application of differentiation (easy).

ANSWER IS 4. WAIT POSTING SOLUTION

yep got the concept..thanks guys..actually posted an incomplete soln earlier..here's how i did it:

let f(x)=y=ax²+bx+c
=>ax²+bx+c-y=0
now
x=[-b±√b²-4a(c-y)]/2a
=> 2ax+b=±√b²-4a(c-y)
=> |2ax+b| = √b²-4a(c-y)

from f(1)=f(-1)=1 and f(0)=-1
a=2, b=0, c=-1

hence substituting these values, we get the answer as 4

yeah, i mean, by basic picture of a quadratic polynomial, we can say that /f'(x)/ is greatest at the end pionts of the domain in which u observe the function...now we can imagine that the derivative of the function at 1 or -1 can be high if, the graph is more dipped, and it is more evenly spaced ( i mean the dip point is not more to the left nor more to the right..) so to strike a combination of both these, we can get it when the dip point is at the origion ie the mid pt of 1 and -1..., and f(1)=f(-1), dis is further favoured if f(1) , f(-1) and the dip ie f(0) are as large as possible in magnitude, (in this case happens to be 1), so we get this way that our eqn. can be 2x²-1....thus f'(x)max=f'(1)=4...what b555 has done is mathamatical, what i have done is purely graphical...
cheers!!

srry but i dint get the logic

post # 5 : i did in brain. so that terrible.

f(x)=ax²+bx+c
|f(x)|≤1

f'(x)=2ax+b

so we need to find |f'(x)|

thts all

thinking on complicated line s! sorry for that

I dont think so. In fact the maximum value will eb far lesser than 24.
You guys are thinking on complicated lines, it seems.

neil.dhruv you're not correct.

√b²-4ac