f+f+x+4x-1

荆喻包15321.已知f(x)是一次函数且f(f(x))=4x - 1.求f(x).2.已知f(x)是二次函数,若f(0)=1且f(x+1)=f(x)+2x,求f(x). -
乜肢之17560675890 ______[答案] (1)设f(X)=aX+b,所以f(f(X))=a(aX+b)+b = 4X-1,得a=2,b=-(1/3)或a=-2,b=1,所以f(X)=2X-(1/3)或f(X)=-2X+1 (2)因为f(0)=1且f(X)为二次函数,所以设f(X)=aX^2+bX+1,又因为 f(x+1)=f(x)+2x,带入f(X)=aX^2+bX+1,化简并比较系数,得a=1,b=-1,所...

荆喻包1532已知f(2x - 1)=4x^2+4x - 1,求f(x)的表达式及定义域和值域,并画出f(x)的图像 -
乜肢之17560675890 ______[答案] f(2x-1)=4x^2+4x-1=[(2x-1)^2+4x-1]+4x-1 =(2x-1)^2+8x-2 =(2x-1)^2+[4(2x-1)+4]-2 =(2x-1)^2+4(2x-1)+2 f(x)=x^2+4x+2=(x+2)^2-2 x的定义域为(-∞,∞) y的值域[-2,∞) 图像为一开口向上的抛物线: 顶点坐标为(-2,-2), 与x轴的两个交点为(-2-根号2,0)...

荆喻包1532函数f(x)= - x2+4x - 1在[t,t+1]上的最大值为g(t),则g(t)的最大值 -
乜肢之17560675890 ______ 首先由配方得: f(x)=-x2+4x-1=-(x-2)^2+3...

荆喻包1532已知,f(2x+1)=4x - 1,求f(x),f(x+1) -
乜肢之17560675890 ______ f(2x+1)=4x-1令t=2x+1,则x=(t-1)/2f(t)=4[(t-1)/2]-1=2t-3f(x)=2x-3f(x+1)=2x-1

荆喻包1532f(4x+1)=x²,求f(x)
乜肢之17560675890 ______ f(4x+1)=(4x+1)^2/16-1/16-8x/18=(4x+1)^2/16-1/16-2(4x+1)/16+2/16=(4x+1)^2/16-2(4x+1)/16+1/16=[(4x+1)-1]^2f(x)=(x-1)^2/16 f(4x+1)=(4x+1)^2/16-1/16-8x/18=(4x+1)^2/16-1/16-2(4x+1)/16+2/16=(4x+1)^2/16-2(4x+1)/16+1/16=[(4x+1)-1]^2/16 f(x)=(x-1)^2/16

荆喻包1532f(x+1)+f(2x+1)=4x方+5x -
乜肢之17560675890 ______ ∵f(x+1)+f(2x+1)=4x²+5x∴假设f(x)=ax²+bx+c∴f(x+1)=a(x+1)²+b(x+1)+c=ax²+(2a+b)x+a+b+c f(2x+1)=a(2x+1)²+b(2x+1)+c=4ax²+(4a+2b)x+a+b+c∴f(x+1)+f(2x+...

荆喻包1532f(x+2)=x^3+2x+1 求f(x) -
乜肢之17560675890 ______ f(x+2)=x^3+2x+1 令t=x+2 那么x=t-2 所以f(t)=(t-2)^3+2(t-2)+1=t^3-6t^2+14t-11 所以f(x)=x^3-6x^2+14x-11 如果不懂,请追问,祝学习愉快!

荆喻包1532f(x - 1)=x2+2x+2求f(x^2+1)
乜肢之17560675890 ______ f(x-1)=(x-1)²+4x+1 f(x²+1)=(x²+1)²+4x+1=x^4+2x²+1+4x+1=x^4+2x²+4x+2 噢我写错了 f(x-1)=(x-1)²+4x+1=(x-1)²+4x-4+5=(x-1)²+4(x-1)+5 所以f(x²+1)=(x²+1)²+4(x²+1)+5 =x^4+2x²+1+4x²+4+5 =x^4+6x²+10

荆喻包1532关于函数1.已知f(x^2 - x)=2x^2+2x+6,求f(x)解析式.(用配凑法)2.已知f(1+2x)=x^2 - 4x - 1,求f(x)解析式.(用换元法)3.已知f(x+1)=ax+b且f(0)=1,f(1)=2,求f(x)解... -
乜肢之17560675890 ______[答案] 问题3:由题得,f(0)=1即当X=-1时,f(-1+1)=-a+b=1 f(1)=2即当x=0时.f(0+1)=b,即b=2 所以a=1 所以f(x+1)=X+2=(x+1)+1 所以f(x)=x+1 问题2:设t=1+2x,则x=(t-1)/2 代入已知f(t)=(t-1)^2 /2 -4{(t-1)/4}-1 化简得f(t)=(t^2-6t+5)/4 所以f(x)=(x^2-6x+5)/4

荆喻包1532设f(x+2)=x^2+1则f(x - 1)=______我也看答案这有步看不懂,为什么f(x)=x^2 - 4x+5呢按理,把f(x+2)代入x^2+1,应该是f(x)=x^2+4x+5 -
乜肢之17560675890 ______[答案] f(x+2)=x^2+1=x^2+4x+4+1-4x-4=(x+2)^2-4(x+2)+5 所以f(x)=x^2-4x+5 所以f(x-1)=(x-1)^2-4(x-1)+5=x^2-6x+10