作业帮函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域上面写错了以这个为准 函数y=4x²-mx+1在(负无穷,-2）上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域
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![函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域上面写错了以这个为准 函数y=4x²-mx+1在(负无穷,-2）上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域](/uploads/image/z/6921794-2-4.jpg?t=%E5%87%BD%E6%95%B0y%3D4x%26%23178%3B-mx%2B1%E5%9C%A8%28%E8%B4%9F%E6%97%A0%E7%A9%B7%2C-2%5D%E4%B8%8A%E9%80%92%E5%87%8F%2C%E5%9C%A8%5B-2%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%29%E4%B8%8A%E9%80%92%E5%A2%9E%2C%E6%B1%82f%28x%29%E5%9C%A8%5B1%2C2%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F%E4%B8%8A%E9%9D%A2%E5%86%99%E9%94%99%E4%BA%86%E4%BB%A5%E8%BF%99%E4%B8%AA%E4%B8%BA%E5%87%86+%E5%87%BD%E6%95%B0y%3D4x%26%23178%3B-mx%2B1%E5%9C%A8%28%E8%B4%9F%E6%97%A0%E7%A9%B7%2C-2%EF%BC%89%E4%B8%8A%E9%80%92%E5%87%8F%2C%E5%9C%A8%5B-2%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%29%E4%B8%8A%E9%80%92%E5%A2%9E%2C%E6%B1%82f%28x%29%E5%9C%A8%5B1%2C2%5D%E4%B8%8A%E7%9A%84%E5%80%BC%E5%9F%9F)
函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域上面写错了以这个为准 函数y=4x²-mx+1在(负无穷,-2）上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域
函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域
上面写错了以这个为准 函数y=4x²-mx+1在(负无穷,-2）上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域
函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域上面写错了以这个为准 函数y=4x²-mx+1在(负无穷,-2）上递减,在[-2,正无穷)上递增,求f(x)在[1,2]上的值域
因为函数y=4x²-mx+1在(负无穷,-2]上递减,在[-2,正无穷)上递增
所以对称轴就是x=-2,即x=m/8=-2解得m=-16
所以原函数就是y=4x²+16x+1
因为区间[1,2]在对称轴右侧,所以单调递增,其最小值为f（1）=21
最大值是f（2）=16+32+1=49
所以其在[1,2]上值域就是[21,49]