Question

Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples? (i)f(x)=[x] for x∈[5,9] (ii)f(x)=[x]for x∈[-2,2](iii)x^-1 for x∈[1,2]

Answer 1

$\textbf{\underline{By Rolle’s Theorem, for a function}}$ f : [a, b] → R, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f(a) = f(b)
Then there exists some c in (a, b) such that f '(c) = 0.

(i) f(x) = [x] for x∈[5,9]
here [ • ] denotes greatest integer.
we know, greatest integer function is a piece-wise function. hence, f(x) = [x] is not continuous for x∈[5,9]
and we know, if function is discontinuous then it can't be differentiable. hence, f(x) = [x] for x∈(5,9) is not differentiable.
also f(5) = 5 and f(9) = 9
e.g., f(5) ≠ f(9)
hence, f(x) does not satisfy the conditions of Rolle’s Theorem.

(ii) it is also a greatest integer function , I mean piece - wise function. hence, f(x) = [x] for x∈[-2,2] is not continuous.
we know, function is differentiable when
$\bf{lim_{h^+\to\ 0}\frac{f(x+h)-f(x)}{h}=lim_{h^-\to 0}\frac{f(x+h)-f(x)}{h}}$

but here, $lim_{h^+\to 0}\frac{[x+h]-[x]}{h}=0$
and $lim_{h^-\to 0}\frac{[x+h]-[x]}{h}=\infty$
it is clear $\bf{lim_{h^+\to\ 0}\frac{f(x+h)-f(x)}{h}\neq lim_{h^-\to 0}\frac{f(x+h)-f(x)}{h}}$ hence, it is not differentiable.
also f(-2) = -2, f(2) = 2
e.g.,. f(-2) ≠ f(2) .
hence, f(x) does not satisfy the conditions of Rolle’s Theorem.

(iii) f(x) = 1/x for x∈[1,2]
we know, f(x) = 1/x is not defined at x = 0
because x = 0 doesn't exist in domain of f(x)=1/x. but given interval [1,2] , f(x) is continuous.

f'(x) = -1/x² , hence f(x) is differentiable in (1,2)

now, f(1) = 1/1² = 1 and f(2) = 1/2² = 1/4
e.g., f(1) ≠ f(2)
Here, f(x) does not satisfy a condition of Rolle’s Theorem.
Rolle’s Theorem is not applicable for f(x) = 1/x for x ∈ [1, 2].