Let f(x)=tan(x) 1. Calculate the linear approximation for f at x=0. Show work 2. Sketch the graph of f and the obtained linear approximation 3. Use this linear approximation to calculate tan(0.1) approximately. Show work
Let f(x)=tan(x) 1. Calculate the linear approximation for f at x=0. Show work 2. Sketch the graph of f and the obtained linear approximation 3. Use this linear approximation to calculate tan(0.1) approximately. Show work
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Linear Approximation of the Tangent Function
Given the function \( f(x) = \tan(x) \):
1. **Calculate the linear approximation for \( f \) at \( x = 0 \). Show work.**
The linear approximation of a function \( f(x) \) around \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
Here, \( a = 0 \). First, we need the function value and the derivative at \( x = 0 \):
\[
f(0) = \tan(0) = 0
\]
\[
f'(x) = \sec^2(x)
\]
\[
f'(0) = \sec^2(0) = 1
\]
Therefore, the linear approximation is:
\[
L(x) = 0 + 1 \cdot (x - 0) = x
\]
So, the linear approximation of \( f(x) = \tan(x) \) at \( x = 0 \) is \( L(x) = x \).
2. **Sketch the graph of \( f \) and the obtained linear approximation.**
On a coordinate system:
- Plot \( f(x) = \tan(x) \).
- Plot the linear approximation \( L(x) = x \).
**Description of the Graph:**
- The tangent function \( \tan(x) \) passes through the origin (0,0) and has asymptotes at \( x = \pm \frac{\pi}{2} \).
- The line \( L(x) = x \) is a straight line passing through the origin with a slope of 1, intersecting points like (1,1) and (-1,-1).
3. **Use this linear approximation to calculate \( \tan(0.1) \) approximately. Show work.**
Using the linear approximation \( L(x) = x \):
\[
L(0.1) = 0.1
\]
Thus, \( \tan(0.1) \approx 0.1 \) using the linear approximation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F63c12736-f43f-4777-9d94-03ebf9ed40f6%2Fa53b8014-8847-4725-9c40-be3842653999%2Fndq13zb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Linear Approximation of the Tangent Function
Given the function \( f(x) = \tan(x) \):
1. **Calculate the linear approximation for \( f \) at \( x = 0 \). Show work.**
The linear approximation of a function \( f(x) \) around \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]
Here, \( a = 0 \). First, we need the function value and the derivative at \( x = 0 \):
\[
f(0) = \tan(0) = 0
\]
\[
f'(x) = \sec^2(x)
\]
\[
f'(0) = \sec^2(0) = 1
\]
Therefore, the linear approximation is:
\[
L(x) = 0 + 1 \cdot (x - 0) = x
\]
So, the linear approximation of \( f(x) = \tan(x) \) at \( x = 0 \) is \( L(x) = x \).
2. **Sketch the graph of \( f \) and the obtained linear approximation.**
On a coordinate system:
- Plot \( f(x) = \tan(x) \).
- Plot the linear approximation \( L(x) = x \).
**Description of the Graph:**
- The tangent function \( \tan(x) \) passes through the origin (0,0) and has asymptotes at \( x = \pm \frac{\pi}{2} \).
- The line \( L(x) = x \) is a straight line passing through the origin with a slope of 1, intersecting points like (1,1) and (-1,-1).
3. **Use this linear approximation to calculate \( \tan(0.1) \) approximately. Show work.**
Using the linear approximation \( L(x) = x \):
\[
L(0.1) = 0.1
\]
Thus, \( \tan(0.1) \approx 0.1 \) using the linear approximation.
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