Example 1
Find the linear approximation of \( f(x) = \tan x \) for \( x \) close to 0.
Solution
First, compute the derivative:
\[
f'(x) = \sec^2 x
\]
Evaluate at \( x=0 \):
\[
f'(0) = \sec^2(0) = 1
\]
Thus, the linear approximation \( f_l(x) \) at \( x=0 \) is:
\[
f_l(x) = f(0) + f'(0)(x - 0) = 0 + 1 \cdot x = x
\]
This means \(\tan x \approx x\) for \( x \) close to 0, where \( x \) is measured in radians.
Try calculating \(\tan x\) and compare it with \(x\) for values:
\[
x = 0, 0.001, 0.01, 0.1, 0.2, 0.3, 0.5
\]
Use your calculator set in radians mode and observe the approximation.
Example 2
Find the linear approximation of \( f(x) = \ln x \) for \( x \) close to 1.
Solution
Compute the derivative:
\[
f'(x) = \dfrac{1}{x}
\]
Evaluate at \( x=1 \):
\[
f'(1) = 1
\]
The linear approximation \( f_l(x) \) at \( x=1 \) is:
\[
f_l(x) = \ln 1 + 1 \cdot (x - 1) = 0 + (x - 1) = x - 1
\]
So,
\[
\ln x \approx x - 1
\]
for \( x \) near 1.
Compare \(\ln x\) and \(x - 1\) for values:
\[
x = 1, 1.001, 1.01, 1.1, 1.5
\]
using your calculator.
Example 3
Find the linear approximation of \( f(x) = e^x \) for \( x \) close to 0.
Solution
The derivative is:
\[
f'(x) = e^x
\]
Evaluate at \( x=0 \):
\[
f'(0) = 1
\]
The linear approximation \( f_l(x) \) at \( x=0 \) is:
\[
f_l(x) = e^0 + 1 \cdot (x - 0) = 1 + x
\]
Compare \( e^x \) and \( 1 + x \) for values:
\[
x = 0, 0.001, 0.01, 0.1, 0.5
\]
using your calculator.
