verify Rolle's theorem for the function
Answers
How do you verify whether rolle's theorem can be applied to the function
f
(
x
)
=
1
x
2
in [-1,1]?
Calculus Differentiable vs. Non-differentiable Functions
1 Answer
Steve M
Dec 24, 2016
Answer:
f
(
x
)
does not satisfy the conditions of Rolle's Theorem on the interval
[
−
1
,
1
]
, as
f
(
x
)
is not continuous on the interval
Explanation:
Rolle's Theorem states that if a function,
f
(
x
)
is continuous on the closed interval
[
a
,
b
]
, and is differentiable on the interval, and
f
(
a
)
=
f
(
b
)
, then there exists at least one number
c
, in the interval such that
f
'
(
c
)
=
0
.
enter image source here
So what Rolle's Theorem is stating should be obvious as if the function is differentiable then it must be continuous (as differentiability
⇒
continuity), and if its is continuous and
f
(
a
)
=
f
(
b
)
then the curve must chge direction (at least once) so it must have at least one minimum or maximum in the interval.
With
f
(
x
)
=
1
x
2
with
x
∈
[
−
1
,
1
]
, then we can see
f
(
1
)
=
1
=
f
(
−
1
)
, so our quest to verify Rolle's Theorem is to find a number
c
st
f
'
(
c
)
=
0
.
Differentiating wrt
x
we have,
f
'
(
x
)
=
−
2
x
−
3
=
−
2
x
3
To find a turning point we require;
f
'
(
x
)
=
0
⇒
−
2
x
3
=
0
Which has no finite solution. We can conclude that
f
(
x
)
does not satisfy the conditions of Rolle's Theorem on the interval
[
−
1
,
1
]
, so it must be that
f
(
x
)
is not continuous in that interval.
We can see that this is the case graphically, as f(x) has a discontinuity when
x
=
0
∈
[
−
1
,
1
]
:
graph{1/x^2 [-10, 10, -2, 10]}
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