作业帮已知函数f(x)=4x²-mx+1在(-∞,-2]上递减,在[-2,﹢∞﹚上递增,则f(x)在[1,2]上的值域为——
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 00:13:05
![已知函数f(x)=4x²-mx+1在(-∞,-2]上递减,在[-2,﹢∞﹚上递增,则f(x)在[1,2]上的值域为——](/uploads/image/z/8624759-23-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D4x%26%23178%3B-mx%2B1%E5%9C%A8%28-%E2%88%9E%2C-2%5D%E4%B8%8A%E9%80%92%E5%87%8F%2C%E5%9C%A8%5B-2%2C%EF%B9%A2%E2%88%9E%EF%B9%9A%E4%B8%8A%E9%80%92%E5%A2%9E%2C%E5%88%99f%28x%29%E5%9C%A8%5B1%2C2%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%BA%E2%80%94%E2%80%94)
已知函数f(x)=4x²-mx+1在(-∞,-2]上递减,在[-2,﹢∞﹚上递增,则f(x)在[1,2]上的值域为——
已知函数f(x)=4x²-mx+1在(-∞,-2]上递减,在[-2,﹢∞﹚上递增,则f(x)在[1,2]上的值域为——
已知函数f(x)=4x²-mx+1在(-∞,-2]上递减,在[-2,﹢∞﹚上递增,则f(x)在[1,2]上的值域为——
f(x)=4x²-mx+1=4(x-m/8)²-m²/16+1
m/8=-2
m=-16
f(x)=4(x+2)²-15
f(1)=21
f(2)=49
值域为：[21,49]