作业帮已知函数f(x)=cosx+½x,x∈[-π/2,π/2] sinx0=1/2,x0∈[-π/2,π/2].那么下面命题中真命题的序号是①f(x)的最大值为f(x0) ②f(x)的最小值为f(x0) ③f(x)在[-π/2,x0]上是增函数 ④f(x)在[x0,π/2]上是增函数,ps:{x0
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/24 02:07:21
![已知函数f(x)=cosx+½x,x∈[-π/2,π/2] sinx0=1/2,x0∈[-π/2,π/2].那么下面命题中真命题的序号是①f(x)的最大值为f(x0) ②f(x)的最小值为f(x0) ③f(x)在[-π/2,x0]上是增函数 ④f(x)在[x0,π/2]上是增函数,ps:{x0](/uploads/image/z/1037090-2-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dcosx%2B%26%23189%3Bx%2Cx%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D+sinx0%3D1%2F2%2Cx0%E2%88%88%5B-%CF%80%2F2%2C%CF%80%2F2%5D.%E9%82%A3%E4%B9%88%E4%B8%8B%E9%9D%A2%E5%91%BD%E9%A2%98%E4%B8%AD%E7%9C%9F%E5%91%BD%E9%A2%98%E7%9A%84%E5%BA%8F%E5%8F%B7%E6%98%AF%E2%91%A0f%28x%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BAf%28x0%29+%E2%91%A1f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BAf%28x0%29+%E2%91%A2f%28x%29%E5%9C%A8%5B-%CF%80%2F2%2Cx0%5D%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0+%E2%91%A3f%28x%29%E5%9C%A8%5Bx0%2C%CF%80%2F2%5D%E4%B8%8A%E6%98%AF%E5%A2%9E%E5%87%BD%E6%95%B0%2Cps%3A%7Bx0)
已知函数f(x)=cosx+½x,x∈[-π/2,π/2] sinx0=1/2,x0∈[-π/2,π/2].那么下面命题中真命题的序号是①f(x)的最大值为f(x0) ②f(x)的最小值为f(x0) ③f(x)在[-π/2,x0]上是增函数 ④f(x)在[x0,π/2]上是增函数,ps:{x0
已知函数f(x)=cosx+½x,x∈[-π/2,π/2] sinx0=1/2,x0∈[-π/2,π/2].那么下面命题中真命题的序号是①f(x)的最大值为f(x0) ②f(x)的最小值为f(x0) ③f(x)在[-π/2,x0]上是增函数 ④f(x)在[x0,π/2]上是增函数,
ps:{x0,0是x的小角标}
已知函数f(x)=cosx+½x,x∈[-π/2,π/2] sinx0=1/2,x0∈[-π/2,π/2].那么下面命题中真命题的序号是①f(x)的最大值为f(x0) ②f(x)的最小值为f(x0) ③f(x)在[-π/2,x0]上是增函数 ④f(x)在[x0,π/2]上是增函数,ps:{x0
答案为1 、3