Part II: The Answers
316
Answers
401–500
Because f
 
(8) – f
 
(4) = 4f
 
'(c), it follows that
8
8
4 24
≤
−
≤
f
f
( ) ( )
489.
2
9 25
12
3
<
<
The given function f x
x
x
( ) =
=
1 3
3
is continuous and differentiable on the given interval. By the mean value theorem, there exists a number c in the interval [8, 9] such that
′
f c
( ) =
−
−
=
−
9
8
9 8
9 2
3
3
3
But because ′
f x
x
( ) =
−
1
3
2 3
, you have ′
f c
c
( ) =
−
1
3
2 3
. Replacing ′ ( )
f c in the
equation ′
f c
( ) =
−
−
=
−
9
8
9 8
9 2
3
3
3
gives you
1
3
9 2
2 3
3
c
−
=
−
Because 1
3
0
2 3
c
−
> , you have 9 2 0
3
− > , so
9 2
3
>
Notice also that f
 
' is decreasing (f
 
" will be negative on the given interval), so the
following is true:
′
′
f c
f
( )
( )
( )
( )
( )
<
=
=
=
=
−
8
1
3
8
1
3 8
1
3 4
1
12
2 3
2 3
Because 1
12
9 2
3
>
=
−
′
f c
( )
, you have 9 2 1
12
25
12
3
< +
= . Therefore, it follows that
2
9 25
12
3
<
<
490.
velocity: 2; acceleration: 2
Begin by taking the derivative of the position function s(t) to find the velocity function:
′
s t
v t
t
( )
( )
=
= −
2 8
Substituting in the value of t = 5, find the velocity:
v( )
( )
5 2 5 8 2
=
− =
Next, take the derivative of the velocity function to find the acceleration function:
′′
′
s t
v t
a t
( )
( )
( )
=
=
= 2
Because the acceleration is a constant at t = 5, the acceleration is a(5) = 2.