Chapter 3: Problem 42
Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ g(x)=\sin ^{-1} x \text { at } a=0,[-1,1] $$
Short Answer
Expert verified
The linear approximation of \( g(x) = \sin^{-1} x \) at \( x = 0 \) is \( L(x) = x \).
Step by step solution
01
Recall the formula for linear approximation
The linear approximation of a function \( f(x) \) at a point \( a \) is given by \( L(x) = f(a) + f'(a)(x - a) \). We will apply this formula to the given function \( g(x) = \sin^{-1} x \) at \( a = 0 \).
02
Evaluate the function at the point a
Given \( g(x) = \sin^{-1} x \), we need to evaluate \( g(0) \). Since \( \sin^{-1}(0) = 0 \), we have \( g(0) = 0 \).
03
Find the derivative of the function
The derivative of \( g(x) = \sin^{-1} x \) is \( g'(x) = \frac{1}{\sqrt{1-x^2}} \). Now, we evaluate \( g'(0) = \frac{1}{\sqrt{1-0^2}} = 1 \).
04
Write the linear approximation
Using the linear approximation formula: \( L(x) = g(0) + g'(0)(x - 0) \), we find \( L(x) = 0 + 1(x - 0) = x \). The linear approximation of \( \sin^{-1}x \) at \( x = 0 \) is \( L(x) = x \).
05
Plot the function and its linear approximation
To visualize the functions, plot \( g(x) = \sin^{-1} x \) and its linear approximation \( L(x) = x \) on the interval \([-1, 1]\). The plot should show how closely \( L(x) \) approximates \( g(x) \) near \( x=0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Functions
Inverse functions help us find the input of a function based on its output. Specifically, an inverse function reverses the effect of the original function. If you have a function, say, \( f(x) \), with an output \( y \), the inverse function, denoted \( f^{-1}(y) \), gives you back \( x \). In simpler terms, if \( f(x) = y \), then \( f^{-1}(y) = x \). This swapping of inputs and outputs is useful when solving for variables that aren't easily isolated. For example, in the exercise, the function \( g(x) = \sin^{-1} x \) illustrates how inverse functions work. Here, \( \sin^{-1} x \) is the inverse of the sine function, providing the angle whose sine is \( x \). Inverse functions can be identified on a graph as they are reflections of the original function across the line \( y = x \). This reflection property aids in visualizing and understanding how an inverse function operates in real contexts.
Derivative
Derivatives reveal the rate at which a function changes at any given point. They are fundamental in understanding the behavior of functions and are frequently used in linear approximations. A derivative gives us the slope of the tangent line to the function at a specific point, which is crucial for building linear models. For the function \( g(x) = \sin^{-1} x \), its derivative is \( g'(x) = \frac{1}{\sqrt{1-x^2}} \). At \( x = 0 \), this derivative equals 1, indicating that the slope of the tangent line at this point is 1.
- This tangent slope helps construct the linear approximation.
- Linear approximations utilize the derivative to ensure the curve of the line closely resembles the curve of the function at a given point.
Plotting Functions
Visualizing functions by plotting them provides insights into their behavior over specified intervals. In the given exercise, plotting \( g(x) = \sin^{-1} x \) and its linear approximation \( L(x) = x \) on the interval \([-1, 1]\) demonstrates the proximity of the linear approximation to the original function near the point \( a = 0 \). Attempting to understand this spatial relationship without visual aid can be challenging.
- By plotting, one can observe where the linear approximation holds closely to the original function (which is usually around the point of approximation \( a \)).
- It also highlights discrepancies further from this point, offering an understanding of approximation limitations.