已知函数f(x)=〔(x+1)lnx〕\/x-1 (x>0且x≠1)。(1)讨论函数f(x)的单...
当x∈(0,1)时,g(x)<g(1)=0,于是f'(x)<0, f(x)单调递减;当x∈(1,+∞)时,g(x)>g(1)=0,于是f'(x)>0, f(x)单调递增;于是得出f(x)min=f(1),由罗比达法则得:lim〔(x+1)lnx〕\/(x-1)x---1 =lim(lnx+(x+1)\/x)=2 x---1 所以 f(x)>2 .
...x?1(x>0且x≠1)(1)讨论函数f(x)的单调性(2)证明:f(x)>2
2x+1+1x2=(x?1x)2由g′(x)≥0恒成立得,g(x)在(0,+∞)单调递增,又∵g(1)=0故当x∈(0,1)时,g(x)<0,f′(x)<0,f(x)单调递减;当x∈(1,+∞)时,g(x)>0,f′(
已知函数f(x)=(x+1)Inx-x+1.(1)若xf'(x)≤x^2+ax+1,求a的取值范围;(2...
f'(x)=lnx+1\/x x>1时,f'(x)>0,f(x)递增,f(x)>f(1)=0 ∴(x-1)f(x)>0 x=1时,(x-1)f(x)=0 0<x<1时,f'‘(x)=1\/x-1\/x^2=(x-1)\/x^2<0 ∴f'(x)递减,f'(x)>f'(1)=1>0,f(x)递增 f(x)<f(1)=0 ∴(x-1)f(x)>0 综上,x>0时,总有(...
已知函数f(x)=(x+1)Inx-x+1 1.若xf(x)≤x^2+ax+1,求a的范围 2、证明(x...
=>f‘(1)是f'(x)的最小值=1\/1+ln1=1>0,推出f(x)单调递增。f(1)=0 =>0<x<1时f(x)<=0,x>1时,f(x)>=0 =>(x-1)*f(x)>=0
函数f(x)=(x+1)lnx-x+1.证明:(x-1)f(x)≥0.
当x≥1,f(x)=(x+1)lnx-x+1,f’(x)=(x+1)*1\/x+lnx-1=1\/x+1nx,因为x≥1,则lnx≥0,1\/x>0,所以f’(x)>0,所以f(x)在[1,+oo)上递增,则f(x) ≥f(1)=0-1+1=0,又(x-1)≥0所以(x-1)f(x)≥0.当1>x>0,f(x)=(x+1)lnx-x+1,f’(x)=(x+1)*1\/x+...
已知函数f(x)=lnx-x+1.求fx的导数,证明当x大于等于1时,f(x)小于...
(1)f(x)=lnx-x+1 求导得到f'(x)=1\/x-1 (2)令f'(x)>0 1\/x-1>0 0<x<1 f'(x)<0 1\/x-1<0 x>1 得到函数在(0,1)上递增,(1,+∞)递减 而f(1)=0 所以当x>=1 时,f(x)<=f(1)=0 故x大于等于1时,f(x)小于等于0 ...
已知函数f(x)=(x\/lnx)-ax(x>0,且x≠1). (1)若函数f(x)在(1,正无穷...
f(x)=(x\/lnx)-ax ===> f'(x)=(lnx-1)\/(lnx)^2-a=[-a(lnx)^2+lnx-1]\/(lnx)^2 令t=lnx>0,g(t)=-at^2+t-1 ①当a>0时,g(t)开口向下,对称轴为t=1\/2a>0 此时要满足在t>0时,g(t)≤0 则,△=1-4a≤0 ===> a≥1\/4 ②当a<0时,g(t)开口向上,...
...=(1+㏑x)\/x,(x≥1) (1)试判断函数f(x)的单调性,并说明理由
f'(x) = (1\/x)x⁻¹ + (1+lnx)(-1)x⁻² = -x⁻²lnx x ≥1时, lnx ≥0, x⁻² > 0, f'(x) ≤ 0, 减函数 (2)令g(x) = (x+1)f(x) = x⁻¹(x+1)(1+lnx)g'(x) = - x⁻²(x+1)(...
已知函数f(x)=ln(x+1)\/(x-1)(Ⅰ)求函数的定义域.并证明f(x)=ln(x+...
=-f(x),故f(x)是奇函数。(2).y=lnx是增函数,故由ln[(x+1)\/(x-1)]>ln[m\/(x-1)(x-7)]得(x+1)\/(x-1)>m\/(x-1)(x-7)移项得(x+1)\/(x-1)-m\/(x-1)(x-7)=[(x+1)(x-7)-m]\/(x-1)(x-7)=(x²-6x-7-m)\/(x-1)(x-7)>0...(1)不等式(1)...
已知函数f(x)=x\/lnx-ax(x>1且x≠1) .若存在x1,x2属于[e,e^2],使f...
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