For mean value theorem
"By the MVT , since f(-1) = -1 and f(1) = 1 , there is c [-1.1] with f'(c) = 1.
Apart that the function is discontinious, how can f'(c) = 1?
Isnt f'(c) = f(b) - f(a) / b-a ?

Question

For mean value theorem
"By the MVT , since f(-1) = -1 and f(1) = 1 , there is c [-1.1] with f'(c) = 1. 
Apart that the function is discontinious, how can f'(c) = 1? 
Isnt f'(c) = f(b) - f(a) / b-a ?

zepdrix · Answer

$$\large \color{purple}{f(-1)=-1}, \qquad \color{forestgreen}{f(1)=1}$$
$$\large f'(c)=\frac{\color{forestgreen}{f(1)}-\color{purple}{f(-1)}}{1-(-1)}$$
$$\large =\frac{1-(-1)}{1-(-1)}=1$$Sorry for the slow response, I wanted to use the colors lolol

anonymous · Answer

LOL, neer know u can change colours.. 
and this statement is wrong because it is not defined at x=0 right?

zepdrix · Answer

Oh a discontinuity? :O hmm

anonymous · Answer

oh forgot to put up the function
f(x) = sqrt (x) if x > or equal to 0
f(x) = -sqrt(x) if x <0

anonymous · Answer

but actually..i think it is lol.. only possible reason that's wrong haha
thx