2 − 36\u003Cbr>= 12 | No, 􀝒 \u003C 0 and 􀜽 > 0 |\n| 3 \u003C 􀝔 \u003C 5 | 􀝔 = 4 | 4 (4)3 − 36 (4)= 112 | 12 (4)2 − 36\u003Cbr>= 156 | Yes, both\u003Cbr>positive. |\n\nSo, the particle is speeding up from 0 to √3 and from 3 to 5.  \n4. (No Calc) Which of the following statements is (are) false for 􀝂(􀝔) = 􀝁 􀯫 sin 􀝔?  \nI. lim 􀝂(􀝔) = 0 II. lim 􀝂′(􀝔) = 1 III. lim 􀝂′′(􀝔) = 2 􀯫→0 􀯫→0 􀯫→0  \nLet’s look at each of these individually.  \nI. lim 􀝁 􀯫 sin 􀝔 = 􀝁 0 sin 0 = 0  \n􀯫→0  \nSo, the first one is true.  \nII. lim (􀝁􀯫 sin 􀝔 + 􀝁􀯫 cos 􀝔) = 􀝁 0 sin 0 + 􀝁 0 cos 0 = 1  \n􀯫→0  \nThis one is also true.  \nIII. lim (􀝁􀯫 sin 􀝔 + 􀝁􀯫 cos 􀝔 + 􀝁􀯫 cos 􀝔 − 􀝁􀯫 sin 􀝔) = 0 + 1 + 1 − 0 = 2  \n􀯫→0  \nThis last one is also true.  \nAll three are true. Be careful when you’re taking the derivative here – did you use the product rule?  \n5. (No Calc) Let 􀝂(􀝔) = (3 + 2􀝔 − 􀝔2)3 be defined for the closed interval −2 ≤ 􀝔 ≤ 3. If 􀜯 is the  \n􀝕 −coordinate of the absolute maximum and 􀝉 is the 􀝕 −coordinate of the absolute minimum, what is |􀜯 + 􀝉|?  \nWe need to get 􀜯 and 􀝉 separately. At the very end, we’ll add them together. Be careful, they’re asking for the 􀝕 −value, not the 􀝔 −value!  \nTake the derivative and set it equal to zero:  \n􀝂 ′(􀝔) = 3 (3 + 2􀝔 − 􀝔2 )2 (2 − 2􀝔)  \n0 = 3 (3 + 2􀝔 − 􀝔2 )2 (2 − 2􀝔)  \nWe need to consider:  \n0 = 3 + 2􀝔 − 􀝔 2  \nand  \n0 = 2 − 2􀝔  \nFor the first one, if we rearrange it and then factor:  \n0 = −􀝔2 + 2􀝔 + 3  \n0 = 􀝔 2 − 2􀝔 − 3  \n0 = (􀝔 − 3)(􀝔 + 1)  \n􀝔 = 3, 􀝔 = −1  \nFrom the other factor, we get:  \n􀝔 = 1  \nRemember, when we’re looking for ABSOLUTE extrema, we also need to consider the interval ends. Notice that the 3 we got is the same as the end ofthe interval.  \n\n| 􀝔 −value | 􀝕 −value |\n| --- | --- |\n| −2 | (3 + 2(−2) − (−2)2)3 = (3 − 4 − 4)3\u003Cbr>= (−5)3\u003Cbr>= −125 |\n| 3 | (3 + 2(3) − (3)2)3 = (3 + 6 − 9)3\u003Cbr>= 0 |\n| −1 | (3 + 2(−1) − (−1)2)3 = (3 − 2 − 1)3\u003Cbr","cbCaib5j1oN2OGGs","https://ap.wps.com/l/cbCaib5j1oN2OGGs","pdf",677259,18,"English","# Problem 1: Limit using L’Hopital’s Rule\n# Problem 2: Tangent line at x = 0\n# Problem 3: Intervals of speeding up\n# Problem 4: True/false limit statements\n# Problem 5: Absolute max/min on a closed interval\n# Problem 6: Radical limit as x → ∞\n# Problem 7: Implicit differentiation for second derivative\n# Problem 8: Differentiability of a piecewise function","[{\"question\":\"How do you evaluate the indeterminate limit in Problem 1?\",\"answer\":\"Substitute the target value to confirm the indeterminate form, then apply L’Hopital’s Rule by differentiating the numerator and denominator separately before plugging in again.\"},{\"question\":\"How do you find the tangent line in Problem 2?\",\"answer\":\"Compute the function value at x=0 for the point, find the derivative using the chain rule to get the slope at x=0, then use the line equation y = m(x-0) + y0.\"},{\"question\":\"When is the particle speeding up in Problem 3?\",\"answer\":\"The particle speeds up when velocity and acceleration have the same sign. The solution finds where each changes sign by setting the velocity and acceleration expressions equal to zero and then checking intervals within 0 to 5.\"}]","Calculus Final Exam Study Guide - Solutions | PDF",1790771086,45]