Thread: maximum value of a polynomial

Let
Prove that

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 . 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.

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

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.

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

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