Answers and Explanations 305
Answers
401–500
473.
no inflection points
Being by finding the first and second derivatives of the function f x
x
x
( )
sin
cos
= +
1
.
The first derivative is
′
=
+
−
−
+
=
+
+
f x
x
x
x
x
x
x
x
x
( )
( cos )cos
sin ( sin )
( cos )
cos
cos
sin
(
1
1
1
2
2
2
+ +
=
+
+
= +
= +
−
cos )
cos
( cos )
cos
( cos )
x
x
x
x
x
2
2
1
1
1
1
1
1
And the second derivative is
′′
f x
x
x
x
x
( )
( cos ) ( sin )
sin
( cos )
= − +
−
= +
−
1 1
1
2
2
Next, find where the second derivative is equal to zero or undefined. Setting the
numerator equal to zero gives you sin x = 0, which has solutions x = 0, x = π, and x = 2π.
Setting the denominator equal to zero gives you 1 + cos x = 0, or cos x = –1, which has
the solution x = π.
To determine the concavity on the intervals (0, π) and (π, 2π), pick a point from each interval and substitute it into the second derivative to see whether it’s positive or negative.
Using the values x = π
2
and x = 3
2
π gives you the following:
′′ ( ) = ( )
+
(
)
>
′′ ( ) = ( )
+
(
)
<
f
f
π
π
π
π
π
π
2
2
1
2
0
3
2
3
2
1
3
2
0
2
2
sin
cos
sin
cos
However, note that at x = π, the original function is undefined, so there are no inflection
points. In fact, if you noticed this at the beginning, there’s really no need to determine
the concavity on the intervals!
474.
(π, 0)
Begin by finding the first and second derivatives of the function f
 
(x) = 3 sin x – sin
3 
x.
The first derivative is
′
=
−
=
−
(
)
=
(
)
=
f x
x
x
x
x
x
x
x
( )
cos
(sin ) (cos )
cos
sin
cos cos
co
3
3
3
1
3
3
2
2
2
s s
3 x