已知f(x-2)=ax^2-(a-3)x=a-2(其中a为负整数)函数f(x)过点(m-2,0)已知f(x-2)=ax^2-(a-3)x+a-2(a为负整数)函数f(x)过点(m-2,0)(m属于R)1.求f(x)解析式设q(x)=f(f(x)),F(x)=p-q(x)+f(x),是否存
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 03:51:19
![已知f(x-2)=ax^2-(a-3)x=a-2(其中a为负整数)函数f(x)过点(m-2,0)已知f(x-2)=ax^2-(a-3)x+a-2(a为负整数)函数f(x)过点(m-2,0)(m属于R)1.求f(x)解析式设q(x)=f(f(x)),F(x)=p-q(x)+f(x),是否存](/uploads/image/z/7241566-22-6.jpg?t=%E5%B7%B2%E7%9F%A5f%28x-2%29%3Dax%5E2-%28a-3%29x%3Da-2%28%E5%85%B6%E4%B8%ADa%E4%B8%BA%E8%B4%9F%E6%95%B4%E6%95%B0%EF%BC%89%E5%87%BD%E6%95%B0f%28x%29%E8%BF%87%E7%82%B9%EF%BC%88m-2%2C0%29%E5%B7%B2%E7%9F%A5f%28x-2%29%3Dax%5E2-%28a-3%29x%2Ba-2%28a%E4%B8%BA%E8%B4%9F%E6%95%B4%E6%95%B0%EF%BC%89%E5%87%BD%E6%95%B0f%28x%29%E8%BF%87%E7%82%B9%EF%BC%88m-2%2C0%29%28m%E5%B1%9E%E4%BA%8ER%291.%E6%B1%82f%EF%BC%88x%EF%BC%89%E8%A7%A3%E6%9E%90%E5%BC%8F%E8%AE%BEq%EF%BC%88x%EF%BC%89%3Df%EF%BC%88f%EF%BC%88x%EF%BC%89%EF%BC%89%EF%BC%8CF%EF%BC%88x%EF%BC%89%3Dp-q%EF%BC%88x%EF%BC%89%2Bf%EF%BC%88x%EF%BC%89%EF%BC%8C%E6%98%AF%E5%90%A6%E5%AD%98)
已知f(x-2)=ax^2-(a-3)x=a-2(其中a为负整数)函数f(x)过点(m-2,0)已知f(x-2)=ax^2-(a-3)x+a-2(a为负整数)函数f(x)过点(m-2,0)(m属于R)1.求f(x)解析式设q(x)=f(f(x)),F(x)=p-q(x)+f(x),是否存
已知f(x-2)=ax^2-(a-3)x=a-2(其中a为负整数)函数f(x)过点(m-2,0)
已知f(x-2)=ax^2-(a-3)x+a-2(a为负整数)函数f(x)过点(m-2,0)(m属于R)
1.求f(x)解析式
设q(x)=f(f(x)),F(x)=p-q(x)+f(x),是否存在实数p(p
已知f(x-2)=ax^2-(a-3)x=a-2(其中a为负整数)函数f(x)过点(m-2,0)已知f(x-2)=ax^2-(a-3)x+a-2(a为负整数)函数f(x)过点(m-2,0)(m属于R)1.求f(x)解析式设q(x)=f(f(x)),F(x)=p-q(x)+f(x),是否存
(1)
函数y=f(x)的图象过点M(m-2,0)
f(m-2)=am^2-(a-3)m+(a-2)=0
m∈R,上面关于m方程有解
△=(a-3)^2-4a(a-2)=-3a^2+2a+9>=0
3a^2-2a-9<=0
(1-2√7)/3<=a<=(1+2√7)/3
a为负整数,a=-1
f(x-2)=-x^2+4x-3
f(x)=-(x+2)^2+4(x+2)-3=-x^2+1
(2)
q(x)=f[f(x)]=-f(x)^2+1=-(-x^2+1)^2+1=-x^4+2x^2
F(x)=p-q(x)+f(x)=x^4+3x^2 +p+1
F'(x)=4x^3 +6x
令F'(x)=0 ,x=0或√6/2或-√6/2
与p无关,所以不存在p.
(是不是你的题目抄错了,可能是F(x)=p*q(x)+f(x))
451