Problem 58 - Calculus Single Variable

Question

Find the linearization of the given function \(f(x)\) at the given point \(c\). $$ f(x)=\arctan (x), c=-1 $$

Step-by-Step Solution

Answer

The linearization of \( f(x) = \arctan(x) \) at \( c = -1 \) is \( L(x) = \frac{1}{2}x - \frac{\pi}{4} + \frac{1}{2} \).

Step 1: Understand the Linearization Formula

The linearization of a function at a given point is expressed by the formula: \( L(x) = f(c) + f'(c)(x - c) \), where \( c \) is the point at which you are linearizing the function.

Step 2: Evaluate the Function at the Given Point

Substitute \( c = -1 \) into the function \( f(x) = \arctan(x) \). Calculate \( f(-1) = \arctan(-1) = -\frac{\pi}{4} \). This is because the arctangent of \(-1\) is \(-\frac{\pi}{4}\).

Step 3: Find the Derivative of the Function

Find \( f'(x) \) of the function \( f(x) = \arctan(x) \), which is \( f'(x) = \frac{1}{1 + x^2} \). This is a standard derivative identity for the arctan function.

Step 4: Evaluate the Derivative at the Given Point

Substitute \( c = -1 \) into the derivative \( f'(x) = \frac{1}{1 + x^2} \). Calculate \( f'(-1) = \frac{1}{1 + (-1)^2} = \frac{1}{2} \).

Step 5: Substitute into the Linearization Formula

Now, use the linearization formula: \( L(x) = -\frac{\pi}{4} + \frac{1}{2}(x + 1) \). This expression simplifies to \( L(x) = -\frac{\pi}{4} + \frac{1}{2}x + \frac{1}{2}\).

Key Concepts

Derivative of inverse trigonometric functions

When we talk about derivatives of inverse trigonometric functions, we are dealing with the calculus concept of finding rates of change. Specifically, for inverse trig functions like the arctangent function (also written as \( \arctan(x) \)), the derivative serves to analyze how the output of this function changes as its input (or argument) changes.

For the arctangent function, the derivative is given by the formula:

• \( f'(x) = \frac{1}{1 + x^2} \)

This derivative formula implies that the rate of change of the arctangent function decreases as the value of \( x \) becomes large, because the term \( 1 + x^2 \) in the denominator increases. This means that the arctangent function grows more slowly as \( x \) moves away from zero.

This derivative is a crucial tool when linearizing the arctangent function at a particular point, as it helps in constructing the tangent line, which serves as the linear approximation of the function at that point.

Evaluating functions at a point

Evaluating a function at a point involves substituting a specific value into the function to determine its output at that particular input. This is like checking a function's result for a certain \( x \)-value. In the case of our linearization exercise, we evaluate the arctangent function \( f(x) = \arctan(x) \) at \( c = -1 \).

Performing this evaluation means substituting \( -1 \) into the function, giving us:

• \( f(-1) = \arctan(-1) \)

The value resulting from this is \(-\frac{\pi}{4}\), since the arctangent of \(-1\) corresponds to an angle of \(-\frac{\pi}{4}\) radians on the unit circle.

In any context requiring the linearization of a function, determining this value is essential because it becomes the intercept of the linear approximation or the \( y \)-coordinate where the tangent line crosses the y-axis at the given point.

Arctangent function

The arctangent function, denoted as \( \arctan(x) \), is one of the six inverse trigonometric functions. It is particularly useful in mathematics for converting a ratio back into an angle. Essentially, it tells us the angle whose tangent is \( x \).

This function is defined for all real numbers, and its range is limited to \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) radians, which makes it suitable for various applications, especially in algebra and geometry. The arctangent function is differentiable everywhere because it smoothly transitions through values, making calculus operations like differentiation straightforward.

In the context of linearization, the arctangent function serves as a model for finding a tangent line at a particular point. By using its derivative and function value at a specific point, we can create a linear model that approximates the function near that point, which is precisely what we achieve with the linearization formula:

• \( L(x) = f(c) + f'(c)(x - c) \)

For the example at \( c = -1 \), this leads to the linear function \( L(x) = -\frac{\pi}{4} + \frac{1}{2} (x + 1) \), which serves as a quick way to estimate the behavior of \( \arctan(x) \) for values of \( x \) near \(-1\).