f
(
x
)
=
x
3
−
9
x
2
+
24
x
−
10
Taking first derivative of
f
(
x
)
f
′
(
x
)
=
3
x
2
−
18
x
+
24
For finding critical points substituting
f
′
(
x
)
=
0
f
′
(
x
)
=
0
⇒
3
x
2
−
18
x
+
24
=
0
⇒
3
(
x
2
−
6
x
+
8
)
=
0
(
x
−
4
)
(
x
−
2
)
=
0
After solving the value of x is
x
=
2
,
4
Thus critical points at
x
=
2
,
4
