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 applied to function f, on the set [-1,1].
First of all, in order to make possible the practicability of Rolle's theorem, we have to consider that the function is continue. For this reason, the left limit of the function has to have the same value with the right limit of the function and has to have the same value with the value of function in the point x=0.
left lim f(x)=left lim (x^2+mx+n)=n
The function has to have derivative in x=0
lim (f(x)-f(0))/(x-0)=lim( x^2+mx+4-4)/x=0/0
We have to deal with an indetermination problem, so we'll apply L'Hospital rule:
lim ( x^2+mx+4-4)/x=lim ( x^2+mx+4-4)' /x'=lim(2x+m)=m
lim (px^2+4x+4-4)/x=0/0=lim (px^2+4x+4-4)' / x'=lim (2px+4)/1
The Rolle's rule says that the interval's heads f(1)=f(-1)
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
Therefore to hold Roll's theorem,
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