设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/28 04:42:19
设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
![设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c](/uploads/image/z/1473979-67-9.jpg?t=%E8%AE%BEf%28x%29%3Dax%26%23178%3B%2Bbx%2Bc+f%28x%2B1%29%2Bf%28x-1%29+%3D2ax%26%23178%3B%2B2bx%2B2a%2B2c)
f(x)=ax²+bx+c
故:f(x+1)+f(x-1)
=a(x+1)²+b(x+1)+c +a(x-1)²+b(x-1)+c
=2ax²+2bx+2a+2c
故:f(x+1)+f(x-1)
=a(x+1)²+b(x+1)+c +a(x-1)²+b(x-1)+c
=2ax²+2bx+2a+2c
设f(x)=ax²+bx+c f(x+1)+f(x-1) =2ax²+2bx+2a+2c
设函数f(x)=ax²+bx+c,且f(1)=-a/2
设函数F(X)=AX^2+BX+C(A>0),满足F(1-X)=F(1+X),
设函数f(x)=ax²+bx+c(a>0),且f(1)=-a/2
设f(x)=3ax+2bx+c,若a+b+c=0,f(0)f(1)>0
设函数f(x)=ax^2+bx+c((a≠0),满足f(x+1)=f(-x-3),且f(-2)>f(2),解不等式f(-
设函数f(x)=ax^2+bx+c(a>0),已知1/2
设函数f(x)=(ax^2+1)/(bx+c)是奇函数,a,b,c都是整数,且f(
设f(x)=(ax²+1)/(bx+c)是奇函数(a、b、c∈Z),且f(1)=2,f(2)
设f(x)=3ax²+2bx+c 若a+b+c=0,f(0)f(1)>0
设f(x)=3ax²+ 2bx+c,若a+b+c=0,f(0)f(1)>0,求证:
设f(x)=ax的平方+bx+c 且f(0)=f(2)