show that f(x) =x^is conditions and differentiate at x=0
Answers
Answered by
0
Answer:
see the explanation below
Step-by-step explanation:
To show that
f
(
x
)
=
|
x
|
is continuous at
0
, show that
lim
x
→
0
|
x
|
=
|
0
|
=
0
.
Use
ε
−
δ
if required, or use the piecewise definition of absolute value.
f
(
x
)
=
|
x
|
=
{
x
if
x
≥
0
−
x
if
x
<
0
So,
lim
x
→
0
+
|
x
|
=
lim
x
→
0
+
x
=
0
and
lim
x
→
0
−
|
x
|
=
lim
x
→
0
−
(
−
x
)
=
0
.
Therefore,
lim
x
→
0
|
x
|
=
0
which is, of course equal to
f
(
0
)
.
To show that
f
(
x
)
=
|
x
|
is not differentiable, show that
f
'
(
0
)
=
lim
h
→
0
f
(
0
+
h
)
−
f
(
0
)
h
does not exists.
Observe that
lim
h
→
0
|
0
+
h
|
−
|
0
|
h
=
lim
h
→
0
|
h
|
h
But
|
h
|
h
=
{
1
if
h
>
0
−
1
if
h
<
0
,
so the limit from the right is
1
, while the limit from the left is
−
1
.
So the two sided limit does not exist.
That is, the derivative does not exist at
x
=
0
.
Answered by
0
Answer:
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