f(x)=tan^2x+2tanx+2,求f(x)的最大值和最小值

解:
当-π/3<=x<=π/4,
tanx的范围是:[-根号3,1]
另tanx=a,a属于[-根号3,1]
y=tan^2x+2tanx+2
=a^2+2a+2
=(a^2+2a+1)+1
=(a+1)^2+1
当a=-1,y有最小值是:1
当a=1,y有最大值是;5

f(x)=(tanx+1)^2+1;
-π/3<=x<=π/4,-根号3<=tanx<=1;
当tanx=1,f(x)有最大值是:5 ；
当tanx=-1,f(x)有最小值是:1；