Question

Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ h(x)=x \sec x \text { at } a=0,(-\pi / 2, \pi / 2) $$

Short answer

The linear approximation of \(h(x) = x\sec x\) at \(x=0\) is \(L(x) = x\).

Step-by-step solution

Identify the Function and Point

The given function is \(h(x) = x \sec x\) and we need to find its linear approximation at the point \(a = 0\). Additionally, the interval is \((-\pi/2, \pi/2)\).

Compute the Derivative of h(x)

First, find the derivative of \( h(x) = x \sec x \). Use the product rule: if \( u = x \) and \( v = \sec x \), then \( h'(x) = u'v + uv' \). This results in \( h'(x) = 1 \cdot \sec x + x \cdot \tan x \sec x = \sec x + x \tan x \sec x \).

Evaluate the Function and Derivative at a=0

Substitute \(x = 0\) into the function and its derivative:1. \(h(0) = 0 \times \sec(0) = 0\)2. \(h'(0) = \sec(0) + 0 \times \tan(0) \sec(0) = 1\)

Construct the Linear Approximation

The formula for the linear approximation (tangent line) of a function \(f(x)\) at \(x = a\) is \( f(a) + f'(a)(x - a) \). Here, it becomes \( h(0) + h'(0)(x - 0) = 0 + 1 \cdot x = x \). Thus, the linear approximation is \( L(x) = x \).

Plot the Function and Linear Approximation

Next, graph \(h(x) = x \sec x\) and its linear approximation \(L(x) = x\) over the interval \((-\pi/2, \pi/2)\). In this graph, \(L(x) = x\) will be a straight line passing through the origin with a slope of 1, while \(h(x) = x \sec x\) will show how the function deviates slightly from the line as \(x\) increases or decreases from 0.

Key concepts

Derivative

The derivative is a fundamental concept in calculus that describes how a function changes as its input changes. It's like a snapshot of the function's rate of change at a specific point. When you find the derivative of a function, you're essentially asking, "How steep is this function at this point?" This helps you understand the behavior of the function in a small neighborhood around that point.

For the function in this exercise, \( h(x) = x \sec x \), we need the derivative to find the linear approximation. The derivative is computed by taking the derivative of each part using the product rule (which will be explained later). The result is \( h'(x) = \sec x + x \tan x \sec x \).

This expression tells us exactly how the function \( h(x) \) is changing at every point \( x \), and helps us approximate \( h(x) \) near \( a = 0 \). Calculating this derivative involves using rules from calculus like the product rule, which makes it a bit more complex than simple derivatives like that of \( x^2 \).

Product Rule

The product rule is a useful tool in calculus when you need to find the derivative of a product of two functions. It's an essential method because many functions can be represented as products of simpler functions.

Suppose you have two functions \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). This rule ensures that both parts of the product, as well as their rates of change, are considered.

In our function \( h(x) = x \sec x \), we apply the product rule with \( u = x \) and \( v = \sec x \). Following this rule, we differentiate as follows:

• The derivative of \( u = x \) is \( u' = 1 \).
• The derivative of \( v = \sec x \) is \( v' = \sec x \tan x \).

Therefore, the derivative using the product rule is \( h'(x) = 1 \cdot \sec x + x \cdot \sec x \tan x \), or simply \( \sec x + x \tan x \sec x \).

This result captures the nature of both \( x \) and \( \sec x \), and how they combine to affect the slope of the tangent to the curve at any point \( x \).

Tangent Line

A tangent line is a straight line that just "touches" a function at a specific point, without crossing it, at least in the immediate vicinity. At this point, the slope of the tangent line is equal to the slope of the function, which is given by the derivative.

This concept is key in linear approximation because the tangent line gives the best straight-line approximation to the function near the point of tangency. In this context, it simplifies the function to something much easier to work with: a line.

For the function \( h(x) = x \sec x \) at the point \( a = 0 \), the linear approximation or tangent line can be calculated using the derivative found earlier. The formula for the tangent line at \( x = a \) is:

• \( f(a) + f'(a)(x - a) \)

Substituting the values we calculated, the tangent line becomes \( L(x) = 0 + 1 \cdot x = x \).

This equation \( L(x) = x \) gives us a line with a slope of 1 that passes through the origin, providing us a simple model of the function around \( x = 0 \).

Interval

Understanding the concept of intervals is crucial when working with functions because it provides boundaries to where a function is examined, in this case, where the approximation and the original function are both examined and compared.

Here, the interval \((-\pi/2, \pi/2)\) represents the range over which we compare the function \( h(x) = x \sec x \) and its linear approximation \( L(x) = x \). Why this interval? It is chosen such that the function \( \sec x \), which involves trigonometric behavior, remains defined and avoids values where it becomes infinite or undefined (like at multiples of \( \pi/2 \)).

This interval, therefore, gives us a practical region to plot and study how well the linear approximation mirrors the original function. Observing within this interval enables us to see the point where the line \( L(x) = x \), which originates from the tangent line at \( a = 0 \), diverges from the curve of \( h(x) = x \sec x \) as \( x \) moves away from zero. It highlights the accuracy and limitations of the linear approximation.