f(x)=1/3x^3-ax^2=4x,y=(x)在点(1,f(1))处的切线倾斜角为π/4,求a.若函数y=f(x)在区间【0,2】上单调递增
问题描述：
f(x)=1/3x^3-ax^2=4x,y=(x)在点(1,f(1))处的切线倾斜角为π/4,求a.若函数y=f(x)在区间【0,2】上单调递增
f(x)=1/3x^3-ax^2+4x,若函数y=f(x)在区间【0,2】上单调递增,求a的取值范围
答
f（x）求导得到：y‘=x^2-2ax+4又因为函数y=f(x)在区间【0,2】上单调递增所以y‘在[0.2]上恒大于0.对称轴为x=a,若a小于等于0,即当x=0,y'大于等于0.==》满足若a大于等于2,即当x=2,y'大于等于0===》a小于1,不符题设.舍...