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Question
Mathematics
Let f: R arrow R be defined by f(x)=x3+3 x+1 and g is the inverse of f then the value of g prime prime(5) is equal to
Q. Let
f
:
R
→
R
be defined by
f
(
x
)
=
x
3
+
3
x
+
1
and
g
is the inverse of
f
then the value of
g
′′
(
5
)
is equal to
1548
117
Continuity and Differentiability
Report Error
A
36
−
1
B
6
−
1
C
6
1
D
36
1
Solution:
g
(
5
)
=
1
and
f
(
1
)
=
5
Now,
g
′′
(
5
)
=
−
[
f
′
(
g
(
x
))
]
3
f
′′
(
g
(
x
))
⇒
g
′′
(
5
)
=
−
[
f
′
(
g
(
5
))
]
3
f
′′
(
g
(
5
))
=
−
[
f
′
(
1
)
]
3
f
′′
(
1
)
f
′
(
x
)
=
3
x
2
+
3
,
f
′
(
1
)
=
6
f
′′
(
x
)
=
6
x
,
f
′′
(
1
)
=
6
g
′′
(
5
)
=
216
−
6
=
36
−
1