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Mathematics
Mathematics
Problem
If
f
(
x
)
=
{
e
x
;
x
≤
0
∣
1
−
x
∣
;
x
>
0
f(x) = \begin{cases} e^x; & x \leq 0 \\ |1-x|; & x > 0 \end{cases}
f
(
x
)
=
{
e
x
;
∣1
−
x
∣
;
x
≤
0
x
>
0
, then (a)
f
(
x
)
f(x)
f
(
x
)
is differentiable at
x
=
0
x=0
x
=
0
(b)
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
0
x=0
x
=
0
(c)
f
(
x
)
f(x)
f
(
x
)
is differentiable at
x
=
1
x=1
x
=
1
(d)
f
(
x
)
f(x)
f
(
x
)
is not continuous at
x
=
1
x=1
x
=
1
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