作业帮设函数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-](/uploads/image/z/13320811-19-1.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx%E5%B9%B3%E6%96%B9%2B1%2Cg%28x%29%3Df%5Bf%28x%29%5D%E5%85%88%E6%8A%8A+g%28x%29+%E7%9A%84%E5%BD%A2%E5%BC%8F%E5%85%B7%E4%BD%93%E5%86%99%E5%87%BA%E6%9D%A5+g%28x%29+%3D+f%5Bf%28x%29%5D+%3D+%5Bf%28x%29%5D%5E2+%2B+1+%3D+%28x%5E2+%2B1%29%5E2+%2B+1+%3D+x%5E4+%2B+2x%5E2+%2B+2+G%28x%29+%3D+g%28x%29-cf%28x%29+%3D+x%5E4+%2B+2x%5E2+%2B+2+-+c%28x%5E2+%2B+1%29+%3D+x%5E4+%2B+%282-c%29x%5E2+%2B+2-c+%E9%85%8D%E6%96%B9+G%28x%29+%3D+x%5E4+%2B+2%2A%5B%282-c%29%2F2%5D+x%5E2+%2B+%5B%282-)
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;