application of differentiation (easy).

b555's method is really nice. Although I had actually solved in a different way (Gordo's proposal.)
As can be seen from the illustration----- the parabola that will contain the maximum slope will have the vertex at (0,1) and will pass through
(1,-1) and (-1,-1) OR will have a vertex at (0,-1) and will pass through
(-1,1) and (1,1)

Let us take the first one ( one could also take the 2nd one)

Equation:-

x² = -4a(y-1)
It passes through (1,-1) and (-1,-1)
Putting the values, we get 4a = 1/2.
So, x² = -(1/2)(y-1)
|dy/dx|=4|x|
|x|≤1
So, maximum of |dy/dx| = |2ax+b|= 4.

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!!

i cant do anything dude. just solved it way i could.

lets wait for metal`s solution

Utna bhi difficult nahin hai-------- try to karo.
1 - 2 lines mein ho jayega , iff you think correctly :-)

f(x)=ax²+bx+c
|f(x)|≤1

f'(x)=2ax+b

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

thts all

√b²-4ac