Mathematics 19A; Fall 2001; V. Ginzburg Practice Final Solutions


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1 Mthemtics 9A; Fll 200; V Ginzburg Prctice Finl Solutions For ech of the ten questions below, stte whether the ssertion is true or flse ) Let fx) be continuous t x Then x fx) f) Answer: T b) Let f be differentible function nd f c) 0 Then fx) necessrily hs locl mximum or locl minimum t x c Answer: F: Counterexmple: fx) x 3, c 0 c) Let fx) x Then f x) x x Answer: F: f x) ln ) x d) Let fx) ln x Then f x) /x Answer: T e) fx) x gx) x fx) x gx), provided tht the its exist nd x gx) 0 Answer: T f) Assume tht f c) 0 nd f c) > 0 Then y fx) hs locl mximum t x c Answer: F The function hs locl minimum t x c g) The function fx) is continuous t x Answer: T { x 2 for x <, x 2 4 for x h) The function fx) x is differentible t x 0 Answer: F i) Assume tht fx) is continuous on [, b] nd differentible on, b) nd f) fb) Then there exists number c in, b) such tht f c) 0 Answer: T This is Rolle s theorem j) Let fx) nd gx) be differentible functions nd g ) 0 Then, by L Hospitl s rule, one necessrily hs tht fx) x gx) f x) x g x) Answer: F: For L Hospitl s rule to pply, x fx) gx) must be n indetermincy of the form 0/0 or /
2 2 2 Find the following its ) t+9 3 t 0 t Solution: t t t t t t t + 9) t t ) t t t ), t for t 0 Hence, t 0 t t t 0 t Alterntively, one cn use L Hospitl s rule b) x ln x 3 x Solution: We hve ln x nd 3 x s x Hence, L Hospitl s rule pplies By L Hospitl s rule, ln x ln x) x 3 x x 3 x) x x 3x 2/3 3x 2/3 x x x 3 3 x 0 cos x c) x 0 We hve cos x 0 nd x 2 0 s x 0 Hence, L Hospitl s x 2 rule pplies By L Hospitl s rule, cos x cos x) sin x x 0 x 2 x 0 x 2 ) x 0 2x Now we pply L Hospitl s rule gin: sin x x 0 2x sin x) x 0 2x) cos x cos 0 x d) x x) tnh x Solution: This is n indetermincy of the type 0 Write it s x) tnh x x tnh x x x), which is n indetermincy of the type 0/0, nd hence L Hospitl s rule pplies
3 3 Thus, x tnh x x) x 3 Find f x) for the following functions ) fx) x2 Solution: +x 2 [tnh x] [ x) ] x 2 ) x x) 2 x) 2 x x 2 x x + x + 0 f x) x2 ) + x 2 ) x 2 + x 2 ) + x 2 ) 2 2x + x2 ) x 2 2x) + x 2 ) 2 2x + 2x3 2x 3 + x 2 ) 2 2x + x 2 ) 2 b) fx) sin ) ln x x Solution: By the chin rule: ) ) ln x ln x f x) cos x x By the quotient rule: ) ln x ln x) x x) ln x Thus, x x 2 x x ln x x 2 ln x x 2 ) ln x f x) cos ln x x x 2 c) fx) x+)2 x Solution: Use logrithmic differentition: 2 +2x fx) x + )2 x x + 2 nd so ln fx) 2 lnx + ) 2 ln x lnx + 2) 2
4 4 Thus nd f x) fx)[ln fx)] [ln fx)] 2 x + 2x 2x + 2) x + )2 2 x 2 + 2x x + ) 2x 2x + 2) d) fx) x tn x 2 Solution: By the product rule: f x) x ) tn x 2 + x tn x 2) tn x 2 Furthermore, by the chin rule, Thus, tn x 2) + x 2 ) 2 x 2) 2x + x 4 f x) tn x 2 x x 4 x 2 + x tn x 2) 4 Find the eqution of the tngent line to the curve x 2 + xy y 2 t the point 2, 3) Solution: Step Using implicit differentition, d dx x2 + d dx xy) d dx y2 0, 2x + y + x dy dy 2y dx dx 0 Now let us plug in the vlues x 2 nd y 3 We get: dy dx 6 dy dx 7 4 dy dx 0 Thus dy dx 7/4 This is the slope m of the tngent Step 2 The eqution of the tngent through the point, b) is y mx + b m Plugging 2 nd b 3, we obtin 5 Let fx) 2x 3 + 3x 2 2x + 7 ) Find f x) nd f x) Solution: nd y 7x x 4 2 f x) 6x 2 + 6x 2 f x) 2x + 6 For wht follows it s lso useful to note tht f x) 6x )x + 2)
5 5 b) Find the locl mxim nd minim of f Solution: Locl mxim nd minim my occur only when f x) 0, ie, f x) 6x 2 + 6x 2 0 which hs solutions 2 nd Then f 2) 8 < 0 nd f ) 8 > 0 By the second derivtive test, f hs locl mximum t x 2 equl to f 2) 27 nd locl minimum t x equl to f) 0 c) Find the intervls of increse nd decrese for f Solution: The intervls of increse re given by the inequlity f x) 6x 2 + 6x 2 > 0 Hence, the intervls of increse re, 2) nd, ) The intervls of decrese re given by the inequlity f x) 6x 2 + 6x 2 < 0 Therefore, the intervl of decrese is 2, ) d) Find the inflection points of f Solution: The inflection points re given by the eqution f x) 0: f x) 2x This eqution hs one solution /2 Thus the inflection point is /2 e) Find the intervls of concvity of f Solution: The intervls of concvity upwrd re given by the inequlity f x) 2x + 6 > 0 Hence, the intervl of concvity upwrd is /2, ) The intervls of concvity downwrd re given by the inequlity f x) 2x + 6 < 0 Hence, the intervl of concvity downwrd is, /2) 6 Find the bsolute mximum nd the bsolute minimum of fx) xe x2 /2 on [0, 2] Solution: Let us first find the criticl numbers of fx), ie, solutions to f x) 0: The solutions to ) f x) x) e x2 /2 + x e x2 /2 e x 2 /2 + xe x2 /2 x 2 /2 ) e x2 /2 x 2 e x2 /2 e x2 /2 x 2 ) f x) e x2 /2 x 2 ) 0 or, equivlently, x 2 0
6 6 re x ± Only one of these numbers, nmely x, is in the intervl [0, 2] To find the bsolute mximum nd the bsolute minimum we need to pick the gretest nd the smllest vlue of fx) mong its vlues t the end points of the intervl nd the criticl numbers within the intervl: f0) 0, f) e /2, criticl number) f2) 2e 2 Note tht 0 < 2e 2 < e /2 Thus the bsolute mximum is f) e /2 nd the bsolute minimum is f0) 0 7 Find the re of the lrgest rectngle tht cn be inscribed in right tringle with legs of length nd b if two sides of the rectngle lie long the legs Solution: Let x be the length of the side long the leg of length nd y be the length of the side long the leg of length b Then b y b x Solving this eqution for y, we obtin y b x ) The re of the rectngle is A xy bx x ) Thus Solving da dx b x ) + bx ) b 2x ) da dx b 2x ) 0,
7 7 we obtin x /2, which corresponds to the re A b/4 As in the previous problem x /2 is criticl number nd to find the mximl re we need to compre A b/4 with A for the extreme vlues of x which re 0 nd For both of these vlues of x, we hve A 0 Hence A b/4 is the lrgest possible re
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