f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/25 23:10:07
f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方f(x)=5x
f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方
f(x)=5x^2-4x-3
f(x)=-3x^2+5x-8
配方
f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方
f(x)=5x²-4x-3=5(x²-4x/5)-3=5[(x-2/5)²-4/25]-3=5(x-2/5)²-4/5-3=5(x-2/5)²-19/5
f(x)=-3x²+5x-8=-3(x²-5x/3)-8=-3[(x-5/6)²-25/36]-8=-3(x-5/6)²+25/12-8=-3(x-5/6)²-71/12.
解:
f(x)=5x^2-4x-3
=5(x^2-4/5x+(4/10)^2-(4/10)^2)-3
=5(x-4/10)^2-5(4/10)^2 -3
f(x)=-3x^2+5x-8
=-3(x^2-5/3x+(5/6)^2-(5/6)^2) -8
=-3(x-5/6)^2 +3(5/6)^2 -8
5（x-2/5）^2-19/5
-3(X-5/6)^2-71/12
f(x)=5x^2-4x-3
= 5(x²-(4/5)x+4/25)-3-4/5
= 5(x-2/5)²-19/5
f(x)=-3x^2+5x-8
= -3(x²-(5/3)x+(25/36))-8+25/12
= -3(x-5/6)²-71/12
① f(x)为一次函数,且f[f(x)]=1+4x,求f(x)② f(x)+2f（-x）=3x+x平方 ,求f(x)③ f(x)为一次函数,且f(x+1)+f(x-1)=2x平方-4x+4,求f(x)④ f(2x-1)定义域（-1,5],求f(2-5x)定义域,求f(x)定义域⑤ f(x)定义域[0,2] ,求f(x平方)
f(x+1)+f(x-1)=4x^3-2x求f(x)
数学f(x)=x(x+1)(x+2)(x+3)(x+4)(x+5),求f'(0)=?
已知f(x)=x(x-1)(x-2)(x-3)(x-4)(x-5),则f‘(0)为
f(x+2)>=f(x)+2,f(x+3)
f(x)=5x+3 f(x)=5x f(x)=x+2x+1 f(x)=5x+3 f(x)=5x f(x)=x+2x+1
若函数f(x)满足f(x)+2f(1/x)=5x+4/x,则f(x)=
f(X)=f(X+2)(x
已知f(x)满足2f(x)+f(1/x)=3x,求f(x)
F(X)满足F(x)+2f(x分之1)=3X,求f(x)
已知f(x)满足2f(x)+f(1/x)=3x,求f(x)?
已知f(x)满足2f(x)+f(-x)=-3x+1,求f(x)
已知f(x)满足f(x)+2f(1/x)=3x,求f(x) ,
x-2 ,X>=0 f(x)=f[f(x+5)],x分段函数f(x)= x-2 ,X>=0 f[f(x+5)],x
设f(x)=(2x+5)^3(3x-10)^4,求f'(x)
设f(x)=(2x+5)^2*(3x-1)^4,求f'(x)
f(x)=5x^2-4x-3f(x)=-3x^2+5x-8配方
已知f(x)=(x-1)(x-2)(x-3)(x-4)(x-5),则f'(1)=?求过程