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?

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- May 2nd 2010, 08:02 AMnahdumamaximum value of a polynomial
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? - May 2nd 2010, 03:36 PMhmmmm
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 - May 2nd 2010, 07:39 PMmaddas
@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. - May 2nd 2010, 07:59 PMhmmmm
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

- May 2nd 2010, 08:05 PMmaddas
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.

- May 2nd 2010, 08:11 PMhmmmm
yeah but i think that it works, its the only way i could think to do it

- May 2nd 2010, 08:28 PMmaddas
Ah, don't thank me, I totally screwed this up ;[