Problem 141 - Thomas Calculus in SI Units

Question

The linearization of \(2^{x}\) a. Find the linearization of \(f(x)=2^{x}\) at \(x=0 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together for \(-3 \leq x \leq 3\) and \(-1 \leq x \leq 1\)

Step-by-Step Solution

Answer

L(x) is approximately 1 + 0.69x.

Step 1: Find the Function and its Derivative

We are given the function to be linearized, which is \(f(x) = 2^{x}\). First, find the derivative of \(f(x)\). The derivative of \(2^{x}\) is \(f'(x) = \ln(2) \cdot 2^{x}\).

Step 2: Evaluate Function and Derivative at the Point

Evaluate both \(f(x)\) and \(f'(x)\) at \(x = 0\). Calculate \(f(0) = 2^{0} = 1\), and \(f'(0) = \ln(2) \cdot 2^{0} = \ln(2)\).

Step 3: Find the Linearization

The linearization of a function \(f(x)\) at a point \(x = a\) can be written as \(L(x) = f(a) + f'(a)(x-a)\). For \(x=0\), the linearization becomes \(L(x) = 1 + \ln(2) \cdot x\). Round \(\ln(2)\) to two decimal places, which is approximately \(0.69\), resulting in \(L(x) \approx 1 + 0.69x\).

Step 4: Graph the Linearization and the Original Function

Graph both the linearization \(L(x) = 1 + 0.69x\) and the original function \(f(x) = 2^{x}\) on the same graph. Plot these over the intervals \(-3 \leq x \leq 3\) and \(-1 \leq y \leq 1\). This visually shows how closely the linearization approximates \(f(x)\) near \(x=0\).

Key Concepts

Differentiation

Differentiation is a mathematical process used to find the derivative of a function. This derivative represents the rate at which the function's value changes as the input changes. For exponential functions, like the function given here, differentiation helps us understand how quickly something grows or decays.
In the problem, the function to be differentiated is \( f(x) = 2^x \). To find its derivative, we use rules specific to exponential functions: the derivative of \( a^x \) is \( \ln(a) \cdot a^x \). Therefore, the derivative of \( 2^x \) is \( f'(x) = \ln(2) \cdot 2^x \).
Understanding differentiation is critical since it allows us to determine the tangent line's slope at any given point. In this exercise, we specifically evaluate the derivative at \( x = 0 \) to aid in finding the linearization of the function.

Tangent Line Approximation

Tangent line approximation is a method used to simplify complex functions into linear forms. This makes it easier to estimate the value of the function near a specific point. Linearization is the primary form of tangent line approximation.
To achieve this, we use the formula for linearization: \( L(x) = f(a) + f'(a)(x - a) \). For the given function \( f(x) = 2^x \) at the point \( x = 0 \), we previously calculated \( f(0) = 1 \) and \( f'(0) = \ln(2) \). Plugging these values into the linearization formula, we get \( L(x) = 1 + \ln(2) \cdot x \), which simplifies to \( L(x) \approx 1 + 0.69x \) after rounding.
This approximation is beneficial since it provides a simpler linear model that closely follows the function's behavior around \( x = 0 \), helping in quick calculations and analyses.

Graphing Functions

Graphing functions is a crucial way to visualize and understand mathematical relationships. It allows us to see how functions behave over different ranges. In this problem, graphing both the original function \( f(x) = 2^x \) and its linear approximation \( L(x) = 1 + 0.69x \) provides visual insight.
The exercise specifies two intervals: \(-3 \leq x \leq 3\) and \(-1 \leq y \leq 1\). By plotting the functions within these ranges, we observe how the linear approximation closely matches the original function near \( x = 0 \).
The graph will show that as \( x \) moves away from zero, the linear approximation will start diverging from the exponential function. This is typical with linearizations and highlights why they are most effective near the tangent point. Such visual comparisons enhance our understanding of the function's behavior and the effectiveness of tangent line approximations.