Question

设定义在R上的函数f(x)=a0x^4+a1x^3+a3x+a4(a0,a1,a3,a4属于R）,当x=-1时,f(x)

Answer 1

当x=-1时,f(x)取极大值,当x=1时,f(x)取极小值

连续,则过原点,a4=0

导数f'=4a0x^3+3a1x^2+a3,极值处=0,得a0=0,3a1+a3=0

又2/3= -a1-a3,得a1=-1/3,a3=1

所以f(x)= -1/3*x^3+x

f'= -x^2+1,设两切点-√2≤x1＜x2≤√2,(-x1^2+1)*(-x2^2+1)= -1

若x1= -2^0.5,则x2=0