函數(shù)y(17+x^2)=(2-5x^2)的圖像示意圖

函數(shù)y=eq\f(2-5x2,17+x2)的圖像示意圖※.函數(shù)的定義域∵分式函數(shù)的分母17+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(2-5x2,17+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)來判斷，先求函數(shù)的一階導(dǎo)數(shù)：∵y=eq\f(2-5x2,17+x2)∴eq\f(dy,dx)=eq\f(-10x(17+x2)-2x(2-5x2),(17+x2)2)=-2*87*eq\f(x,(17+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(2-0,17+0)=eq\f(2,17).※.函數(shù)的值域把函數(shù)y=eq\f(2-5x2,17+x2)變形為x2的二次方程為：(17+x2)y=2-5x2(5+y)x2=2-17y則有：x2=eq\f(2-17y,5+y)≥0，解出：-5＜y≤eq\f(2,17)，所以函數(shù)y的值域?yàn)椋?-5,eq\f(2,17)]?！?函數(shù)的凸凹性∵eq\f(dy,dx)=-2*87*eq\f(x,(17+x2)2),∴eq\f(d2y,dx2)=-2*87*eq\f((17+x2)2-4x2(17+x2),(17+x2)4)eq\f(d2y,dx2)=2*87*eq\f(3x2-17,(17+x2)3),令eq\f(d2y,dx2)=0，則3x2-17=0，則x1=-eq\f(\r(51),3)，x2=eq\f(\r(51),3)，此時(shí)函數(shù)的凸凹性為：(1).當(dāng)x∈(-∞，-eq\f(\r(51),3))∪(eq\f(\r(51),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(\r(51),3)，eq\f(\r(51),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(2-5x2,17+x2),∴f(-x)=eq\f(2-5(-x)2,17+(-x)2)=eq\f(2-5x2,17+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x01.192.383.574.762-5x22-5.08-26.32-61.72-111.2917+x21718.4222.6629.7439.66y0.12-0.28-1.16-2.08-2.81x-4.76-3.57-2.38-1.1902-5x2-111.29-61.72-26.32-5.08217+x239.6629.7422.6618.4217y-2.81-2.08-1.16-0.280.12.函數(shù)的圖像示意圖y=eq\f(2-5x2,17+x2) y（0,0.12）(-1.19,-0.28)(1.19,-0.28)(-2.38,-1.16)(2.38,-1.16)(-3.57,-2.08)(3.57,-2.08)