若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值
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![若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值](/uploads/image/z/8632186-34-6.jpg?t=%E8%8B%A5-%CF%80%2F2%E2%89%A4x%E2%89%A4%CF%80%2F2%2C%E6%B1%82f%28x%29%3D%E2%88%9A3sinx%2Bcosx%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E5%92%8C%E6%9C%80%E5%B0%8F%E5%80%BC%2C%E5%B9%B6%E6%B1%82%E5%87%BA%E6%AD%A4%E6%97%B6%E7%9A%84x%E7%9A%84%E5%80%BC)
若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值
若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值
若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值
f(x)=2(√3/2sinx+1/2cosx)
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
-π/2<=x<=π/2
-π/3<=x+π/6<=2π/3
所以x+π/6=-π/3,sin(x+π/6)最小=-√3/2
x+π/6=π/2,sin(x+π/6)最大=1
所以
x=-π/2,f(x)最小=-√3
x=π/3,f(x)最大=2