设函数f(x)=x平方+1,g(x)=f[f(x)]先把 g(x) 的形式具体写出来 g(x) = f[f(x)] = [f(x)]^2 + 1 = (x^2 +1)^2 + 1 = x^4 + 2x^2 + 2 G(x) = g(x)-cf(x) = x^4 + 2x^2 + 2 - c(x^2 + 1) = x^4 + (2-c)x^2 + 2-c 配方 G(x) = x^4 + 2*[(2-c)/2] x^2 + [(2-
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设函数f(x)=x平方+1,g(x)=f[f(x)]先把 g(x) 的形式具体写出来 g(x) = f[f(x)] = [f(x)]^2 + 1 = (x^2 +1)^2 + 1 = x^4 + 2x^2 + 2 G(x) = g(x)-cf(x) = x^4 + 2x^2 + 2 - c(x^2 + 1) = x^4 + (2-c)x^2 + 2-c 配方 G(x) = x^4 + 2*[(2-c)/2] x^2 + [(2-
设函数f(x)=x平方+1,g(x)=f[f(x)]
先把 g(x) 的形式具体写出来
g(x) = f[f(x)] = [f(x)]^2 + 1 = (x^2 +1)^2 + 1
= x^4 + 2x^2 + 2
G(x) = g(x)-cf(x)
= x^4 + 2x^2 + 2 - c(x^2 + 1)
= x^4 + (2-c)x^2 + 2-c
配方
G(x) = x^4 + 2*[(2-c)/2] x^2 + [(2-c)/2]^2 - [(2-c)/2]^2 + (2-c)
= [x^2 + (2-c)/2]^2 + ……
这是一个偶函数.关于y轴对称.
这个二次函数的对称轴怎么确定?
设函数f(x)=x平方+1,g(x)=f[f(x)]先把 g(x) 的形式具体写出来 g(x) = f[f(x)] = [f(x)]^2 + 1 = (x^2 +1)^2 + 1 = x^4 + 2x^2 + 2 G(x) = g(x)-cf(x) = x^4 + 2x^2 + 2 - c(x^2 + 1) = x^4 + (2-c)x^2 + 2-c 配方 G(x) = x^4 + 2*[(2-c)/2] x^2 + [(2-
偶函数.关于y轴对称则对称轴为x=0;