Playing with Graphs - 8

$L(x)=h(|x|)+|h(x)| \\ L(x)=2-g(|x|)+|2-g(x)| \\ L(x)=2-|1-f(|x|)|+|2-|1-f(x)|| \\ L(x)=2-|1-|x||+|2-|1-x||$

---

$\text{It is rare to use the Derivative to } |x| , \text{This is how it's done} \\ \text{Let , } \quad v=|x| \implies v^2=x^2 \\ 2v\dfrac{dv}{dx}=2x \implies \dfrac{dv}{dx}=\dfrac{x}{v} \\ \boxed{\dfrac{dv}{dx}=\dfrac{x}{|x|}} \\ \text{: 2nd Method :} \\ \text{Use the defenition of } |x| = \sqrt{x^2} \\ f(x)=\sqrt{x^2} \implies f^{'}(x)=\dfrac{1}{2\sqrt{x^2}}\cdot2x \\ \boxed{f^{'}(x)=\dfrac{x}{|x|}}$

---

$\text{Now we calculate the derivative of } L(x) \\ L^{'}(x)=0-\dfrac{(1-|x|)}{|1-|x||}\cdot\dfrac{x}{|x|}\cdot -1+\dfrac{2-|1-x|}{|2-|1-x||}\cdot\dfrac{-(1-x)}{|1-x|}\cdot-1 \\ L^{'}(x)=\dfrac{x(1-|x|)}{|x|\cdot|1-|x||}+\dfrac{(1-x)\cdot(2-|1-x|)}{|1-x|\cdot|2-|1-x||} \\ \text{Possible Points where Derivative is not defined : } \quad x=-1,0,1,3 \\ \text{LHD: Left hand derivative} \\ \text{RHD: Right hand derivative} \\ x=-1 \quad : \quad \text{LHD }= \displaystyle\lim_{ x \to -1^{-ve} }L^{'}(x)=0 \quad , \quad \text{RHD }= \displaystyle\lim_{ x \to -1^{+ve} }L^{'}(x)=0 \\ x=0 \quad : \quad \text{LHD }= \displaystyle\lim_{ x \to 0^{-ve} }L^{'}(x)=0 \quad , \quad \text{RHD }= \displaystyle\lim_{ x \to 0^{+ve} }L^{'}(x)=2 \\ x=1 \quad : \quad \text{LHD }= \displaystyle\lim_{ x \to 1^{-ve} }L^{'}(x)=2 \quad , \quad \text{RHD }= \displaystyle\lim_{ x \to -1^{+ve} }L^{'}(x)=-2 \\ x=3 \quad : \quad \text{LHD }= \displaystyle\lim_{ x \to 3^{-ve} }L^{'}(x)=-2 \quad , \quad \text{RHD }= \displaystyle\lim_{ x \to 3^{+ve} }L^{'}(x)=0 \\ \text{Points where function is non-differentiable are : } x=0,1,3$

---

$\text{I loved This problem, Keep posting :) }$

$\text{In my opinion , this Question should belong to the class of Level 5 }$

At x = 0,1,3 ..... :)