Answers and Explanations 299
Answers
401–500
465.
local maxima at −
−
+
11
6
11
6
3
π
π
,
(
) , π π
6 6
3
, +
(
) ; local minima at − − −
7
6
7
6
3
π
π
,
(
) ,
5
6
5
6
3
π π
,
−
(
)
Begin by finding the derivative of the function f
 
(x) = x + 2 cos x:
′
f x
x
( )
sin
= −
1 2
Next, find the critical numbers by setting the function equal to zero and solving for x
on the given interval:
1 2
0
1
2
11
6
7
6 6
5
6
−
=
=
= −
−
sin
sin
,
, ,
x
x
x
π
π π π
Then determine whether the function is increasing or decreasing on the intervals
− −
2
11
6
π
π
,
(
) , − −
11
6
7
6
π
π
,
(
) , − 7
6 6
π π
,
( ) , π π
6
5
6
,
( ) , and 5
6
2
π π
,
( ) by taking a point inside
each interval and substituting it into the derivative. Using the value x = –6.27, which is
slightly larger than –2π, gives you f
 
'(–6.27) ≈ f
 
'(–2π) = 1 > 0. Likewise, using the points
−3
2
π , 0, π
2
, and 3
2
π from each of the remaining four intervals gives you the following:
′ −
( ) = − −
( ) = − <
′
= −
= >
′ ( ) = −
f
f
f
3
2
1 2
3
2
1 2 0
0 1 2
0 1 0
2
1 2
π
π
π
sin
( )
sin( )
sin n
sin
π
π
π
2
1 2 0
3
2
1 2
3
2
3 0
( ) = − <
′ ( ) = − ( ) = >
f
Therefore, the function has local maxima when x = − 11
6
π and x = π
6
and local minima
when x = 7
6
π and x = 5
6
π . To find the points on the original function, substitute these x
values into the original function:
f −
= −
+
−
= −
+
= −
+
11
6
11
6
2
11
6
11
6
2
3
2
11
6
3
π
π
π
π
π
( )
( )






cos
f −
= − +
−
= − +
−
= − −
7
6
7
6
2
7
6
7
6
2
3
2
7
6
3
π
π
π
π
π
( )
( )






cos