已知函数f(x)=xΛ2+2/x+alnx,a∈R,若a=-4,求函数f(x)的单调区间
2个回答
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f(x)=x^2+2/x-4lnx,
f'(x)=2x-2/x^2-4/x
=2(x^3-2x-1)/x^2
=2[(x^3+x^2)-(x^2+2x+1)]/x^2
=2[x^2(x+1)-(x+1)^2]/x^2
=2(x+1)(x^2-x-1)/x^2
=2(x+1)[x-(1-√5)/2][x-(1+√5)/2]
f(x)的递减区间是(-无穷,-1)和((1-√5)/2,(1+√5)/2),递增区间是(1,(1-√5)/2)和((1+√5)/2,+无穷)
f'(x)=2x-2/x^2-4/x
=2(x^3-2x-1)/x^2
=2[(x^3+x^2)-(x^2+2x+1)]/x^2
=2[x^2(x+1)-(x+1)^2]/x^2
=2(x+1)(x^2-x-1)/x^2
=2(x+1)[x-(1-√5)/2][x-(1+√5)/2]
f(x)的递减区间是(-无穷,-1)和((1-√5)/2,(1+√5)/2),递增区间是(1,(1-√5)/2)和((1+√5)/2,+无穷)
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