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
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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
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
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
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
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.舍去
当a大于0,小于2,即x=a时,y’大于等于0===》a大于0,小于2
综上:a属于(负无穷,2)
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,
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