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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54
174 Chapter 2 • The Derivative and Its Properties
EXAMPLE 7 Finding an Equation of a Tangent Line
√
(a) Find the derivative of f (x) = 2x at x = 8.
(b) Use the derivative f (8) to find an equation of the tangent line to the graph
′
of f at the point (8, 4).
Solution
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(a) The derivative of f at 8 is
√ √ √
f (x) − f (8) 2x − 4 2x − 4 2x + 4
f (8) = lim = lim = lim √
′
x→8 x − 8 ↑ x→8 x − 8 ↑ x→8 (x − 8) 2x + 4
√
y f (8) = 2 · 8 = 4 Rationalize
the numerator.
2x − 16 2(x − 8) 2 1
6 1
y !x " 2 = lim √ = lim √ = lim √ =
4 x→8 (x − 8) 2x + 4 x→8 (x − 8) 2x + 4 x→8 2x + 4 4
1
(8, 4) f (x) 2x (b) The slope of the tangent line to the graph of f at the point (8, 4) is f (8) = . Using
′
4
4
the point-slope form of a line, we get
′
y − 4 = f (8)(x − 8) y − y = m(x − x 1 )
2 1
1 1
′
y − 4 = (x − 8) f (8) =
4 4
1
4 8 12 x y = x + 2
4
Figure 6 The graphs of f and the tangent line to the graph of f at (8, 4) are shown in Figure 6.
R
NOW WORK Problem 15 and AP Practice Problem 4.
Approximating the Derivative of a Function Represented
EXAMPLE 8
by a Table
The table below lists several values of a function y = f (x) that is continuous on the
interval [−1, 7] and has a derivative at each number in the interval (−1, 7). Approximate
the derivative of f at 2.
x 0 1 2 4 6
f (x) 0 3 12 32 72
Solution
There are several ways to approximate the derivative of a function defined by a table.
Each uses an average rate of change to approximate the rate of change of f at 2, which
is the derivative of f at 2.
• Using the average rate of change from 2 to 4, we have
f (4) − f (2) 32 − 12
= = 10
4 − 2 2
With this choice, f (2) is approximately 10.
′
• Using the average rate of change from 1 to 2, we have
f (2) − f (1) 12 − 3
= = 9
2 − 1 1
With this choice, f (2) is approximately 9.
′
• A third approximation can be found by averaging the above two approximations.
10 + 9 19
Then f (2) is approximately = .
′
2 2
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