函數(shù)y=(9x+1)²(5x+20)³的主要性質(zhì)及圖像示意圖

函數(shù)y=(9x+1)2(5x+20)3的圖像示意圖及其性質(zhì)主要內(nèi)容：通過(guò)函數(shù)的定義域、值域、單調(diào)性、凸凹性及極限的性質(zhì)，并通過(guò)函數(shù)導(dǎo)數(shù)知識(shí)求解函數(shù)的單調(diào)區(qū)間和凸凹區(qū)間，并簡(jiǎn)要畫(huà)出函數(shù)y=(9x+1)2(5x+20)3示意圖的過(guò)程與步驟。※.函數(shù)定義域根據(jù)函y=(9x+1)2(5x+20)3特征，可知函數(shù)自變量x可以取全體實(shí)數(shù)，即函數(shù)的定義域?yàn)椋海?∞，+∞）※.函數(shù)一階導(dǎo)數(shù)：※.函數(shù)乘積求導(dǎo)法?！遹=(9x+1)2(5x+20)3，∴y'=18(9x+1)(5x+20)3+(9x+1)2*15*(5x+20)2，=(9x+1)(5x+20)2(90x+360+135x+15),=(9x+1)(5x+20)2(225x+375)※.取對(duì)數(shù)求導(dǎo)法?！遹=(9x+1)2(5x+20)3，取導(dǎo)數(shù)有：∴l(xiāng)ny=ln(9x+1)2(5x+20)3,即：lny=2ln(9x+1)+3ln(5x+20),兩邊同時(shí)對(duì)x求導(dǎo)：eq\f(y',y)=eq\f(18,9x+1)+eq\f(15,5x+20),y'=y(eq\f(18,9x+1)+eq\f(15,5x+20)),y'=(9x+1)2(5x+20)3(eq\f(18,9x+1)+eq\f(15,5x+20)),y'=(9x+1)(5x+20)2[18(5x+20)+15(9x+1)],y'=(9x+1)(5x+20)2(225x+375).令y'=0，有9x+1=0，225x+375=0，即：x1=-eq\f(1,9)，x2=-eq\f(5,3).(1)當(dāng)x∈(-∞，-eq\f(5,3))，(-eq\f(1,9),+∞)時(shí)，eq\f(dy,dx)>0,此時(shí)函數(shù)為增函數(shù)。(2)當(dāng)x∈[-eq\f(5,3)，-eq\f(1,9)]時(shí)，eq\f(dy,dx)<0,此時(shí)函數(shù)為減函數(shù)。※.函數(shù)的凸凹性∵y'=(9x+1)(5x+20)2(225x+375)∴y''=9(5x+20)2(225x+375)+(9x+1)[10(5x+20)(225x+375)+225(5x+20)2]=9(5x+20)2(225x+375)+(9x+1)(5x+20)[10(225x+375)+225(5x+20)]=(5x+20)[9(5x+20)(225x+375)+10(9x+1)(225x+375)+225(9x+1)(5x+20)]=(5x+20){(225x+375)[9(5x+20)+10(9x+1)]+225(9x+1)(5x+20)}=(5x+20)(40500x2+135000x+75750)=750(5x+20)(54x2+180x+101).令y''=0，則5x+20=0，或54x2+180x+101=0，即：x3=-4,x4=eq\f(-30-7\r(6),18)≈-2.61,x5=eq\f(-30+7\r(6),18)≈-0.71,此時(shí)函數(shù)的凸凹性性及凸凹區(qū)間為：(1)當(dāng)x∈(-∞,-4),(-2.61,-0.714)時(shí)，y''<0,此時(shí)函數(shù)y為凸函數(shù)。(2)當(dāng)x∈[-4,-2.61],[-0.714,+∞)時(shí)，y''>0,此時(shí)函數(shù)y為凹函數(shù)。.函數(shù)的極限lim(x→-∞)(9x+1)2(5x+20)3=-∞；lim(x→+∞)(9x+1)2(5x+20)3=+∞；lim(x→0)(9x+1)2(5x+20)3=12*203。.函數(shù)的五點(diǎn)圖x-9-4-2.61-1.66-0.714-0.110.3888(9x+1)264001225509.54919629.450020.25(5x+20)3-15620329.041587.94434.87351.610567.y-999680001676623112281306050213982※.函數(shù)的示意圖y=(9x+1)2(5x+20)3y(-1.66,311228)(-2.61,167662)(-0.714,130605)