Учебно-методическое пособие (1017796), страница 4
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Ïîñòðîèòü ãðàôèê ôóíêöèè.√31y = (2x + 3)e−2x − 22y=x2 e−x32y = xe−x4y=5y=xln x6y = xe−x7y = e1/x − x8y = x2 − ln |x|9( 2)1/x−4x+4y=e10 y = (1 + x)e1/xÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ√x3 ln xx−1x−311 y = e−x sin x12 y = 3 lne2 − x13 y =2−x14 y = 3 − 3 ln15 y = xe1/(x − 1)16 y = ex cos xe2x + 217 y =2x + 2ln x18 y = √x19 y = (3 − x)ex − 220 y = ln221 y = e1/x22 y = xe1/(2 − x)xx+41+x1−x()423 y = (2x + 5)e−2x − 4 24 y = 2 ln 1 −−3x5225 y = 4 lnx+1x+226 y = x (2 − ln x)2√xln x28 y =x+329 y = 2 ln−3xex − 330 y =2x + 7ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ27 y = x3 e−xÇàäà÷à 23.
Ïîñòðîèòü ëèíèþ, çàäàííóþ óðàâíåíèåì ρ = f (φ)â ïîëÿðíûõ êîîðäèíàòàõ (ρ ≥ 0, 0 ≤ φ ≤ 2π). f (φ) f (φ) f (φ)147101316192225282581114172023262936912151821242730cos(3φ + π/4)4 tg(φ/2)1 + cos2 2φsin2 2φ4 cos 2φ3(2 − sin φ)√4 cos φ2 + cos φ5(2 − cos φ)3 cos2 2φ1 + cos φ2 cos 3φ5(1 − cos 2φ)√cos(π + φ)sin(φ/2)2 + sin2 2φ5(1 + cos 3φ)2 sin φ√sin(−2φ)3 + 2 cos φ2 + sin φ√3 cos 2φ2(1 + sin 3φ)7(1 + sin φ)2 sin 3φ1 − sin 2φ4(1 − cos 4φ)cos(φ/2)2 tg φ4 sin2 3φÇàäà÷à 24.
Âû÷èñëèòü ÷àñòíûå ïðîèçâîäíûå ïåðâîãî ïîðÿä-êà. z = f (x, y)1z = x3 + y 3 + 3x/y z = f (x, y)2z=√x2 − y 253(z = arctg57)4z = sin(x + cos y)z = x2 ln(x + y)6z = ex/yz = ln cos (y/x)8z = ln tg(x − y)( 3)2 2x+yz=e10 z = x3 + 4x2 y 2 − y 4ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ3x+y1 − xy911 z = x sin(2x + 3y)(213 z = ln x + y)( )12 z = cos x2 /y14 z = xy + y/x15 z = x2 sin4 y16 z = ln(1 + x/y)17 z = cos(y + sin x)18 z = ln tg(y/x)19 z = xy sin(xy)20 z = x4 cos2 y21 z = esin(y/x)22 z = √23 z = x cos(x + y)24 z = ex cos yxyx2 + y 225 z = x3 + 2y 2 − 2y 3 x2 26 z = y x27 z = ln tg (x/y)28 z = x3 sin y + y 3 cos x()29 z = tg y 2 /x()30 z = y ln x2 − y 254Çàäà÷à 25. Âû÷èñëèòü ñìåøàííûå ïðîèçâîäíûå âòîðîãî ïî-ðÿäêà è ïðîâåðèòü, ÷òî îíè ðàâíû. z = f (x, y)35722z = exy(x + y )(2z = x +y2)· ex + yx2 − y 2z= 2x + y2z = arcsin √4z = x ln(x3 y 2 )()yz = arctg1 + x2z=6xyx+y8x+yx−y( )11 z = sin x2 /y9x2x2 + y 2ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ1 z = f (x, y)z=(213 z = tg x /y15 z = ln√)x2 + y 217 z = x2 /y 2 − y/x19 z =√2xy + y 2z = ex (x sin y + y 2 )()10 z = arcctg x/y 2)(y12 z = x arctgy−x()x+y14 z = arctg1 − xy16 z = 2x2 y + 3xy 2 + x318 z = √xx2 + y 2( 2)220 z = y ln x − y21 z = arctg (y/x) + arcctg (x/y) 22 z = ex (cos y + x sin y)23 z = xy24 z = x ln (y/x)5525 z = ex + y (x cos y + y sin x)()26 z = ln x2 + xy + y 2()27 z = sin x2 − y 32228 z = e3x + 2y − xy30 z = xy sin x − y2)ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ29 z = xy + x/y(Çàäà÷à 26.
Íàéòè è èññëåäîâàòü òî÷êè ýêñòðåìóìà1u = 2x2 + y 2 + z 2 − xy − 3x + 4y + 2z2u = x2 + 3y 2 + z 2 − xz − 2x + 3y3u = −9x2 − 6y 2 − 11z 2 + 3xy + 5xz − 8yz4u = −4x2 − 3y 2 − z 2 − 2xy + yz5u = x2 + y 2 + 2z 2 + yz + 2z − 3y63u = 2x2 + y 2 + z 2 − xy + 2xz + yz − y + 2z27u = −5x2 − y 2 − 3z 2 + xz + yz − 2xy + 6z8u = 2x2 + y 2 + 3z 2 + xy + xz − 4x − 2y + z3u = x2 + y 2 + 2z 2 + xy + xz − 4x − 2y + z2110 u = −x2 − y 2 − 5z 2 + xy + xz − 2x + 4y2956311 u = 2x2 + y 2 + z 2 − xz + 2xy + yz − 3y2112 u = 5x2 + y 2 + 5z 2 + 2xy − xz − yz − 10x213 u = 5x2 + y 2 + 5z 2 + 2xy − xz − yz − 4x + 2yÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ14 u = −4x2 − y 2 − 5z 2 + xy + xz − 2yz − 2y15 u = −2x2 − y 2 − z 2 + xy − 2z + x − 4y16 u = −x2 − 3y 2 − z 2 + xz + x − 6y + z17 u = 2x2 + y 2 + 3z 2 + 2xz + 2x − 4y18 u = −3x2 − 4y 2 − 5z 2 + 2yz − 2xy + 6x19 u = x2 + 5y 2 + z 2 + 2xy − xz − yz − 10y20 u = x2 + 2y 2 + 5z 2 + xy + 4yz − 4z21 u = −2x2 − 5y 2 − 5z 2 + 2xz + 4yz − 2xy − 4x22 u = −5x2 − 6y 2 − 4z 2 + 2xy + 2xz − 8z23 u = 3x2 + 4y 2 + 5z 2 + 2xy − 2yz − 8y24 u = 4x2 + 6y 2 + 5z 2 − 2xz − 2yz + 10z25 u = 9x2 + 6y 2 + 11z 2 − 3xy − 5xz + 8yz5726 u = x2 + 17y 2 + 3z 2 + 2xy − xz − 7yz27 u = −x2 − 17y 2 − 3z 2 + xz + 7yz − 2xy28 u = −9x2 − 6y 2 − 11z 2 + 3xy + 5xz − 8yzÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ29 u = 2x2 + y 2 + 3z 2 + xy + xz − 4x − 2y + z130 u = 5x2 + y 2 + z 2 + 2xy − xz − yz − 10x2ÏÐÈËÎÆÅÍÈÅ äàííîì ïðèëîæåíèè èçëàãàåòñÿ êðàòêàÿ òåîðèÿ è ìåòîäèêàðåøåíèÿ òèïîâûõ çàäà÷ ïî ñëåäóþùèì òåìàì, óêàçàííûì íèæå.Èçó÷åíèå ìàòåðèàëà ýòîãî Ïðèëîæåíèÿ íåîáõîäèìî äëÿ óñïåøíîãî âûïîëíåíèÿ êîíòðîëüíûõ ðàáîò è òèïîâîãî ðàñ÷åòà, à òàêæåÿâëÿåòñÿ ïîëåçíûì ïðè ïîäãîòîâêå ê ýêçàìåíó (çà÷åòó).Òåìà 1.
Òåîðèÿ ïðåäåëîâ.1.1. Îïðåäåëåíèå ïðåäåëà ôóíêöèè.1.2. Îñíîâíûå òåîðåìû î ïðåäåëàõ.1.3. Ýëåìåíòàðíûå ìåòîäû âû÷èñëåíèÿ ïðåäåëà.1.4. Ïåðâûé è âòîðîé çàìå÷àòåëüíûå ïðåäåëû.1.5. Áåñêîíå÷íî ìàëûå ôóíêöèè.1.6. Ýêâèâàëåíòíûå áåñêîíå÷íî ìàëûå ôóíêöèè.1.7. Ïðèìåíåíèå ýêâèâàëåíòíûõ áåñêîíå÷íî ìàëûõ ê âû÷èñëåíèþïðåäåëîâ.58Òåìà 2. Íåïðåðûâíîñòü ôóíêöèè.2.1. Íåïðåðûâíîñòü ôóíêöèè.2.2. Îäíîñòîðîííèå ïðåäåëû.2.3. Òî÷êè ðàçðûâà.Òåìà 3. Äèôôåðåíöèðîâàíèå ôóíêöèè îäíîé ïåðåìåííîé.ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀ3.1. Ïðîèçâîäíàÿ ôóíêöèè.3.2. Òàáëèöà ïðîèçâîäíûõ.3.3. Äèôôåðåíöèðîâàíèå ñëîæíîé ôóíêöèè.3.4.
Âû÷èñëåíèå ëîãàðèôìè÷åñêîé ïðîèçâîäíîé.3.5. Âû÷èñëåíèå ïðîèçâîäíîé ôóíêöèè, çàäàííîé ïàðàìåòðè÷å-ñêè.3.6. Âû÷èñëåíèå ïðîèçâîäíîé ôóíêöèè, çàäàííîé íåÿâíî.3.7. Ïðîèçâîäíûå âûñøèõ ïîðÿäêîâ3.8. Äèôôåðåíöèàë ôóíêöèè.Òåìà 4. Ïðèëîæåíèÿ ïðîèçâîäíîé.4.1. Óðàâíåíèå êàñàòåëüíîé è íîðìàëè ê êðèâîé.4.2. Ïðèìåíåíèå äèôôåðåíöèàëà â ïðèáëèæåííûõ âû÷èñëåíèÿõ.4.3. Ïðèêëàäíûå çàäà÷è íà èñïîëüçîâàíèå ïðîèçâîäíîé.4.4.
Ïðàâèëî Ëîïèòàëÿ.Òåìà 5. Èññëåäîâàíèå ôóíêöèè: âîçðàñòàíèå, óáûâàíèå, ýêñòðåìóìû.5.1. Ïðèçíàêè âîçðàñòàíèÿ è óáûâàíèÿ ôóíêöèè íà èíòåðâàëå.5.2. Ýêñòðåìóìû ôóíêöèè.59Òåìà 6. Èññëåäîâàíèå ôóíêöèè: âûïóêëîñòü è âîãíóòîñòü,àñèìïòîòû.6.1. Âûïóêëîñòü è âîãíóòîñòü ãðàôèêà ôóíêöèè.6.2. Òî÷êè ïåðåãèáà.6.3. Àñèìïòîòû ãðàôèêà ôóíêöèè.ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀÒåìà 7. Îáùàÿ ñõåìà èññëåäîâàíèÿ ôóíêöèè è ïîñòðîåíèåãðàôèêà.Òåìà 8. Ôîðìóëà Òåéëîðà.8.1. Ìíîãî÷ëåí Òåéëîðà.8.2. Îñòàòî÷íûé ÷ëåí â ôîðìóëå Òåéëîðà.8.3.
Ôîðìóëà Òåéëîðà äëÿ íåêîòîðûõ ýëåìåíòàðíûõ ôóíêöèé.8.4. Ïðèìåíåíèå ôîðìóëû Òåéëîðà.Èçó÷åíèå òåì 1,2 ïîìîãàåò ïðè ïîäãîòîâêå ê êîíòðîëüíîé ðàáîòå 1. Ñîäåðæàíèå òåì 3, 4 è 8 ìîæíî ðàññìàòðèâàòü â êà÷åñòâåîñíîâû ïðè ïîäãîòîâêå ê êîíòðîëüíîé ðàáîòå 2. Òåîðåòè÷åñêèéìàòåðèàë òåì 5 - 7 ïîëåçåí ïðè âûïîëíåíèè òèïîâîãî ðàñ÷åòà.1. Òåîðèÿ ïðåäåëîâ1.1. Îïðåäåëåíèå ïðåäåëà ôóíêöèèÎáîçíà÷åíèÿ, èñïîëüçóåìûå â ïðèëîæåíèè:∀ äëÿ ëþáîãî“ , äëÿ êàæäîãî“ ;””∃ ñóùåñòâóåò“ , íàéäåòñÿ“ ;””∃ íå ñóùåñòâóåò“ ;”x ∈ A x ïðèíàäëåæèò A;x∈/ A x íå ïðèíàäëåæèò A;60B ⊂ A B ÿâëÿåòñÿ ïîäìíîæåñòâîì A;⇒ ñëåäîâàòåëüíî“ ;”⇔ òîãäà è òîëüêî òîãäà“ ;”: òàêîé ÷òî“ ;”x → a x ñòðåìèòñÿ ê a.Îïðåäåëåíèå 1.1 ×èñëî b íàçûâàåòñÿ ïðåäåëîì ôóíêöèè y =ÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀf (x) ïðè x ñòðåìÿùåìñÿ ê a, åñëè äëÿ ëþáîãî ε > 0 íàéäåòñÿ òàêîå δ = δ(ε) > 0, ÷òî ïðè 0 < |x − a| < δ âûïîëíÿåòñÿíåðàâåíñòâî |f (x) − b| < ε.Äëÿ îáîçíà÷åíèÿ ïðåäåëà ôóíêöèè èñïîëüçóþò ñëåäóþùóþñèìâîëèêólim f (x) = bx→aÈñïîëüçóÿ ââåäåííûå âûøå îáîçíà÷åíèÿ, îïðåäåëåíèå 1.1 ìîæíî ïåðåïèñàòü â âèäå:b = lim f (x) ⇔ ∀ε > 0 ∃ δ = δ(ε) > 0 : ∀x : 0 < |x − a| < δ ⇒x→a⇒ |f (x) − b| < εÇàìå÷àíèå.
Ïðè îïðåäåëåíèè ïðåäåëà íå ñóùåñòâåííî, êàê âåäåò ñåáÿ ôóíêöèÿ â ñàìîé òî÷êå a.  ÷àñòíîñòè, çíà÷åíèå f (a)ìîæåò áûòü íå îïðåäåëåíî.Îïðåäåëåíèå 1.2lim f (x) = ∞ ⇔ ∀k > 0 ∃ δ = δ(k) > 0 : ∀x : 0 < |x − a| < δ ⇒x→a⇒ |f (x)| > kÎïðåäåëåíèå 1.3lim f (x) = b ⇔ ∀ε > 0 ∃ k = k(ε) > 0 : ∀x > k(ε) ⇒x→+∞⇒ |f (x) − b| < ε61Çàäà÷à. Ñôîðìóëèðîâàòü ñëåäóþùèå îïðåäåëåíèÿ1. lim f (x) = b,x→−∞2. lim f (x) = +∞,x→+∞4. lim f (x) = −∞,3.
lim f (x) = ∞,x→−∞5. lim f (x) = +∞.x→ax→a1.2. Îñíîâíûå òåîðåìû î ïðåäåëàõÊàÌôåäÃÒðÓàÌ ÂÌÈ-2ÐÝÀÏðèâîäèìûå íèæå òåîðåìû ñïðàâåäëèâû äëÿ ñëó÷àåâ êîãäà a ÷èñëî è êîãäà a = ±∞.Òåîðåìà 1.1 lim C = C ,ãäå C ïîñòîÿííàÿ (C const)x→aÒåîðåìà 1.2 lim [f (x) ± g(x)] = lim f (x) ± lim g(x)x→ax→ax→aÒåîðåìà 1.3 lim [f (x) · g(x)] = lim f (x) · lim g(x)x→ax→ax→aÒåîðåìà 1.4 lim [C · f (x)] = C · lim f (x),ãäå C ïîñòîÿííàÿx→ax→alim f (x)f (x) x→aÒåîðåìà 1.5 lim=, åñëè lim g(x) ̸= 0x→a g(x)x→alim g(x)x→af (x)= ∞, åñëè lim f (x) = C ̸= 0, lim g(x) = 0x→a g(x)x→ax→aÒåîðåìà 1.6 limf (x)= 0, åñëè |f (x)| < C , lim g(x) = ∞x→a g(x)x→a[] lim g(x)g(x)Òåîðåìà 1.8 lim f (x)= lim f (x) x→aÒåîðåìà 1.7 limx→ax→a()Òåîðåìà 1.9 lim f (x) = f lim x , åñëè f (x) ýëåìåíòàðíàÿx→ax→aôóíêöèÿ è a ïðèíàäëåæèò îáëàñòè îïðåäåëåíèÿ ôóíêöèè fÒåîðåìà 1.10 Ïóñòü lim g(x) = b, lim f (x) = f (b).x→aÒîãäà lim f (g(x)) = f (b)x→ay→b621.3.
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