Firstly, we know that f(x)=ax2+bx+1.
The conditions can be rewritten as inequalities for f(−1) and f(2):
−1≤a(−1)2+b(−1)+1≤1
−2≤a(2)2+b(2)+1≤2
Let's simplify these inequalities:
1. For f(−1):
−1≤a−b+1≤1
This simplifies to:
−2≤a−b≤0
2. For f(2):
−2≤4a+2b+1≤2
This simplifies to:
−3≤4a+2b≤1
Next, we examine both inequalities together. We need to find constraints that satisfy both conditions simultaneously:
{−2≤a−b≤0−3≤4a+2b≤1
We can add these inequalities together by eliminating b, which gives us:
(−2)+(−3)≤(a−b)+(4a+2b)≤0+1
−5≤5a+b≤1
Now, we can add the second inequality from the original set of constraints to obtain a new inequality:
−3≤4a+2b≤1
−5≤5a+b≤1
Add these two together:
−8≤(5a+b)+(4a+2b)≤2
−8≤9a+3b≤2
Finally, to find the range of f(3), we plug in x=3 into our function:
f(3)=a(3)2+b(3)+1
f(3)=9a+3b+1
Knowing that −8≤9a+3b≤2, we can find the range for f(3):
−8+1≤9a+3b+1≤2+1
−7≤9a+3b+1≤3
Thus, the range of f(3) is within the interval [−7,3].