We have the function f:[-1,1]->R, f(x)=x^2+mx+n, for x in [-1,0) or f(x)=px^2+4x+4, for x in [0,1].Find m,n,p real, so that Rolle's theorem to be...

Roll's theorem: If f(x) is a continuous and derivable function  in a closed interval (a,b) and f(a)=f(b), then there exists a  c  in the interval (a, b) for which f'(c) = 0

For the given function;

f(-1) = (-1)^2+m(-1)+n=1-m+n.

Condition of continuity holds everywhere, but it should hold at the special point, 0. So,

f(-0) = n= f(0+)=4.  Therefore, n=4

For the condition of differentiablity  holds good every where but at the special point ,0 it should satisfy:

f '(-0)= f '(0+): (x^2+mx+4)'=9px^2+4x+4) gives :m=4

f(1)=p(1)^2+4(1)+4 =p+8.

Therefore to hold Roll's theorem,

f(-1)=f(1):

1-m+4 =1-4+4= p+8  or p=-7.

Now, f'(c) = 2c+m=0 or 2pc+4 =0 for some c in [-1,1] to reduces to:

2x+4= 0=>c=-4/2=-2. But c = -2 is not in [-1,  1]

2(-7)c+4=0 implies c= 4/14=2/7. c =2/7 belongs to [-1,1].

Therefore, m=4,n=5 and p=-7 are the real values that makes Roll's theorem applicable to the given function f(x) defined  in [-1, 1]  and there exists a   c in accordance with Roll such that c=2/7 in  [-1, 1] for which f '(2/7)=0