Question

Is the function defined by

$f(x) = \begin{cases} x+5, & \quad \text{if } x{\leq 1}\\\ x-5, & \quad \text{if } x \text{>1} \end{cases}$

a continuous function?

Answer 1

The given function is
$f(x) = \begin{cases} x+5, & \quad \text{if } x{\leq 1}\\\ x-5, & \quad \text{if } x \text{>1} \end{cases}$
The given function f is defined at all the points of the real line.
Let c be a point on the real line.

Case (i):
If c<1, then f(c) = c+5 and $\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ f(x+5) = c+5
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x<1

Case (ii):
If c = 1, then f(1) = 1+5 = 6
The left hand limit of f at x = 1 is
$\lim\limits_{x \to 1^-}$f(x) = $\lim\limits_{x \to 1^-}$(x+5) = 1+5 = 6
The right hand limit of f at x = 1 is,
$\lim\limits_{x \to 1^+}$f(x) = $\lim\limits_{x \to 1^+}$(x-5) = 1-5 = -4
It is observed that the left and right hand limit of f at x=1 do not coincide.
Therefore,f is not continuous at x=1

** Case(III):**
Ifc>1, then f(c) = c-5 and
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (x-5) = c-5
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x>1

Thus, from the above observation, it can be concluded that x=1 is the only point of discontinuity of f.