ISI B.Math Entrance(6)

The sign or signum function takes on three values as follows:

$\text{sgn }(x) = \begin{cases} -1 & \text{for } x < 0 \\ 0 & \text{for } x = 0 \\ 1 & \text{for } x > 0 \end{cases}$

Therefore,

$\begin{aligned} f(x) & = (x+1)\text{ sgn }(x^2-1) = (x+1) \times \begin{cases} 1 & \text{for } x < - 1 \\ 0 & \text{for } x = -1 \\ -1 & \text{for } -1 < x < 1 \\ 0 & \text{for } x = 1 \\ 1 & \text{for } x > 1 \end{cases} = \begin{cases} x+1 & \text{for } x < - 1 \\ 0 & \text{for } x = -1 \\ -(x+1) & \text{for } -1 < x < 1 \\ 0 & \text{for } x = 1 \\ x+1 & \text{for } x > 1 \end{cases} \end{aligned}$

Now, let us check the continuity of $f(x)$ at $x=-1$ and $x=1$ .

$\begin{cases} \displaystyle \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (x+1) = 0 \\ \displaystyle \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} -(x+1) = 0 \end{cases} \implies \displaystyle \lim_{x \to -1^-} f(x) \ {\color{#3D99F6} = } \lim_{x \to -1^+} f(x) \implies f(x)$ is $\color{#3D99F6}\text{continous}$ at $x=-1$ .

$\begin{cases} \displaystyle \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} -(x+1) = -2 \\ \displaystyle \lim_{x \to 1^+} f(x) = \lim_{x \to -1^+} (x+1) = 2 \end{cases} \implies \displaystyle \lim_{x \to 1^-} f(x) \ {\color{#D61F06} \ne } \lim_{x \to 1^+} f(x) \implies f(x)$ is $\color{#D61F06}\text{discontinous}$ at $x=1$ .

Therefore, there is only $\boxed{1}$ discontinuity in $f(x)$ .