已知fx=x+2/x-1.证明fx在（1.+∞上是减函数，当x属于[2.5]求fx的最大值和最小值
\u5df2\u77e5\u51fd\u6570f(x)=-x^2+2x 10\u8bc1\u660e\uff1bf(x)\u5728[1\uff0c+00]\u4e0a\u662f\u51cf\u51fd\u6570\uff1b 2\uff09\u5f53x\u5c5e\u4e8e[2,5]\u65f6\uff0c\u6c42f(x)\u7684\u6700\u5927\u503c\u548c\u6700\u5c0f\u503c\u8bc1\u660e\uff1a
\u8bbex1>x2>=1
\u5219 f(x1)-f(x2)=-x1^2+2*X1-(-x2^2+2*x2)
=-x1^2+2*X1+x2^2-2*x2
= x2^2- x1^2 +2*X1-2*x2
=(x2+x1)(x2-x1)-2(x2-x1)
=(x2-x1)(x2+x1-2)
\u56e0\u4e3ax1>x2
\u6240\u4ee5\u4e0a\u5f0f\u4e2dx2-x1<0
\u53c8\u56e0\u4e3ax1,x2>=1
\u6240\u4ee5x1+x2>=2 \u5373x1+x2-2>=0
\u7efc\u4e0a\uff1af(x1)-f(x2)=(x2-x1)(x2+x1-2)<=0
\u6240\u4ee5f(x)\u5728[1,+00]\u662f\u51cf\u51fd\u6570\u3002
2\uff09\u3001 \u75311\uff09\u53ef\u77e5x\u5c5e\u4e8e\u30102,5\u3011\u65f6\uff0cf(x)\u662f\u51cf\u51fd\u6570
\u6240\u4ee5f(x)\u7684\u6700\u5927\u503c\u7b49\u4e8ef(2)=-2^2+2*2=0
f(x)\u7684\u6700\u5c0f\u503c\u7b49\u4e8ef(5)=-5^2+2*5=-15
(1).\u8bbe1\uff1cX₁\uff1cX₂
f(X₁)-f(X₂)=x₁/(x₁-1)-x₂/(x₂-1)=(x₂-x₁)/\u3010(x₁-1)(x₂-1)\u3011>0
\u6240\u4ee5f(x)\u5728\u5b9a\u4e49\u57df\u5185\u662f\u51cf\u51fd\u6570
(2).\u56e0\u4e3af(x)\u5728\u5b9a\u4e49\u57df\u5185\u662f\u51cf\u51fd\u6570
f(x)max=f(2)=2
f(x)min=f(5)=5/4
=((x-1)+3)/(x-1)
=1+3/(x-1)
令x1＞x2＞1
f(x1)-f(x2)
=3/(x1-1)-3/(x2-1)
=(3x2-3-3x1+3)/(x1-1)(x2-1）
=3(x2-x1)/(x1-1)(x2-1)
∵x1＞x2＞1
∴x2-x1＜0
x1-1＞0
x2-1＞0
即3(x2-x1)/(x1-1)(x2-1)＜0
所以f（x）在（1.+∞〕上是减函数
因f（x）在（1.+∞〕上是减函数，且x属于[2.5]
所以
当x=2时，函数有最大值f(x)=1+3/(2-1)=4
当x=5时，函数有最小值f(x)=1+3/(5-1)=7/4
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