[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-detail-394184-en":3,"doc-seo-394184-105":31,"detail-sidebar-cat-0-en-105":98},{"code":4,"msg":5,"data":6},0,"success",{"doc_id":7,"user_id":8,"nickname":9,"user_avatar":10,"doc_module":4,"category_id":11,"category_name":12,"doc_title":13,"doc_description":14,"doc_content":15,"file_id":16,"file_url":17,"file_type":18,"file_size":19,"view_count":20,"is_deleted":4,"is_public":21,"is_downloadable":21,"audit_status":21,"page_count":22,"language":23,"language_code":24,"site_id":25,"html_lang":24,"table_of_contents":26,"faqs":27,"seo_title":28,"seo_description":14,"update_tm":29,"read_time":30,"preview_page_count":4},394184,5909890329169,"McQueen","https://ap-avatar.wpscdn.com/davatar_9964176cb1d06d4a9deccf72a44ae3dc",4,"Exam","Math 111 - Midterm I Practice Exam 1 Solutions","Math 111 Introduction to the Calculus Fall 2011 Midterm I Practice Exam 1 Solutions provides model solutions and detailed comments for five calculus problems. It includes graphing a transformed sine function using scaling and translation, determining where the function is continuous, computing a square-root limit via algebraic manipulation and continuity, proving a limit using the precise epsilon definition, and finding a derivative at a point from the derivative definition. It also guides sketching functions from a given derivative graph and explaining how vertical shifts preserve derivatives.","Math 111, Introduction to the Calculus, Fall 2011  \nMidterm I Practice Exam 1 Solutions  \nFor each question, there is a model solution (showing you the level of detail I expect on the exam) and then below that are further comments.  \n1. (a) Use scaling and translation to sketch a graph of the following function for 􀀀3􀀙 􀀔 x 􀀔 3􀀙  \nf (x) = sin 􀀐 x2 + 􀀙􀀑 :  \n(You should explain how you arrived at your answer. Note that it is not su􀀎cient to create a table of values to plot this graph.)  \n(b) At what points in R is this function continuous? (No explanation necessary.) Solution: (a) The graph of sin(x) is:  \nTranslating by 􀀙 units to the left we get the graph of sin(x + 􀀙):  \nStretching by a factor of 2 in the horizontal direction, we get the graph of sin 􀀀~~x~~2 + 􀀙 􀀁 :  \n(b) f (x) is continuous at all real numbers.  \nComments: It's easy to make the mistake here of 􀀌rst stretching, and then translating, because it seems like the formula ~~x~~2 + 􀀙 means we should deal with the ~~x~~2 and then the +􀀙 . But if you do that you get a di􀀋erent graph and plugging in some x-values you see that it is wrong.  \nThere are two ways to think about this. Firstly, when transforming the graph based on changing what is `inside' the sin(􀀀), everything is backwards. That is, dividing by 2 stretches the graph and adding 􀀙 moves the graph to the left, not the right. The order you do the transformations is also backwards to what you think. First you deal with the +􀀙 and then the dividing by 2 .  \nSecondly, think about what you do to the formula when you do each step. In particular, stretching horizontally by a factor of 2 corresponds to replacing x with ~~x~~2 , and translating left by 􀀙 units correspond to replacing x with x + 􀀙 . Therefore, if you 􀀌rst stretch, you get the graph of sin( ~~x~~2), but when you then translate, you get the graph of sin( ~~x+~~2~~􀀙~~ ) which is not what you want.  \n2. Calculate the following limit: px2 + 1 􀀀p5  \nlim ~~ ~~ :  \nx!2 x 􀀀 2  \nSolution: For x  2 we have:  \npx2 + 1 􀀀p 5 ( px2 + 1 􀀀p 5)( px2 + 1 +p 5) ~~ ~~ = ~~      ~~x 􀀀 2 (x 􀀀 2)( px2 + 1 +p 5)  \n(x2 􀀀 4)  \n= ~~     ~~(x 􀀀 2)( px2 + 1 +p 5)  x + 2   \n=     px2 + 1 +p 5  \nThis function is continuous at x = 2, so:  \npx2 + 1 􀀀p5 x + 2  \nlim ~~ ~~ = lim ~~    ~~  \nx!2 x 􀀀 2 x!2 px2 + 1 +p 5 =  2 + 2   \np 22 + 1 +p 5  \n=  4   \n2 p 5  \n 2   \n=    \np 5  \nComments: This is a standard trick for dealing with square-roots: make sure you know it. Ultimately it relies on the fact that there are two di􀀋erent ways to factor x2 􀀀 4. The more familiar one is  \nx2 􀀀 4 = (x 􀀀 2)(x + 2)  \nbut we also have  \nx2 􀀀 4 = (x2 + 1) 􀀀 5 = ( px2 + 1 􀀀 p5)( px2 + 1 +p5):  \nYou can imagine that there are lots of other questions that work the same way.  \nAlso, you should say that you are using the fact that the function is continuous when you calculate the limit by substituting in x = 2 .  \n3. Prove, using the precise de􀀌nition of limit, that:  \nlim(1 􀀀 5x) = 􀀀4:  \nx! 1  \nSolution: Scratch work (you do not *need* to write this but I recommend it):  \n􀀏 > j(1 􀀀 5x) 􀀀 (􀀀4)j = j5 􀀀 5xj = 5jx 􀀀 1j  \nso 􀀎 = 􀀏=5.  \nProof: Given 􀀏 > 0, let 􀀎 = 􀀏=5 > 0. Then, if 0 \u003C jx 􀀀 1j \u003C 􀀎, we have  \n5jx 􀀀 1j \u003C 􀀏  \nand so  \nj5x 􀀀 5j \u003C 􀀏  \nand so  \nj5 􀀀 5xj \u003C 􀀏  \nand so  \nj(1 􀀀 5x) 􀀀 (􀀀4)j \u003C 􀀏  \nas required.  \nComments: The most common error here is to write j5􀀀5xj = j (􀀀5)(x􀀀1)j = (􀀀5)jx􀀀1jand then to try to divide by 􀀀5 to get 􀀎 = 5 . There are two problems with that: (i) 􀀎 always has to be positive, so you cannot have 􀀎 = 􀀏= 􀀀 5 as that would be negative (since 􀀏 is positive), and (ii) it is not true that  \nj (􀀀5)(x 􀀀 1)j = (􀀀5)jx 􀀀 1j :  \nThe left-hand side is positive (since the absolute value of anything is positive) but the right hand-side is negative. What is true instead is that  \nj (􀀀5)(x 􀀀 1)j = 5jx 􀀀 1j :  \n(More generally, jabj = jajj bj .) Therefore, the right thing to get is 􀀎 = ~~􀀏~~5 . This also avoids the problem that if you divide by a negative number, the direction on the inequality would reve","cbCaiepbPZZv27rf","https://ap.wps.com/l/cbCaiepbPZZv27rf","pdf",145388,19,1,6,"English","en",105,"# Midterm I Practice Exam 1 Solutions\n## Problem 1: Graphing and continuity\n## Problem 2: Limit using algebraic trick\n## Problem 3: Epsilon definition of limit\n## Problem 4: Derivative from definition\n## Problem 5: Reconstructing a function from f'","[{\"question\":\"How should you sketch the graph when f(x)=sin(x^2+?) uses scaling and translation?\",\"answer\":\"Translate first in the correct order for “inside the sin” changes, then apply the horizontal stretch via the replacement x→x^2 pattern. The comments warn that doing stretch and translation in the wrong order gives an incorrect graph.\"},{\"question\":\"At which real numbers is the transformed function continuous?\",\"answer\":\"The solution states that f(x) is continuous at all real numbers.\"},{\"question\":\"How is the limit near x=2 computed in Problem 2?\",\"answer\":\"Factor x^2−4 appropriately to cancel expressions, then use continuity to substitute x=2 into the remaining continuous expression. The solution ultimately evaluates the limit to 4.\"}]","Math 111 - Midterm I Practice Exam 1 Solutions | PDF",1790399862,15,{"code":4,"msg":32,"data":33},"ok",{"site_id":25,"language":24,"slug":34,"title":13,"keywords":35,"description":14,"schema_data":36,"social_meta":92,"head_meta":94,"extra_data":96,"updated_unix":97},"math-111-midterm-i-practice-exam-1-solutions","",{"@graph":37,"@context":91},[38,54,74],{"@type":39,"itemListElement":40},"BreadcrumbList",[41,45,49,52],{"item":42,"name":43,"@type":44,"position":21},"https://docshare.wps.com","Home","ListItem",{"item":46,"name":47,"@type":44,"position":48},"https://docshare.wps.com/document/","Document",2,{"item":50,"name":12,"@type":44,"position":51},"https://docshare.wps.com/document/exam/",3,{"item":53,"name":13,"@type":44,"position":11},"https://docshare.wps.com/document/math-111-midterm-i-practice-exam-1-solutions/394184/",{"url":53,"name":13,"@type":55,"image":56,"author":61,"headline":13,"publisher":63,"fileFormat":66,"inLanguage":24,"description":14,"dateModified":67,"datePublished":68,"encodingFormat":66,"isAccessibleForFree":69,"interactionStatistic":70},"DigitalDocument",{"url":57,"@type":58,"width":59,"height":60},"https://docshare.wps.com/thumbnails/math-111-midterm-i-practice-exam-1-solutions/394184.png","ImageObject",300,407,{"name":9,"@type":62},"Person",{"url":42,"name":64,"@type":65},"DocShare","Organization","application/pdf","2026-11-08","2026-09-26",true,{"@type":71,"interactionType":72,"userInteractionCount":20},"InteractionCounter",{"@type":73},"ViewAction",{"@type":75,"mainEntity":76},"FAQPage",[77,83,87],{"name":78,"@type":79,"acceptedAnswer":80},"How should you sketch the graph when f(x)=sin(x^2+?) uses scaling and translation?","Question",{"text":81,"@type":82},"Translate first in the correct order for “inside the sin” changes, then apply the horizontal stretch via the replacement x→x^2 pattern. The comments warn that doing stretch and translation in the wrong order gives an incorrect graph.","Answer",{"name":84,"@type":79,"acceptedAnswer":85},"At which real numbers is the transformed function continuous?",{"text":86,"@type":82},"The solution states that f(x) is continuous at all real numbers.",{"name":88,"@type":79,"acceptedAnswer":89},"How is the limit near x=2 computed in Problem 2?",{"text":90,"@type":82},"Factor x^2−4 appropriately to cancel expressions, then use continuity to substitute x=2 into the remaining continuous expression. 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