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  • The set of all points, where the function  <img alt="\mathrm{f(x)=\frac{x}{1+|x|}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%28x%29%3D%5Cfrac%7Bx%7D%7B1&plus

The set of all points, where the function  \mathrm{f(x)=\frac{x}{1+|x|}}   differentiable, is

Option: 1

(-\infty, \infty)


Option: 2

(0, \infty)


Option: 3

(-\infty, 0) \cup(0, \infty)


Option: 4

none of these.


Answers (1)

best_answer

We have,  \mathrm{ f(x)=\frac{x}{1+|x|}=\frac{g(x)}{h(x)}} , where  \mathrm{h(x)=1+|x|}  Clearly g(x) is differentiable on \mathrm{ (-\alpha ~domain ~of ~h(x)~is ~(-\infty, \infty)} . Since |x| is not different x=0, therefore h(x) is not differntiable at x=0. Thu: differntiable on \mathrm{(-\infty, 0) \cup(0, \infty)} . For x=0, we have

\mathrm{\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{h-0}=\lim _{h \rightarrow 0} \frac{\frac{1+|x|}{x}}{x}=\lim _{h \rightarrow 0} \frac{1}{1+|x|}=1 .}

Therefore, f(x) is differntiable at x=0. Hence, f is differentiable on \mathrm{(-\infty, \infty) .} 

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manish

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