Problem 2 - Thomas Calculus

Question

Find the linearization \(L(x)\) of \(f(x)\) at \(x=a.\) \(f(x)=\sqrt{x^{2}+9}, \quad a=-4\)

Step-by-Step Solution

Answer

The linearization is \( L(x) = \frac{9 - 4x}{5} \).

Step 1: Understand the Problem

To find the linearization of a function at a given point, we need the tangent line of the function at that point. The formula for linearization is \( L(x) = f(a) + f'(a)(x-a) \). Here, \( f(x) = \sqrt{x^2 + 9} \) and \( a = -4 \).

Step 2: Calculate the Function Value at x=a

Find \( f(a) \). Here, \( f(-4) = \sqrt{(-4)^2 + 9} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step 3: Find the Derivative of the Function

Use the chain rule to differentiate \( f(x) = \sqrt{x^2 + 9} \). Let \( u = x^2 + 9 \). Then \( f(x) = \sqrt{u} \), and by the chain rule, \( f'(x) = \frac{d}{dx}(u^{1/2}) = \frac{1}{2}(u^{-1/2})(2x) \). Simplify to get \( f'(x) = \frac{x}{\sqrt{x^2 + 9}} \).

Step 4: Calculate the Derivative at x=a

Substitute \( x = -4 \) into the derivative to find \( f'(-4) \). Thus, \( f'(-4) = \frac{-4}{\sqrt{(-4)^2 + 9}} = \frac{-4}{\sqrt{25}} = \frac{-4}{5} \).

Step 5: Substitute into the Linearization Formula

Now, substitute \( f(a) \), \( f'(a) \), and \( a \) into the linearization formula: \( L(x) = f(a) + f'(a)(x-a) = 5 + \left(\frac{-4}{5}\right)(x + 4) \).

Step 6: Simplify the Expression

Simplify \( L(x) = 5 - \frac{4}{5}(x + 4) \). Distribute to get \( L(x) = 5 - \frac{4}{5}x - \frac{16}{5} \). Simplify further: \( L(x) = \frac{25}{5} - \frac{4}{5}x - \frac{16}{5} = \frac{9}{5} - \frac{4}{5}x \). So, \( L(x) = \frac{9 - 4x}{5} \).

Key Concepts

Tangent Line Approximation

The tangent line approximation, also known as linearization, is an essential concept in calculus. It involves approximating a function near a given point using the tangent line at that point. This method helps to estimate the value of a function by using a linear equation, which is typically easier to handle than the original nonlinear function.

To find this tangent line, we use the linearization formula:

• \( L(x) = f(a) + f'(a)(x-a) \)

Here, \( f(a) \) is the function value at the point \( a \) and \( f'(a) \) is the derivative at that point.
The approximation works well when \( x \) is close to \( a \), because the tangent line closely follows the curve. Linearization is invaluable in simplifying complex calculations by providing a linear model to work with.

Chain Rule

The chain rule is a vital technique in differentiation, applicable when dealing with composite functions. It allows us to differentiate functions nested within other functions accurately.

In essence, the chain rule states:

• If you have a composite function \( y = f(g(x)) \), then its derivative \( y' \) is given by: \( y' = f'(g(x)) \cdot g'(x) \)

This rule requires differentiating the outer function and then multiplying by the derivative of the inner function.
For our exercise, when finding \( f'(x) \) of \( f(x) = \sqrt{x^2 + 9} \), the inner function is \( u = x^2 + 9 \).
We calculate \( f'(x) = \frac{1}{2}(u^{-1/2})(2x) \), which simplifies to \( \frac{x}{\sqrt{x^2 + 9}} \). This step is crucial for finding the slope of the tangent line.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function concerning its variable. They are used to find slopes of tangent lines, and in this context, help us understand how a function behaves locally.

To find a derivative:

• Identify the function, such as \( f(x) = \sqrt{x^2 + 9} \)
• Apply differentiation rules, like the power rule or chain rule.
• Simplify your result to get the derivative, \( f'(x) = \frac{x}{\sqrt{x^2 + 9}} \).

Differentiating provides insight into how functions grow or shrink, and by calculating \( f'(-4) = \frac{-4}{5} \), we determine the slope at a specific point.
This slope is then used in the linearization process to approximate the function's value.

Function Evaluation

Function evaluation involves determining the value of a function at a specific point, which is a basic yet important skill in calculus.

The process is straightforward:

• Substitute the point into the function.
• Calculate to find the result.

In the example exercise, we found \( f(-4) \) by calculating \( \sqrt{(-4)^2 + 9} = \sqrt{25} = 5 \).
This evaluation provides the y-coordinate of the function at \( x = a \), necessary for the linearization formula.
Evaluations reveal specific information about the function's behavior at certain points, making them instrumental in various mathematical applications. By using this alongside derivatives, they help construct precise linear models that approximate functions effectively.