f(x)=(x-1) (x2+2)最大值
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函数在x=1处连续则lim(x→1+)(x^2+2x-3)/(x-1)=a*1+1=a+1lim(x→1+)(x^2+2x-3)/(x-1)=lim(x→1+)(x+3)(x-1)/(x-1)=lim
f(2)=3/5,f(1/2)=-3/5,则f(2)/f(1/2)=-1.
已知f(x)是偶函数,g(x)是奇函数,且f(x)+g(x)=x2+x-2.求f(x),g(x)的解析式.因为:f(x)是偶函数所以:f(-x)=f(x)因为:g(x)是奇函数所以:f(-x)=-f(
换元而已,函数的左边你换成了f(t),那么右边自然也要换,x=(t-1)/2就是要代入右边的等式中换掉x
f(x-1)=x2-3x+2=(x-1)(x-2)=(x-1)(x-1-1)所以f(x)=x(x-1)所以f(x+1)=(x+1)(x+1-1)=x(x+1)=x²+x
令X=1则:f(3)=1+1=2设2x+1=t则,x=(t-1)/2于是,f(t)=[(t-1)/2]+(t-1)/2=t/4-1/4∴f(x)=x/4-1/4f(x+1)=(x+1)/4-1/4=x
f(x)=-2x+1(x2),根据分段函数一层一层求解.f(5/2)=2*5/2-1=4(f(x)=2x-1(x>2))4-5=-1f(-1)=-3(f(x)=-3(-1≤x≤2))f(-3)=-2*
设:(x+1)/(x)=t,则:x=1/(t-1)则:f(t)=1+(1/x)+(1/x)².=1+(t-1)+(t-1)².=t²-t+1即:f(t)=t²-
令a=2x-1x=(a+1)/2所以f(a)=[(a+1)/2]²+8=(a²+2a+33)/4所以f(x)=(x²+2x+33)/4
f(1+2x)=(1+x^2)/x^2=1/x^2+1令t=1+2x,则x=(t-1)/2代入上式得:f(t)=4/(t-1)^2+1因此有f(x)=4/(x-1)^2+1
f(x-1)=x2+2x,令x-1=t,x=1+t则f(t)=(1+t)^2+2(1+t)=t^2+4t+3f(1/x)=(1/x)^2+4/x+3再答:不客气,望采纳
令x/(x+1)=tx=xt+tx(1-t)=tx=t/(1-t)则f(x/(x+1))=f(t)=[t/(1-t)]2-2t/(1-t)-1当t=0时f(0)=[0/(1-0)]2-2x0/(1-0
不等式左边=[2^x1+2^x2]/2>2根号(2^x1*2^x2)/2=根号2^(x1+x2){因为x1不等于x2,所以等号取不到}不等式右边=2^[(x1+x2)/2]=根号2^(x1+x2)得证
不妨设x>0,则-x
是奇函数首先它的定义域是全体实数R,关于原点对称.对任意的x属于R,f(-x)=-2x/(x^2+1)=-f(x),所以是奇函数.
f(x)=1/3x-2f(x2)=1/3x²-2f(x+1)=1/3(x+1)-2=1/3x-5/3
f(x)=x2-2x用X/2=T得出X=2T带入就可以了,然后再把T换成X
f(2x-1)=4x^2-10x+4f[f(x)]=x^4-6x^3+6x^2+9x
1若m=0F(X)=-8X1克(X)=0,显然有问题意思不一致若m0为x=0米2的范围内的溶液.乘用(2^1/32+1)=原来的公式(2^1/321)和(2^1/32-1)*(2^1/16+1)*(2
2f(x^2)+f(1/x^2)=x则2f(t)+f(1/t)=根号t用t=1/x^2带入得到2f(1/x^2)+f(x^2)=1/x与2f(x^2)+f(1/x^2)=x联立得3f(1/x^2)=2