Question

求y=tan²x+2tanx,x∈[-π/6,π/4]的值域

Answer 1

设t=tanx,
x∈[-π/6,π/4],
则t=tanx∈[-√3/3,1]y=t²+2t=(t+1)²-1,
是关于t的二次函数,开口向上,对称轴t=-1∴t∈[-√3/3,1]时,y是单调增的t=-√3/3时,y=1/3-2√3/3=(1-2√3)/3;
t=1时,
y=1+2=3∴值域为[(1-2√3)/3,3]