Question

Let we have a function f\left( x \right)=\left\\{ \begin{aligned} & -1,-2\le x< 0 \\\ & {{x}^{2}}-1,0\le x\le 2 \\\ \end{aligned} \right. and $g\left( x \right)=\left| f\left( x \right) \right|+f\left( \left| x \right| \right)$. Then in the interval $\left( -2,2 \right)$,
g is:-
(a) Differentiable at all points
(b) Not differentiable at two points
(c) Not continuous
(d) Not differentiable at one point

Answer 1

Here since the interval breaks at ‘0’ we have to check the continuity and differentiability at ‘0’ only. Also, here there is a modulus of function, define $f\left( -x \right)$in the starting of the solution to avoid confusion in the middle. Then we use continuity formulas in the left and right neighbourhoods of ‘0’ to check the continuity of the function. And for the differentiability checking we use ${g}'\left( x \right)=\underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\dfrac{f\left( x \right)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}$ and check differentiability in both left and right neighbourhoods of ‘0’.

Complete step-by-step solution
In the question we have modulus of the function. So, let us define $f\left( -x \right)$.
We are given that

& -1,-2\le x< 0 \\\ & {{x}^{2}}-1,0\le x\le 2 \\\ \end{aligned} \right.$$ By substituting $$-x$$ in place of $$x$$ we get $$f\left( -x \right)=\left\\{ \begin{aligned} & -1,-2\le -x< 0 \\\ & {{x}^{2}}-1,0\le -x\le 2 \\\ \end{aligned} \right.$$ By multiplying negative sign in the limits the inequality changes as $$f\left( -x \right)=\left\\{ \begin{aligned} & -1,0\le x< 2 \\\ & {{x}^{2}}-1,-2\le x\le 0 \\\ \end{aligned} \right.$$ As we defined the function $$f\left( -x \right)$$, Let us go for continuity of $$g\left( x \right)$$ Let us take left neighbourhood of ‘0’ we will get $$\begin{aligned} & \Rightarrow g\left( {{0}^{+}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right) \\\ & \Rightarrow g\left( {{0}^{+}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left| f\left( x \right) \right|+\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( \left| x \right| \right).............equation(i) \\\ \end{aligned}$$ Here as $$x$$ is approaching to ‘0’ from left we can write $$\left| x \right|=-x$$. Here also from the left side of ‘0’ $$f\left( x \right)=-1$$. By substituting these values in equation (i) we get $$\Rightarrow g\left( {{0}^{-}} \right)=\left| -1 \right|+\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( -x \right)$$ From left of ‘0’ we have$$f\left( -x \right)={{x}^{2}}-1$$. By substituting in above equation we get $$\begin{aligned} & \Rightarrow g\left( {{0}^{-}} \right)=1+\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right) \\\ & \Rightarrow g\left( {{0}^{-}} \right)=1+\left( 0-1 \right) \\\ & \Rightarrow g\left( {{0}^{-}} \right)=0 \\\ \end{aligned}$$ Now let us go for right neighbourhood of ‘0’ we get $$\begin{aligned} & \Rightarrow g\left( {{0}^{+}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right) \\\ & \Rightarrow g\left( {{0}^{+}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left| f\left( x \right) \right|+\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( \left| x \right| \right)......equation(ii) \\\ \end{aligned}$$ Here as $$x$$ is approaching to ‘0’ from right we can write $$\left| x \right|=x$$. Here also from the right side of ‘0’ $$f\left( x \right)={{x}^{2}}-1$$. By substituting these values in equation (ii) we get $$\Rightarrow g\left( {{0}^{+}} \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left| {{x}^{2}}-1 \right|+\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$$ From the right of ‘0’ we have $$f\left( x \right)={{x}^{2}}-1$$. By substituting in above equation we get $$\begin{aligned} & \Rightarrow g\left( {{0}^{+}} \right)=1+\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{x}^{2}}-1 \right) \\\ & \Rightarrow g\left( {{0}^{+}} \right)=1+\left( 0-1 \right) \\\ & \Rightarrow g\left( {{0}^{+}} \right)=0 \\\ \end{aligned}$$ Here, we can say that right and left neighbourhoods of ‘0’ are defined and equals to finite value and also both are equal. So, we can say that $$g\left( x \right)$$ is continuous everywhere. Now let us go for left hand derivative of $$g\left( x \right)$$ as follows $$\Rightarrow LHD=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( 0 \right)}{x-0}.....equation(iii)$$ Let us find $$g\left( 0 \right)$$ as follows $$\begin{aligned} & \Rightarrow g\left( 0 \right)=\left| f\left( 0 \right) \right|+f\left( \left| 0 \right| \right) \\\ & \Rightarrow g\left( 0 \right)=\left| -1 \right|+1=0 \\\ \end{aligned}$$ By substituting this value in equation (iii) we get $$\Rightarrow LHD=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left| f\left( x \right) \right|+f\left( \left| x \right| \right)}{x}$$ Here as $$x$$ is approaching to ‘0’ from left we can write $$\left| x \right|=-x$$. Here also from the left side of ‘0’ $$f\left( x \right)=-1$$ and from left of ‘0’ we have $$f\left( -x \right)={{x}^{2}}-1$$. By substituting these values in above equation we get $$\begin{aligned} & \Rightarrow LHD=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{\left| -1 \right|+f\left( -x \right)}{x} \\\ & \Rightarrow LHD=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{1+{{x}^{2}}-1}{x} \\\ & \Rightarrow LHD=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,x=0 \\\ \end{aligned}$$ Now let us go for right hand derivative of $$g\left( x \right)$$ as follows $$\Rightarrow RHD=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{g\left( x \right)-g\left( 0 \right)}{x-0}.....equation(iv)$$ Let us find $$g\left( 0 \right)$$ as follows $$\begin{aligned} & \Rightarrow g\left( 0 \right)=\left| f\left( 0 \right) \right|+f\left( \left| 0 \right| \right) \\\ & \Rightarrow g\left( 0 \right)=\left| -1 \right|+1=0 \\\ \end{aligned}$$ By substituting this value in equation (iv) we get $$\Rightarrow RHD=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| f\left( x \right) \right|+f\left( \left| x \right| \right)}{x}$$ Here as $$x$$ is approaching to ‘0’ from right we can write $$\left| x \right|=x$$. Here also from the right side of ‘0’ $$f\left( x \right)={{x}^{2}}-1$$. By substituting these values in above equation we get $$\begin{aligned} & \Rightarrow RHD=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left| {{x}^{2}}-1 \right|+f\left( x \right)}{x} \\\ & \Rightarrow RHD=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\left( {{x}^{2}}-1 \right)+\left( {{x}^{2}}-1 \right)}{x} \\\ & \Rightarrow RHD=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,2x=0 \\\ \end{aligned}$$ **Here, as LHD and RHD are finite and equal we can say that $$g\left( x \right)$$ is differentiable everywhere. So, option (a) is the correct answer.** **Note:** We can check the differentiability by direct differentiation of$$g\left( x \right)$$ as $${g}'\left( x \right)=\dfrac{\left| f\left( x \right) \right|}{f\left( x \right)}.{f}'\left( x \right)+{f}'\left( \left| x \right| \right).\dfrac{\left| x \right|}{x}$$ and then check the continuity of $${g}'\left( x \right)$$. It will have the same result as before. Checking the differentiability with definition or by differentiating directly and checking continuity of that derivative will have the same result. Any method can be correct.