Let f(x) = |x-3| + |x+5| and A = {a in mathbb | vedclass

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(C) The function $f(x) = |x-3| + |x+5|$ is differentiable everywhere except at the points where the expressions inside the absolute values are zero,wh

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$1$

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(C) The function $f(x) = |x-3| + |x+5|$ is differentiable everywhere except at the points where the expressions inside the absolute values are zero,which are $x = 3$ and $x = -5$. Thus,the set $A$ is $\mathbb{R} \setminus \{-5, 3\}$. We are looking for real numbers $x$ such that $x \in (-\infty, -3) \cup (5, \infty)$ and $x 
otin A$. Since $A = \mathbb{R} \setminus \{-5, 3\}$,the condition $x 
otin A$ implies $x \in \{-5, 3\}$. Checking the intervals: For $x = -5$,$-5$ is not in $(-\infty, -3)$ because the interval is open at $-3$ and $-5 For $x = 3$,$3$ is not in $(-\infty, -3) \cup (5, \infty)$ because $3$ lies between $-3$ and $5$. Thus,the only value in the set $\{-5, 3\}$ that lies in $(-\infty, -3) \cup (5, \infty)$ is $-5$. Therefore,the number of such real numbers is $1$.

Let $f(x) = |x-3| + |x+5|$ and $A = \{a \in \mathbb{R} \mid \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \text{ exists} \}$. Then the number of real numbers which are in $(-\infty, -3) \cup (5, \infty)$ but not in $A$ is

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