Let $\displaystyle f(x)=ax^2+bx+c, \ \ a,b,c \in \mathbb{R} \ \ \ f(-1),f(0),f(1) \in [-1,1]. $
Prove that $\displaystyle |f(x)|\leq\frac{3}{2} \ \forall \ x \in [-1,1]. $
Any ideas?
well im not to sure how you would do this without differentiation,
however the max values of the function are found at the global extrama, which are a member of [-1,1] so if we find the turning point by differentiation, it should be at x= -b/2a and then if we assume this is in [-1,1] and then show that the corresponding value of f is less than or equal to 3/2, i think this works
@hmmmm
There's no need for calculus, just write f as $\displaystyle a(x+\frac b{2a})^2 - \frac {b^2}{4a^2}+c$. But note that this is insufficient, since you are considering the absolute value of f and also it might get bigger at the endpoints (if a is positive, say). I don't know how to do this though.