函數(shù)y(20+x^2)=(2-5x^2)的圖像示意圖

函數(shù)y=eq\f(2-5x2,20+x2)的圖像示意圖※.函數(shù)的定義域∵分式函數(shù)的分母20+x2>0,對(duì)任意的x都成立，則自變量x可以取全體實(shí)數(shù)，∴函數(shù)y=eq\f(2-5x2,20+x2)的定義域?yàn)?-∞，+∞）?！?函數(shù)的單調(diào)性本步驟用導(dǎo)數(shù)知識(shí)來(lái)判斷，先求函數(shù)的一階導(dǎo)數(shù)：∵y=eq\f(2-5x2,20+x2)∴eq\f(dy,dx)=eq\f(-10x(20+x2)-2x(2-5x2),(20+x2)2)=-2*102*eq\f(x,(20+x2)2),則函數(shù)的單調(diào)性為：(1).當(dāng)x≥0時(shí)，eq\f(dy,dx)≤0，此時(shí)函數(shù)y在定義域上為減函數(shù)；(2).當(dāng)x＜0時(shí)，eq\f(dy,dx)＞0，此時(shí)函數(shù)y在定義域上為增函數(shù)。此時(shí)x=0處，函數(shù)y有最大值，即：ymax=f(0)=eq\f(2-0,20+0)=eq\f(1,10).※.函數(shù)的值域把函數(shù)y=eq\f(2-5x2,20+x2)變形為x2的二次方程為：(20+x2)y=2-5x2(5+y)x2=2-20y則有：x2=eq\f(2-20y,5+y)≥0，解出：-5＜y≤eq\f(1,10)，所以函數(shù)y的值域?yàn)椋?-5,eq\f(1,10)]?！?函數(shù)的凸凹性∵eq\f(dy,dx)=-2*102*eq\f(x,(20+x2)2),∴eq\f(d2y,dx2)=-2*102*eq\f((20+x2)2-4x2(20+x2),(20+x2)4)eq\f(d2y,dx2)=2*102*eq\f(3x2-20,(20+x2)3),令eq\f(d2y,dx2)=0，則3x2-20=0，則x1=-eq\f(2\r(15),3)，x2=eq\f(2\r(15),3)，此時(shí)函數(shù)的凸凹性為：(1).當(dāng)x∈(-∞，-eq\f(2\r(15),3))∪(eq\f(2\r(15),3),+∞）時(shí)，eq\f(d2y,dx2)＞0，此時(shí)函數(shù)y的圖像為凹曲線；(2).當(dāng)x∈[-eq\f(2\r(15),3)，eq\f(2\r(15),3)]時(shí)，eq\f(d2y,dx2)≤0，此時(shí)函數(shù)y的圖像為凸曲線?！?函數(shù)的奇偶性∵f(x)=eq\f(2-5x2,20+x2),∴f(-x)=eq\f(2-5(-x)2,20+(-x)2)=eq\f(2-5x2,20+x2)，即f(-x)=f(x),則函數(shù)為偶函數(shù)，其圖像關(guān)于y軸對(duì)稱。.函數(shù)的五點(diǎn)圖表x01.292.583.875.162-5x22-6.32-31.28-72.88-131.1320+x22021.6626.6634.9846.63y0.10-0.29-1.17-2.08-2.81x-5.16-3.87-2.58-1.2902-5x2-131.13-72.88-31.28-6.32220+x246.6334.9826.6621.6620y-2.81-2.08-1.17-0.290.10.函數(shù)的圖像示意圖y=eq\f(2-5x2,20+x2) y（0,0.10）(-1.29,-0.29)(1.29,-0.29)(-2.58,-1.17)(2.58,-1.17)(-3.87,-2.08)(3.87,-2.08)