Consider the function f : R $\to$ R defined by
$$f(x) = \left\{ \matrix{
\left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr
0,\,\,x = 0 \hfill \cr} \right.$$. Then f is :
Solution
$$f(x) = \left\{ {\matrix{
{ - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \cr
0 & , & {x = 0} \cr
{\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \cr
} } \right.$$<br><br>$$f'(x) = \left\{ \matrix{
- x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \hfill \cr
x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0 \hfill \cr} \right.$$<br><br>= $$\left\{ \matrix{
- {1 \over x}\cos {1 \over x} + \sin {1 \over x} - 2,x < 0 \hfill \cr
{1 \over x}\cos {1 \over x} - \sin {1 \over x} + 2,x > 0 \hfill \cr} \right.$$
<br><br>$\therefore$ f'(x) is an oscillating function which is non-monotonic on ($-$$\infty$, 0) and (0, $\infty$).
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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