已知函数f(x)=1/2ax^2-(a+1)x+lnx当a>0时求f(x)的单调区间
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已知函数f(x)=1/2ax^2-(a+1)x+lnx当a>0时求f(x)的单调区间
已知函数f(x)=1/2ax^2-(a+1)x+lnx当a>0时求f(x)的单调区间
已知函数f(x)=1/2ax^2-(a+1)x+lnx当a>0时求f(x)的单调区间
f(x)=1/2ax^2-(a+1)x+lnx ---> f'(x)=ax+1/x-(a+1)
令f'(x)=0,则 ax+1/x-(a+1)=0, 解得:x1=1 ,x2=1/a
定义域x∈(0,∞)
若0f'(1)=0 ∴f(x)单调上升;
当x∈[1,1/a],f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x≤0 ∴f(x)单价下降;
当x∈(1/a,∞), f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x>0 ∴f(x)单调上升;
若a=1,则 f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x=(x-1)^2/x≥0 ∴f(x)在整个定义域单调上升;
若a>1,则 当x∈(0,1/a), f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x>0 ∴f(x)单调上升;
当x∈[1/a,1],f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x≤0 ∴f(x)单价下降;
当x∈(1,∞), f'(x)=ax+1/x-(a+1)=(ax-1)(x-1)/x>0 ∴f(x)单调上升;