若x∈[-π/4,π/4],y=sec^2x+tanx+1的最大值和最小值

设y=(tan^2x-tanx+1)/(tan^2x+tanx+1)
另tanx=a,则a属于R
y=(a^2-a+1)/(a^2+a+1)[a属于R]
而a^2+a+1>0恒成立
则由判别式法有;
y(a^2+a+1)=a^2-a+1
(y-1)a^2+(y+1)a+(y-1)=0
当y不等于1
关于a的方程的判别式有;
(y+1)^2-4(y-1)(y-1)>=0
y^2+2y+1-4y^2+8y-4>=0
3y^2-10y+3

令tanx=u
y=u^2+u+1
=u^2+u+1/4+3/4
=(u+1/2)^2+3/4
平方项≥0
所以y≥3/4
值域就是[3/4,+∞)

令a=tanx,a属于R
y=(a^2-a+1)/(a^2+a+1)
ya^2+ya+y=a^2-a+1
(y-1)a^2+(y+1)a+(y-1)=0
判别式大于等于0
(y+1)^2-4(y-1)^2>=0
(y+1+2y-2)(y+1-2y+2)>=0
(3y-1)(-y+3)>=0
(3y-1)(y-3)