Question

带绝对值的不等式证明f(x)=ax^2+bx+c 当|x|小于等于1,总有|f(1)|

Answer 1

f(x)=ax^2+bx+c

a+b+c=f(1)

a-b+c=f(-1)

c=f(0)

解得

a=[f(1)+f(-1)]/2-f(0)

b=[f(1)-f(-1)]/2

c=f(0)

|4a+2b+c|

=|2f(1)+2f(-1)-4f(0)+f(1)-f(-1)+f(0)|

=|3f(1)+f(-1)-3f(0)|

≤|3f(1)|+|f(-1)|+|3f(0)|

≤7