Question

The set points where $f(x)=\dfrac{x}{1+|x|}$ is differentiable is
(a) $(-\infty ,0)\cup (0,\infty )$
(b) $(-\infty ,1)\cup (-1,\infty )$
(c) $(-\infty ,\infty )$
(d)$(0,\infty )$

Answer 1

Hint: Separate the modulus part. Then find out the left hand derivative and right hand derivative. And compare them.

Complete step-by-step answer:
The given function is $f(x)=\dfrac{x}{1+|x|}.$
Here we can observe a modulus function in the denominator, so we can rewrite this function as,

\dfrac{x}{1+x},x\ge 0 \\\ \dfrac{x}{1-x},x<0 \\\ \end{matrix} \right.$$ Now we will check the continuity and differentiability at $$x=0$$. First of all let us check the continuity at $$x=0$$. We know for a function f(x) to be continuous at $$x=0$$ its left hand limit (LHL) should be equal to right hand limit (RHL). So, let us consider the LHL first. $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{1-x}$$ Applying the limits, we have $$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1-0}=0$$ So, $LHL=0........(i)$ Now, we will find the RHL. $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{x}{1+x}=\dfrac{0}{1+0}=0$$ Applying the limits, we have $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\dfrac{0}{1+0}=0$$ So, $RHL=0........(ii)$ From equation (i) and (ii), we have LHL = RHL So, the given function f(x) is continuous at $$x=0$$. Now, we shall check the differentiability. We know for a function to be differentiable at $$x=0$$ its left hand derivative (LHD) should be equal to its right hand derivative (RHD). First, we can find LHD. $${{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{d}{dx}\left( \dfrac{x}{1-x} \right)$$ For differentiating this function we will use the quotient rule, i.e., $$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$$ So the LHD becomes, $$\begin{aligned} & {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-x)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1-x)}{{{(1-x)}^{2}}} \\\ & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1-x)(1)-x(-1)}{{{(1-x)}^{2}}} \\\ & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1-x+x}{{{(1-x)}^{2}}} \\\ & \Rightarrow {{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{(1-x)}^{2}}} \\\ \end{aligned}$$ Now, applying the limit, we have $${{\underset{x\to {{0}^{-}}}{\mathop{\lim f}}\,}^{'}}(x)=\dfrac{1}{{{(1-0)}^{2}}}=1$$ $$\therefore LHD=1.......(iii)$$ Now, we will find the RHD. $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{d}{dx}\left( \dfrac{x}{1+x} \right)$$ Again, applying the quotient rule, we have $$\begin{aligned} & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1+x)\dfrac{d}{dx}(x)-x\dfrac{d}{dx}(1+x)}{{{(1+x)}^{2}}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{(1+x)(1)-x(1)}{{{(1+x)}^{2}}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1+x-x}{{{(1+x)}^{2}}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{1}{{{(1+x)}^{2}}} \\\ \end{aligned}$$ Now, by applying the limits we get, $$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}(x)=\dfrac{1}{{{(1+0)}^{2}}}=1$$ $$\therefore RHD=1.......(iv)$$ So, from equation (iii) and (iv), we see that $LHD=RHD$ . Hence, f(x) is differentiable at $x=0$. Hence, we conclude that the given function is differentiable from $$+\infty $$ to $$-\infty $$. Hence, the correct answer is option (c). Answer is Option (c). Note: For finding the left hand and right hand derivative we can use the formula, $\begin{aligned} & LHD=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a-x)-f(a)}{-h} \\\ & RHD=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(a+x)-f(a)}{h} \\\ \end{aligned}$ Using these formulas also we will get the same result.