来自翟玉庆的问题
y=tan²x+tanx+1,x∈[-π/4,π/4],求最值!
y=tan²x+tanx+1,x∈[-π/4,π/4],求最值!
1回答
2020-03-19 16:24
y=tan²x+tanx+1,x∈[-π/4,π/4],求最值!
y=tan²x+tanx+1,x∈[-π/4,π/4],求最值!
因为x∈[-π/4,π/4],tanx∈[-1,1]而y=tan²x+tanx+1=(tanx+1/2)²+3/4tanx+1/2∈[-1/2,3/2],(tanx+1/2)²∈[0,9/4],(tanx+1/2)²+3/4∈[3/4,3]即最小值是3/4,最大值是3...