Question

求y的极值。当x为多少，y有最大值或最小值？怎么求，要过程。

不好意思，忘了x的取值范围了：0<x<a

Answer 1

应该有条件是0<x<a.
[1/x^2+1/(a-x)^2]•a^2
=[1/x^2+1/(a-x)^2]•[x+(a-x)]^2
=[1/x^2+1/(a-x)^2]•[x^2+2x(a-x)+(a-x)^2]
=1+2x(a-x)/x^2+(a-x)^2/x^2+ x^2/(a-x)^2+2x(a-x) /(a-x)^2+1
=1+2(a-x)/x+(a-x)^2/x^2+ x^2/(a-x)^2+2 x /(a-x) +1
=2+[2(a-x)/x+2 x /(a-x) ]+[(a-x)^2/x^2+ x^2/(a-x)^2]
≥2+2√[2(a-x)/x•2 x /(a-x) ]+ 2√[(a-x)^2/x^2•x^2/(a-x)^2]
=2+4+2=8，
∴1/x^2+1/(a-x)^2≥8/ a^2,
即y=1/x^2+1/(a-x)^2的最小值是8/ a^2,

Answer 2

X趋向无穷大时，Y趋向无穷小。
X趋向无穷小时，Y趋向无穷大。