Rolle\'s Theoreme Suppose : a) f is continuous on a closed interval [a; b] b) f
ID: 2861572 • Letter: R
Question
Rolle's Theoreme
Suppose :
a) f is continuous on a closed interval [a; b]
b) f is di erentiable on the open interval (a; b)
c) f (a) = f (b)
Then, there is at least one number c in the open interval (a; b) such
that f'(c) = 0
Case 1: f (x) = f (a) for all x in (a; b)
Since f (a) = f (x) = f (b) for all x in (a; b), then f is a constant
function, so f'(c) = 0 for every c in (a; b)
Case 2: f (x) > f (a) for at least some x in (a,b):
Since f is continuous on [a,b] , f has a maximum on [a,b] , and
because f (x) > f (a) = f (b) for some x in (a,b) by assumption, f has
a maximum in (a,b).
So, f has a local maximum on (a,b). Because f is di erentiable on
(a,b), f'(c) = 0 for some c in (a; b)
Case 3: f (x) < f (a) for at least some x in (a,b):
Since f is continuous on [a,b] , f has a minimum on [a,b] , and
because f (x) < f (a) = f (b) for some x in (a,b) by assumption, the
minimum occurs in (a,b).
So, f has a local minimum on (a,b). Because f is diferentiable on
(a,b), f'(c) = 0 for some c in (a; b) as in case 2.
Based on this proof answer the following questions. Note: Your explanations must be more detailed than the proof themselves! Use your own words!
1) Why are there only three cases to consider (and not more)?
(b) why can't f (a) or f (b) be the maximum?
(c) why must f'(c) = 0 for some c in (a,b)?
Explanation / Answer
A)Because f(x) is a real number and by law of triconami there are only 3 cases.
2 a)by borel's theorem every continuous function on closed bounded interval attains its boundaries.
So our f is continuous on closed [a b], so it has to attain its boundaries.
Hence f has maximum in [a b].
B) uniqueness of boundaries f(a) or f(b) can't be maximum.
C) if f(c) is a maximum locally, c must be a critical point of f in (a b).
Hence it follows.
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