The given function is f(x)=|x|-|x+1|.
The two functions,g and h,are defined as
g(x)=|x| and h(x)=|x+1|
Then,f =g−h
The continuity of g and h is examined first
g(x)=|x| can be written as
\(g(x)=\left\{\begin{matrix} -x &if\,x<0 \\ x&if\,x\geq 0 \end{matrix}\right.\)
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I:
If c<0,then g(c)=-c and \(\lim_{x\rightarrow c}\) g(x)=\(\lim_{x\rightarrow c}\)(-x)=-c
∴\(\lim_{x\rightarrow c}\)g(x)=g(c)
Therefore,g is continuous at all points x, such that x<0
Case II:
If c>0,then g(c)=c and\(\lim_{x\rightarrow c}\) g(x)=\(\lim_{x\rightarrow c}\)x=c
∴\(\lim_{x\rightarrow c}\)g(x)=g(c)
Therefore,g is continuous at all points x, such that x>0
Case III:
If c=0,then g(c)=g(0)=0
\(\lim_{x\rightarrow 0^-}\) g(x)=\(\lim_{x\rightarrow 0^-}\)(-x)=0
\(\lim_{x\rightarrow 0^+}\) g(x)=\(\lim_{x\rightarrow 0^+}\)(x)=0
∴\(\lim_{x\rightarrow 0^-}\)g(x)=\(\lim_{x\rightarrow 0^+}\)(x)=g(0)
Therefore, g is continuous at x = 0
From the above three observations, it can be concluded that g is continuous at all points.
h(x)=|x+1| can be written as
h(x)={-(x+1),if x<-1
x+1,if x≥-1
Clearly,h is defined for every real number
Let c be a real number.
Case I:
If c<-1,then h(c)=-(c+1) and \(\lim_{x\rightarrow c}\) h(x)=\(\lim_{x\rightarrow c}\)[(-x+1))=-(c+1)
∴\(\lim_{x\rightarrow c}\)h(x)=h(c)
Therefore,h is continuous at all points x, such that x<-1
Case II:
If c>-1,then h(c)=c+1 and \(\lim_{x\rightarrow c}\) h(x)=\(\lim_{x\rightarrow c}\)(x+1)=c+1
∴\(\lim_{x\rightarrow c}\)h(x)=h(c)
Therefore,h is continuous at all points x,such that x>-1
Case III:
If c=-1,then h(c)=h(-1)=-1+1=0
\(\lim_{x\rightarrow 1^-}\) h(x)=\(\lim_{x\rightarrow 1^-}\)(-(x+1))=-(-1+1)=0
\(\lim_{x\rightarrow 1^+}\) h(x)=\(\lim_{x\rightarrow 1^+}\)(x+1)=(-1+1)=0
∴\(\lim_{x\rightarrow 1^-}\)h(x)=\(\lim_{x\rightarrow 1^+}\)h(x)=h(-1)
Therefore,h is continuous at x=-1
From the above three observations, it can be concluded that h is continuous at all points of the real line.
g and h are continuous functions. Therefore, f=g−h is also a continuous function. Therefore,f has no point of discontinuity.
Sports car racing is a form of motorsport which uses sports car prototypes. The competition is held on special tracks designed in various shapes. The equation of one such track is given as 
(i) Find \(f'(x)\) for \(0<x>3\).
(ii) Find \(f'(4)\).
(iii)(a) Test for continuity of \(f(x)\) at \(x=3\).
OR
(iii)(b) Test for differentiability of \(f(x)\) at \(x=3\).
Let $\alpha,\beta\in\mathbb{R}$ be such that the function \[ f(x)= \begin{cases} 2\alpha(x^2-2)+2\beta x, & x<1 \\ (\alpha+3)x+(\alpha-\beta), & x\ge1 \end{cases} \] is differentiable at all $x\in\mathbb{R}$. Then $34(\alpha+\beta)$ is equal to}