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The document states constraints on a twice-differentiable function f, including f'(1)=0 and f''(x)=x^2 cos(x^2+π) on -1≤x≤3, then answers four parts (a)–(d). It determines where f is concave up using the sign of f''(x), applies the Second Derivative Test at x=1, uses the Mean Value Theorem to rule out a possible value of f'(-1), and concludes whether an inflection point occurs at x=0 based on concavity behavior.","AP Calculus Practice Test  \n# Unit 5 FRQ Solution\n\nA GRAPHING CALCULATOR IS REQUIRED FOR THIS QUESTION.  \nYou are permitted to use your calculator to solve an equation,find the derivative of a function at a point,or calculate the value of adefinite integral.However,you must clearly indicate the setup of your question,namely the equation,function,or integral you areusing.If you use other built-in features or programs,you must show the mathematical steps necessary to produce your results.Yourwork must be expressed in standard mathematical notation rather than calculator syntax.  \nShowall of your work,even though the question may not explicitly remind you to do so.Clearly label any functions,graphs,tables,orother objects that you use.Justifications require that you give mathematical reasons,and that you verify the needed conditions underwhich relevant theorems,properties,definitions,or tests are applied.Your work wil be scored on the correctness and completenessof your methods as well as your answers.Answers without supporting work will usually not receive credit.  \nUnless otherwise specified,answers(numeric or algebraic)need not be simplifed.If your answer is given as a decimalapproximation,it should be correct to three places after the decimal point.  \nUnless otherwise specified,the domain of a function fis assumed to be the set of all eal numbers a for which f(x)is a realnumber.  \nLet f be a twice-differentiable function such that f'(1)=0.The second derivative of fis given by f\"(z)=x²cos(x²+π)for  \n-1≤x≤3.  \n(a)On what open intervals contained in-1\u003Cz\u003C3is the graph of f concave up?Give a reason for your answer.  \n(b)Does f have a relative minimum,a relative maximum,or neither at x=1?Justify your answer.  \n(c)Use the Mean Value Theorem on the closed interval[-1,1]to show that f'(-1)cannot equal 2.5.  \n(d)Does the graph of f have a point of inflection at z=0?Give a reason for your answer.  \nAP Calculus Practice Test  \n# SOLUTION:\n\n## Part A\n\nA maximum of 1 out of 2 points is earned for one correct interval with reason and no incorrect intervals.Select a point value to view scoring criteria,solutions,and/or examples to score the response.  \nThe student response accurately includes both of the criteria below.  \n□ intervals  \n□ reason  \nSolution:  \nThe graph of fis concave up on(1.253,2.170(or 2.171)and(2.802,3)because f\"(z)is positive on those intervals.  \n## Part B\n\nNote:Sign charts are a useful tool to investigate and summarize the behavior of a function.Byitself a sign chart for f'(z)or f\"(z)is nota sufficient response for a justification.  \nSelect a point value to view scoring criteria,solutions,and/or examples and to score the response.  \n2  \n0  \nThe student response accurately includes both of the criteria below.  \n□ answer  \n□ justification  \nSolution:  \nBecausef'(1)=0and f\"(1)=-0.540302\u003C0,by the Second Derivative Test fhas a relative maximum at z=1.  \nAP Calculus Practice Test  \nPart B  \nPart C  \nPart D  \nPart A  \nThe first point requires use of average rate of change of f'on[-1,1].  \nOne point is earned for using MVTto conclude that there would be a number cin the interval (-1,1)such that f\"(c)=-1.25.Two points are earned for showing that this is not possible.  \nSelect a point value to view scoring criteria,solutions,and/or examples and to score the response.  \n1  \n2  \nThe student response does not accurately include any of the criteria below.  \n□ considers average rate of change  \n□ conclusion using Mean Value Theorem  \nSolution:  \nIff'(-1)=2.5,then the average rate of change of f'on the closed interval[-1,1]would be  \nSince f'is continuous on[-1,1]and differentiable on(-1,1),by the Mean Value Theorem there would be a number cin the  \ninterval (-1,1)such that f\"(c)=-1.25.  \nHowever,since f\"(z)=z²cos(z²+π),itfollows that-1≤f\"(z)≤1on the interval (-1,1).Therefore,there is no number  \ncin the interval(-1,1)such that f\"(c)=-1.25.  \nThus,it is not possible that f'(-1)=2.5.  \nPart D  \nNote:Sign charts are a useful tool t invest","cbCaiagt9ek7eCCK","https://ap.wps.com/l/cbCaiagt9ek7eCCK","pdf",270221,3,"English","en",105,"# Unit 5 FRQ Solution\n## Part A\n## Part B\n## Part C\n## Part D","[{\"question\":\"How is concavity on the open intervals determined for Part (a)?\",\"answer\":\"The solution finds where f''(x) is positive on -1\\u003cx\\u003c3, since f is concave up exactly on those intervals.\"},{\"question\":\"What does the Second Derivative Test show at x=1 in Part (b)?\",\"answer\":\"Because f'(1)=0 and f''(1)=-0.540302\\u003c0, the solution concludes that f has a relative maximum at x=1.\"},{\"question\":\"How does Part (c) use the Mean Value Theorem to restrict f'(-1)?\",\"answer\":\"Assuming f'(-1)=2.5 leads to an average-rate-of-change value that, via the Mean Value Theorem, would require a point c in (-1,1) with f''(c)=-1.25. The solution then shows this cannot happen because f''(z)=z^2 cos(z^2+π) stays between -1 and 1 on (-1,1).\"},{\"question\":\"Why does Part (d) conclude whether there is a point of inflection at x=0?\",\"answer\":\"The solution checks the sign behavior of f''(z) near x=0 and concludes there is no inflection point because f''(z) does not change sign at x=0.\"}]","Unit 5 FRQ Solution | PDF",1790747733,{"code":4,"msg":81,"data":82},"ok",{"site_id":75,"language":74,"slug":83,"title":65,"keywords":84,"description":66,"schema_data":85,"social_meta":142,"head_meta":144,"extra_data":146,"updated_unix":147},"unit-5-frq-solution","",{"@graph":86,"@context":141},[87,100,120],{"@type":88,"itemListElement":89},"BreadcrumbList",[90,94,96,98],{"item":91,"name":92,"@type":93,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":95,"name":9,"@type":93,"position":14},"https://docshare.wps.com/document/",{"item":97,"name":20,"@type":93,"position":72},"https://docshare.wps.com/document/exam/",{"item":99,"name":65,"@type":93,"position":19},"https://docshare.wps.com/document/unit-5-frq-solution/456959/",{"url":99,"name":65,"@type":101,"image":102,"author":107,"headline":65,"publisher":109,"fileFormat":112,"inLanguage":74,"description":66,"dateModified":113,"datePublished":114,"encodingFormat":112,"isAccessibleForFree":115,"interactionStatistic":116},"DigitalDocument",{"url":103,"@type":104,"width":105,"height":106},"https://docshare.wps.com/thumbnails/unit-5-frq-solution/456959.png","ImageObject",300,407,{"name":63,"@type":108},"Person",{"url":91,"name":110,"@type":111},"DocShare","Organization","application/pdf","2026-10-19","2026-09-30",true,{"@type":117,"interactionType":118,"userInteractionCount":52},"InteractionCounter",{"@type":119},"ViewAction",{"@type":121,"mainEntity":122},"FAQPage",[123,129,133,137],{"name":124,"@type":125,"acceptedAnswer":126},"How is concavity on the open intervals determined for Part (a)?","Question",{"text":127,"@type":128},"The solution finds where f''(x) is positive on -1\u003Cx\u003C3, since f is concave up exactly on those intervals.","Answer",{"name":130,"@type":125,"acceptedAnswer":131},"What does the Second Derivative Test show at x=1 in Part (b)?",{"text":132,"@type":128},"Because f'(1)=0 and f''(1)=-0.540302\u003C0, the solution concludes that f has a relative maximum at x=1.",{"name":134,"@type":125,"acceptedAnswer":135},"How does Part (c) use the Mean Value Theorem to restrict f'(-1)?",{"text":136,"@type":128},"Assuming f'(-1)=2.5 leads to an average-rate-of-change value that, via the Mean Value Theorem, would require a point c in (-1,1) with f''(c)=-1.25. The solution then shows this cannot happen because f''(z)=z^2 cos(z^2+π) stays between -1 and 1 on (-1,1).",{"name":138,"@type":125,"acceptedAnswer":139},"Why does Part (d) conclude whether there is a point of inflection at x=0?",{"text":140,"@type":128},"The solution checks the sign behavior of f''(z) near x=0 and concludes there is no inflection point because f''(z) does not change sign at x=0.","https://schema.org",{"og:url":99,"og:type":143,"og:title":65,"og:site_name":110,"og:description":66},"article",{"robots":145,"canonical":99},"index,follow",{"doc_id":61,"site_id":75},1790825309]