5 Steps to a 5 AP Calculus AB 2019 - William Ma

(Marvins-Underground-K-12) #1

MA 3972-MA-Book May 9, 2018 10:9


Differentiation 125

Method 3: Use the slope of the line segment joining the points atx=2 andx=4.


f(2)=2 and f(4)= 5

m=
f(4)− f(2)
4 − 2

=


5 − 2


4 − 2


=


3


2


Note that

3


2


is the average of the results from methods 1 and 2.

Thus,f′(3)≈1, 2, or

3


2


depending on which line segment you use.

Example 2


Let f be a continuous and differentiable function. Selected values of f are shown below.
Find the approximate value off′atx=1.


x − 2 − 1 0 1 2 3
f 1 0 1 1.59 2.08 2.52

You can use the difference quotient


f(a+h)− f(a)
h
to approximatef′(a).

Leth=1; f′(1)≈
f(2)− f(1)
2 − 1


2. 08 − 1. 59


1


≈ 0. 49.


Leth=2; f′(1)≈
f(3)− f(1)
3 − 1


2. 52 − 1. 59


2


≈ 0. 465.


Or, you can use the symmetric difference quotient
f(a+h)− f(a−h)
2 h


to approximate
f′(a).

Leth=1; f′(1)≈
f(2)− f(0)
2 − 0


2. 08 − 1


2


≈ 0. 54.


Leth=2; f′(1)≈
f(3)− f(−1)
3 −(−1)


2. 52 − 0


4


≈ 0. 63.


Thus,f′(3)≈ 0 .49, 0.465, 0.54, or 0.63 depending on your method.
Note thatf is decreasing on (−2,−1) and increasing on (−1, 3). Using the symmetric
difference quotient withh=3 would not be accurate. (See Figure 7.4-2.)

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