# Thread: maximum value of a polynomial

1. ## maximum value of a polynomial

Let $f(x)=ax^2+bx+c, \ \ a,b,c \in \mathbb{R} \ \ \ f(-1),f(0),f(1) \in [-1,1].$
Prove that $|f(x)|\leq\frac{3}{2} \ \forall \ x \in [-1,1].$

Any ideas?

2. 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

3. @hmmmm

There's no need for calculus, just write f as $a(x+\frac b{2a})^2 - \frac {b^2}{4a^2}+c$. 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.

4. ok but why does this show that |f(x)| is always less than 3/2, however if you used the turning point is -b/2a and sub this in for x we can show that the inequality holds, and we have our endpoints from the question

5. Oh, I see. Sorry, I missed the condition on the endpoints. Its not a pretty substitution though. You still have to use the conditions to show that the value at the vertex will be less than 3/4.

6. yeah but i think that it works, its the only way i could think to do it

7. Ah, don't thank me, I totally screwed this up ;[