Verify Rolle,s theorem on f(x) = x sin (1/x), x ∈ [-1,1].

1.	Verify Rolle,s theorem on f(x) = x sin (1/x), x ∈ [-1,1].
Answer» Given function is f(x) = x sin (1/x), x ∈ [-1,1] f(x) = 0 Now, lim_{x → 0}f(x) = lim_{x → 0} x sin (1/x) = (0 x a) a number between 1 and -1 = 0 = f'(0) ∴ f(x) is continuous at x = 0 Again, left hand derivative L f'(0) = lim_{h → 0} (f(0 - h) - f(0))/-h = lim_{h → 0} (-h sin(-1/h) - 0)/-h = lim_{h → 0}(-sin(1/h)) Which does not exist. ∴ The given function is not derivable at x = 0 So, Rolle's theorem is not applicable.

Answer»

Given function is

f(x) = x sin (1/x), x ∈ [-1,1]

f(x) = 0

Now, lim_{x → 0}f(x) = lim_{x → 0} x sin (1/x) = (0 x a) a number between 1 and -1

= 0 = f'(0)

∴ f(x) is continuous at x = 0

Again, left hand derivative L f'(0) = lim_{h → 0} (f(0 - h) - f(0))/-h

= lim_{h → 0} (-h sin(-1/h) - 0)/-h = lim_{h → 0}(-sin(1/h))

Which does not exist.

∴ The given function is not derivable at x = 0

So, Rolle's theorem is not applicable.

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