356
Chapter 4 | Applications of Derivatives
f ( x ) = x ⇒ f (9) = 9=3 f ′( x ) = 1 2 x ⇒ f ′(9) = 1 2 9 = 1 6
Therefore, the linear approximation is given by Figure 4.8 . L ( x ) =3+ 1 6 ( x −9) Using the linear approximation, we can estimate 9.1 by writing 9.1= f (9.1) ≈ L (9.1) =3+ 1 6
(9.1 − 9) ≈ 3.0167.
Figure 4.8 The local linear approximation to f ( x ) = x at x =9 provides an approximation to f for x near 9.
Analysis Using a calculator, the value of 9.1 to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate x , at least for x near 9. At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate 9.1. However, how does the calculator evaluate 9.1? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.
Find the local linear approximation to f ( x ) = x 3 at x =8. Use it to approximate 8.1 3
4.5
to five decimal
places.
Example 4.6 Linear Approximation of sin x Find the linear approximation of f ( x ) = sin x at x = π 3
and use it to approximate sin(62°).
Solution First we note that since π 3
rad is equivalent to 60°, using the linear approximation at x = π /3 seems
reasonable. The linear approximation is given by L ( x ) = f ⎛ ⎝ π 3 ⎞ ⎠ + f ′ ⎛ ⎝ π 3
⎞ ⎠
⎛ ⎝ x − π 3
⎞ ⎠ .
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