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Mathematics
Mathematics
Problem
Let
f
(
x
)
=
{
x
+
2
,
x
<
1
∣
x
−
1
∣
+
3
,
x
≥
1
.
f(x)=\begin{cases}x+2,&x<1\\|x-1|+3,&x\ge 1\end{cases}.
f
(
x
)
=
{
x
+
2
,
∣
x
−
1∣
+
3
,
x
<
1
x
≥
1
.
Determine the
x
x
x
-value(s) where
f
′
(
x
)
f'(x)
f
′
(
x
)
does not exist. a)
x
=
1
x=1
x
=
1
b)
x
=
−
1
x=-1
x
=
−
1
c) No real values of
x
x
x
d)
x
=
0
x=0
x
=
0
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