Part II: The Answers
298
Answers
401–500
Setting the first factor equal to zero gives you cos x = − 1
2
, which has the solutions
x = 2
3
π and x = 4
3
π . Setting the second factor equal to zero gives you cos x = 1, which
has the solutions x = 0 and 2π.
Next, determine whether the function is increasing or decreasing on the intervals
0 2
3
, π
( )
, 2
3
4
3
π
π
,
(
) , and 4
3
2
π π
,
(
) by taking a point inside each interval and substituting it
into the derivative to see whether it’s positive or negative. Using the points π
3
, π, and
3
2
π gives you
′ ( ) = − ( ) + ( ) +
= − + ( ) + >
f π
π
π
3
4
3
2
3
2
1 2 1
2
2 0
2
cos
c os
′
= −
+
+
= − − +
= − <
f ( )
(cos )
cos
π
π
π
4
2
2
4 2 2
4 0
2
′ ( ) = − ( ) + 3
( ) +
= >
f 3
2
4
3
2
2
2
2
2 0
2
π
π
π
cos
c os
So f
 
(x) has a local maximum when x = 2
3
π because the function changes from increasing
to decreasing at this value, and f
 
(x) has a local minimum when x = 4
3
π because the
function changes from decreasing to increasing at this value.
To find the points on the original function, substitute in these values. If x = 2
3
π , then
f 2
3
2
2
3
2 2
3
2 3
2
3
2
3 3
2
π
π
π
( ) ( ) ( )

 

 

 

 

 

 
=
−
=
− −
=
sin
s in
So the local maximum occurs at 2
3
3 3
2
π ,





 . If x = 4
3
π , then
f 4
3
2
4
3
2 4
3
2
3
2
3
2
3 3
2
π
π
π
( ) ( ) ( )

 

 

 

 
=
−
=
−
−
= −
sin
s in
Therefore, the local minimum occurs at 4
3
3 3
2
π , −





 .