设函数f(x)=[(x+1)²+sinx]/x²+1 求函数的最大值和最小值
人气:281 ℃ 时间:2019-10-19 04:18:22
解答
f`(x)=((2(x+1)+cosx)x^2-((x+1)^2+sinx)2x)/(x^4)+1=(2x+2+cosx-(x+1)^2-sinx+x^2)/x^2=(1+cosx-sinx)/x^2.
f`(x)=0,得当x=2kπ+π时,f(x)可取到最大值1+2/π+1/π^2.
当x=2kπ+3π/2时,f(x)可取到最小值2+4/3π.
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