MATH 147
Lab Key #7
3/29/2016
(1) Consider the function f ( x ) = 1x on the domain x ∈ [1, 2] .
(a) Find the average rate of change of f from x = 1 to x = 2.
(b) Find a value c ∈ (1, 2) such that f 0 (c) is the average rate of change and
explain why such a number must exist.
Solution:
(a) The average rate of change of f from x = 1 to x = 2 is given by
f (2) − f (1)
=
2−1
1
2
−
1
1
1
=
1
1
−1 = − .
2
2
(b) We want to know where f 0 (c) = − 12 . Now, f 0 ( x ) = − x12 . Hence, we
need to find a solution to
1
1
− 2 =− .
2
c
c2 = 2.
√
c = ± 2.
However, we’re only√interested in a value on the interval (1, 2) . Therefore, the
correct value is c = 2.
Note that f is continuous on [1, 2] and is differentiable on (1, 2) . Therefore,
by the Mean Value Theorem, there is a c ∈ (1, 2) such that f 0 (c) is the average
rate of change of f ( x ) on [1, 2] .
(2) Estimate the value of (3.9)3 using a linear approximation. Express your
answer in decimal form.
Solution: We want to estimate f (3.9) , where f ( x ) = x3 . We will use linear
approximation at x = 4, to do this. Note that f (4) = 43 = 64. Also, f 0 ( x ) =
3x2 . Therefore, f 0 (4) = 3 · 42 = 48. Hence,
f ( x ) ≈ 48 ( x − 4) + 64.
∴ f (3.9) ≈ 48 (−0.1) + 64 = −4.8 + 64.
∴ (3.9)3 ≈ 59.2.
1
(3) Find all absolute extrema of f ( x ) = 14 x4 + 13 x3 on [−2, 2] .
Solution: We will need to test the values of f at the endpoints and at critical
values. First, we will find the critical values of f . Note that f 0 ( x ) = x3 + x2 .
The critical values are where f 0 = 0 or where it does not exist, but since it is a
polynomial, it exists everywhere. Therefore, we need to solve
x3 + x2 = 0.
x2 ( x + 1) = 0.
x = 0 or x = −1.
Both of these critical values are on [−2, 2] . Hence, we need to know the values
of f (−2) , f (−1) , f (0) , f (2) .
x
f (x)
8
8
12
8
4
-2 14 (−2)4 + 13 (−2)3 = 16
4 − 3 = 4− 3 = 3 − 3 = 3
4
3
1
1
3
4
1
1
1
-1
4 (−1) + 3 (−1) = 4 − 3 = 12 − 12 = − 12
4
3
1
1
0
4 (0) + 3 (0) = 0
4
3
1
8
8
12
8
20
1
16
2
4 (2) + 3 (2) = 4 + 3 = 4 + 3 = 3 + 3 = 3
Hence, f has an absolute maximum of 20
3 at x = 2 and an absolute mini1
mum of − 12
at x = −1 on [−2, 2] .
2
(4) The growth rate of a plant depends on the amount of resources available.
A simple and frequently used model for resource-dependent growth is the
Monod model, according to which the growth rate is equal to
f ( R) =
aR
,
k+R
R ≥ 0,
where R denotes the resource level and a and k are positive constants. Determine the intervals on which the growth rate is increasing or decreasing.
Solution: For this we need to first find the critical values.
f 0 ( R) =
d
dR
[ aR] (k + R) − aR ·
(k + R)
d
dR
[k + R]
2
=
a (k + R) − aR
(k + R)
2
=
ak
( k + R )2
.
Now, we need to know where f 0 ( R) = 0 and where it doesn’t exist. It doesn’t
exist, when R = −k, but R ≥ 0 and k > 0, so f 0 ( R) exists on R ≥ 0. Now, we
need to solve:ak = 0. But this is not the case for any R. Hence, the equation has
no critical points. Hence, we need to calculate the sign the derivative at any
x-value in the interval.
Note that
ak
a
f 0 (0) = 2 = > 0.
k
k
Hence, f 0 (0) is positive. Therefore, f is increasing on [0, ∞) and decreasing
nowhere.
3