Problem 10 - University Calculus: Early Transcendentals

Question

Find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$f(x)=1+x, \quad a=8.1$$

Step-by-Step Solution

Answer

The linearization at \( a = 8 \) is \( L(x) = x + 1 \).

Step 1: Choose a Suitable Integer

The function given is \( f(x) = 1 + x \) and we need to find the linearization at an integer value close to \( a = 8.1 \). The closest integer to 8.1 is 8.

Step 2: Find the Derivative of the Function

The function is \( f(x) = 1 + x \). To find its derivative, use the rule \( \frac{d}{dx}(c + x) = 1 \). Thus, \( f'(x) = 1 \).

Step 3: Use the Linearization Formula

The linearization of a function at a point \( a \) is given by: \[ L(x) = f(a) + f'(a)(x-a) \] For \( a = 8 \), we have \( f(8) = 1 + 8 = 9 \) and \( f'(8) = 1 \). So, the linearization at \( x = 8 \) is: \[ L(x) = 9 + 1 \cdot (x - 8) \]

Step 4: Simplify the Linearization Equation

Simplify \( L(x) = 9 + 1 \cdot (x - 8) \) to get \( L(x) = 9 + (x - 8) = x + 1 \).

Key Concepts

Derivative

The derivative is a vital concept in calculus, representing an instantaneous rate of change. It provides information on how a function changes at any given point. If we think of a function as a journey, the derivative is like speed at a particular moment—showing how fast or slow the journey continues.
For the function given in the exercise, which was simple, the derivative was just 1. This is because the derivative of a linear function, like our example, with the form \( f(x) = 1 + x \), is straightforward. Applying the basic derivative rule, the differentiation of \( c + x \) (where \( c \) is a constant) is simply \( 1 \). This simplicity stems from the fact that the slope of a straight line is constant throughout. Here, it reaffirms that the rate of change of the function \( f(x) = 1 + x \) is uniform across its curve.

Function Approximation

In many situations, instead of working with complex functions, we prefer simpler approximations. Function approximation allows us to replace complicated functions with simpler ones, which are easier to handle while retaining the essential behavior of the original function.
Linearization is one such method of function approximation. It makes use of a linear function to approximate a more complicated one, centering around a particular point. For instance, in the exercise, we were tasked with linearizing near \( a = 8.1 \). Thus, by choosing the nearest integer, 8, we simplified our calculations while staying sufficiently close to the point of interest. The linearization formula, \( L(x) = f(a) + f'(a)(x-a) \), is central to this approach. By calculating \( f(a) \) and \( f'(a) \) at this close integer, we crafted a simple linear approximation that mirrors the behavior of the original function in that vicinity.

Tangent Line Approximation

The tangent line approximation builds upon the concept of derivatives and function approximation. Essentially, when we linearize a function at a certain point, we are developing its tangent line at that point, which best approximates the function nearby. Imagine touching a curve with a straight line—a tangent at that spot.This tangent line serves as a close approximation of the function for values near the point they're touching.
In the step-by-step solution, the tangent line was computed for the function \( f(x) = 1 + x \) at \( x = 8 \). Using derivatives, we calculated that the slope of this line, \( f'(8) \), remained a steady 1. Gathering this along with the function's value at 8, which was 9, we formed the equation of the tangent line: \( L(x) = 9 + (x - 8) \).
This approximation can be especially useful when dealing with complex functions, giving us a simpler polynomial (in our case, linear) representation to work accurately for nearby values.