Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If f(x)=x4-x3+7 x2+14, then what is the value of f'(5) ?
Q. If
f
(
x
)
=
x
4
−
x
3
+
7
x
2
+
14
, then what is the value of
f
′
(
5
)
?
2046
180
AP EAMCET
AP EAMCET 2020
Report Error
A
594
B
549
C
954
D
495
Solution:
Since,
f
(
x
)
=
x
4
−
x
3
+
7
x
2
+
14
Then,
f
′
(
x
)
=
4
x
3
−
3
x
2
+
14
x
∴
f
′
(
5
)
=
500
−
75
+
70
=
495