#### Question

f (x) = sin \[\frac{1}{x}\] for −1 ≤ x ≤ 1 Discuss the applicability of Rolle's theorem for the following function on the indicated intervals ?

#### Solution

The given function is \[f\left( x \right) = \sin\frac{1}{x}\] .

The domain of f is given to be \[\left[ - 1, 1 \right]\] .

It is known that \[\lim_{x \to 0} \sin\frac{1}{x}\] does not exist.

Thus, \[f\left( x \right)\] is discontinuous at *x* = 0 on \[\left[ - 1, 1 \right]\] .

Hence, Rolle's theorem is not applicable for the given function.

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Solution F(X) = Sin 1 X for −1 ≤ X ≤ 1 Discuss the Applicability of Rolle'S Theorem for the Following Function on the Indicated Intervals ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.