5 Steps to a 5 AP Calculus BC 2019

Differentiation 91

Method 3: Using 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 6.4-2.)