If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0.

Question

f(x) = x2 − 2x − 3

x + 4

,    [−1, 3]

c = 

 

Answer 1

Rolle's Theorem states there is a value 'c' in an open interval (a,b) such that if f(a) = f(b), then f'(c) = 0 (critical point or stationary point). So in this case f(x) = x² - 2x - 3.

Computation of f'(x) = 2x - 2 = 2(x-1)

Therefore, if f'(c) = 0 --> 2(c-1) = 0 ---> c = 1

Our interval is [-1,3] where c = 1 is within the interval in question

If we look at the second function, f(x) = x + 4 then we do have f(-1) != f(3) and as per the theorem, therefore there is no stationary point "c" to be found.