Problem 14 - 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. Show that the linearization of \(f(x)=(1+x)^{k}\) at \(x=0\) is \(L(x)=1+k x\)

Step-by-Step Solution

Answer

The linearization of \(f(x)=(1+x)^k\) at \(x=0\) is \(L(x)=1+kx\).

Step 1: Identify the Function

The given function is \( f(x) = (1 + x)^k \). We need to find the linearization of this function at a certain point, usually at a point where both the function and its derivative are simple to compute.

Step 2: Choose the Point for Linearization

Linearization is typically done around a value where both the function and its derivative are easy to evaluate. In this exercise, the point chosen is \( x = 0 \).

Step 3: Calculate the Function Value at \(x=0\)

Calculate \(f(0)\) which is \((1+0)^k = 1\).

Step 4: Find the Derivative of the Function

Using the power rule, the derivative of \(f(x) = (1+x)^k\) is \(f'(x) = k(1+x)^{k-1}\).

Step 5: Calculate the Derivative at \(x=0\)

Substitute \(x = 0\) into the derivative: \(f'(0) = k(1+0)^{k-1} = k \times 1 = k\).

Step 6: Write the Linearization Formula

The linearization of a function \(f(x)\) around \(x = a\) is given by \(L(x) = f(a) + f'(a)(x-a)\). In this case, \(a = 0\).

Step 7: Substitute into Linearization Formula

Substitute \(a = 0\), \(f(0) = 1\), and \(f'(0) = k\) into the linearization formula: \[ L(x) = 1 + k(x-0) = 1 + kx \].

Key Concepts

Differentiation

Differentiation is an essential concept in calculus used to calculate the rate at which a function is changing at any given point. When differentiating, you're essentially finding the derivative of the function. This is exceptionally useful for understanding various behaviours of functions, like finding local maxima, minima, and points of inflection.
The process of differentiation involves several rules, one of which is the Power Rule. This rule is fundamental when you're dealing with polynomial functions.
To differentiate a function, you determine how the output value changes as the input changes, providing insights into the slope of the tangent line at any given point.

Power Rule

The Power Rule is one of the most straightforward differentiation rules used when you're working with functions in the form of a power of a variable, such as \( x^n \). The Power Rule states that if you have a function \( f(x) = x^n \), the derivative of the function is \( f'(x) = nx^{n-1} \).
This rule simplifies the process of finding derivatives, especially for polynomial functions.

• It helps you identify the derivative quickly by bringing the exponent down as a coefficient.
• The exponent is then decreased by one.

For example, applying the Power Rule to \( f(x) = (1 + x)^k \) gives the derivative \( f'(x) = k(1+x)^{k-1} \). Here, \( k \) is treated as a coefficient, and the power is reduced by one.
The Power Rule is integral to the process of linearization, as it provides the necessary derivative required in the tangent line approximation.

Tangent Line Approximation

Tangent Line Approximation, often called linearization, involves approximating a function near a given point using the tangent line at that point. Linearization is grounded in the idea that curves closely resemble their tangent lines over small intervals.
The formula for the linearization of a function \( f(x) \) at a point \( a \) is given by:

• \( L(x) = f(a) + f'(a)(x-a) \)

Here, \( f(a) \) is the function's value at point \( a \) and \( f'(a) \) is the derivative at that point.
This formula helps you create the equation of the tangent line. It approximates the value of the function at points near \( a \).
In this exercise, the linearization of \( f(x) = (1 + x)^k \) at \( x = 0 \) resulted in \( L(x) = 1 + kx \), demonstrating how the tangent line provides a simple linear expression for otherwise more complex expressions over small intervals.