Question

当x趋近于无穷，y=(x²-1)/(x²+3)趋近于1，问X等于多少，使当|x|＞X时，|y-1|

当x趋近于无穷，y=(x²-1)/(x²+3)趋近于1，问X等于多少，使当|x|＞X时，|y-1|＜0.1

Answer 1

学过导数吗，导数是求极值的基本方法，一函数的一价导数=0时，取极值。xE[0,无穷大)y=x+3sqr((x-1)²+2)y'=1+3*1/2*1*2(x-1)/(sqr((x-1)²+2)y'=1+3(x-1)/(sqr((x-1)²+2))=0时,取极值sqrt((x-1)^2+2)=3(1-x)x^2-2x+1+2=9-18x+9x^28x^2-16x+6=04x^2-8x+3=0x=3/2orx=1/2取极值f(1/2)=1/2+3sqrt(2.25)=1/2+3*3/2=5f(3/2)=3/2+3*sqrt(2.25)=1.5+3*1.5=6ymin=f(1/2)=5如果没学过导数：请这么做！y=x+3sqr((x-1)²+2)y-x=3sqr((x-1)²+2)y^2-2yx+x^2=9(x^2-2x+3)8x^2-(18-2y)x+27-y^2=0deta=(18-2y)^2-32(27-y^2)=4[(9-y)^2+8y^2-216]=4[(-135-18y+9y^2]>=036(-15-2y+y^2)>=0(y+3)(y-5)>=0y>=5ory0y=x+3sqr((x-1)²+2)>0所以最小值为y=5