作业帮已知向量a=(√3sinx/2,cosx/2),b=(cosx/2,cosx/2)令f(x)=向量a乘向量b,求f(x)+f`(x)的值域为什么答案是【3/2,√2+1/2】求完整过程谢谢了
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已知向量a=(√3sinx/2,cosx/2),b=(cosx/2,cosx/2)令f(x)=向量a乘向量b,求f(x)+f`(x)的值域为什么答案是【3/2,√2+1/2】求完整过程谢谢了
已知向量a=(√3sinx/2,cosx/2),b=(cosx/2,cosx/2)令f(x)=向量a乘向量b,求f(x)+f`(x)的值域
为什么答案是【3/2,√2+1/2】求完整过程谢谢了
已知向量a=(√3sinx/2,cosx/2),b=(cosx/2,cosx/2)令f(x)=向量a乘向量b,求f(x)+f`(x)的值域为什么答案是【3/2,√2+1/2】求完整过程谢谢了
f(x)=向量a.向量b.
=√3sinx/2cosx/2+cos^2(x/2).
=(√3/2)sinx+(1/2)cosx+1/2.
∴ f(x)=sin(x+π/6)+1/2.
f'(x)=cos(x+π/6).
f(x)+f'(x)=sin(x+π/6)+cos(x+π/6)+1/2.
∴ f(x)+f'(x)=√2sin(x+π/6+π/4)+1/2.
∵|sin(x+π/6+π/4)|≤1,
∴当sin(x+π/6+π/4)=1时,函数f(x)+f'(x)取得最大值（√2+1/2);
当sin(x+π/6+π/4)=-1时,函数f(x)+f'(x)取得最小值（-√2+1/2).
∴ 函数f(x)+f'(x)的值域为[ -√2+1/2,√2+1/2]√
所给参考答案[3/2,√2+1/2]是错的.∵|√2sin(x+π/6+π/4)|≤√2.不会等于1.
f(x) =a.b
= (√3sinx/2,cosx/2).(cosx/2,cosx/2)
=√3sinx/2cosx/2+cosx/2cosx/2
= (√3/2)sinx + (1/2)cosx + 1/2
= sin(x+π/6) +1/2
f'(x) = cos(x+π/6)
f(x) + f...
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f(x) =a.b
= (√3sinx/2,cosx/2).(cosx/2,cosx/2)
=√3sinx/2cosx/2+cosx/2cosx/2
= (√3/2)sinx + (1/2)cosx + 1/2
= sin(x+π/6) +1/2
f'(x) = cos(x+π/6)
f(x) + f'(x) = sin(x+π/6) +1/2 + cos(x+π/6)
consider
g(x)= sin(x+π/6) + cos(x+π/6)
g'(x) = cos(x+π/6) - sin(x+π/6) =0
tan(x+π/6) = 1
x = π/12(max) or 13π/12(min)
max f(x) +f'(x) = sin(π/4) +1/2 + cos(π/4)
= √2 +1/2
min f(x) +f'(x) = sin(5π/4) +1/2 + cos(5π/4)
= -√2 +1/2
域值=[-√2 +1/2,√2 +1/2]
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