If f(x)=|x-1|+|x+3|, then discuss the continuity and differentiability of

If $f\left(x\right)=|x-1|+|x+3|$, then discuss the continuity and differentiability of the function at x=-3 and x=1.
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au4gsf
Consider the given function
$f\left(x\right)=|x-1|+|x+3|$
First, check the continuity of the function at x=1
Since a function f(x) is continuous at x=a if
$\underset{x\to {a}^{-}}{lim}f\left(x\right)=\underset{x\to {a}^{+}}{lim}f\left(x\right)=f\left(a\right)$
Now check the continuity at the point x=1, then
$\underset{x\to {1}^{-}}{lim}f\left(x\right)=\underset{x\to {1}^{-}}{lim}\left(|x-1|+|x+3|\right)$
$=-\underset{x\to {1}^{-}}{lim}\left(x-1\right)+\underset{x\to {1}^{-}}{lim}\left(|x+3|\right)$
$=0+4$
$=4$
And
$\underset{x\to {1}^{+}}{lim}f\left(x\right)=\underset{x\to {1}^{+}}{lim}\left(|x-1|+|x+3|\right)$
$=\underset{x\to {1}^{+}}{lim}\left(x-1\right)+\underset{x\to {1}^{+}}{lim}\left(|x+3|\right)$
$=0+4$
$=4$
Therefore, the function is continuous at x=1
Now, check the continuity at the point x=-3, then
$\underset{x\to -{3}^{-}}{lim}f\left(x\right)=\underset{x\to -{3}^{-}}{lim}\left(|x-1|+|x+3|\right)$
$=\underset{x\to -{3}^{-}}{lim}\left(|x-1|\right)-\underset{x\to -{3}^{-}}{lim}\left(x+3\right)$
$=\left(|-3-1|\right)-\left(-3+3\right)$
$=|-4|-0$
$=4+0$
$=4$
And
$\underset{x\to -{3}^{+}}{lim}f\left(x\right)=\underset{x\to -{3}^{+}}{lim}\left(|x-1|+|x+3|\right)$
$=\underset{x\to -{3}^{+}}{lim}|x-1|+\underset{x\to -{3}^{+}}{lim}|x+3|$
$=\left(|-3-1|\right)+\left(-3+3\right)$
$=4+0$
$=4$
And
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