dse 2010 doubt

[Nam]

Suppose the function f: R-> R is given by f(x)=x^3-3x+b. Find the number of points in closed interval [-1,1] at which f(x)=0.
1)none
2)at most one
3)one
4)at least one

[Akshay Jain]

In the interval [-1,1] the derivative of the function 3x^2-3 is less then equal to 0 which means that there can be atmost 1 x belongs the interval such that f(x)=0 (atmost 1 root)
the existence will depend on the value of constant b

[Arushi :))]

Also  b cant be 0 because then value of function at -1 and 1 would be 0 and the function is decreasing in given interval  so it can't reach b = 0
so the it depends on the value of b and the function an have atmost one root

[Akshay Jain]

the graph of function x^3-3x is like this....the constant b will just shift the graph up along y axis when b>0 and will shift the graph downward along y axis when b<0. when b does not belong to [-2,2] then there will be no root in interval [-1,1]

[Arushi :))]

ohh sorryyy f (1) < 0  and f (-1) > 0 so only root at  x= 0 whn b = 0

[Nam]

thnx guys for such a prompt reply...