Problem 18 - Thomas Calculus

Question

Find the linearization of \(f(x)=\sqrt{x+1}+\sin x\) at \(x=0 .\) How is it related to the individual linearizations of \(\sqrt{x+1}\) and \(\sin x\) at \(x=0 ?\)

Step-by-Step Solution

Answer

The linearization of \( f(x) \) is \( L(x) = 1 + \frac{3}{2}x \), obtained by summing individual linearizations.

Step 1: Understanding Linearization

The linearization of a function at a point is the approximation of the function by its tangent line at that point. For a function \( f(x) \), the linearization at \( x = a \) is given by \( L(x) = f(a) + f'(a)(x-a) \). We'll apply this to each component of \( f(x) = \sqrt{x+1} + \sin x \) individually.

Step 2: Linearization of \( \sqrt{x+1} \) at \( x=0 \)

First, compute \( \sqrt{x+1} \) at \( x = 0 \). The function is \( \sqrt{0+1} = 1 \). Next, compute the derivative: \( \frac{d}{dx}(\sqrt{x+1}) = \frac{1}{2\sqrt{x+1}} \). At \( x = 0 \), this derivative is \( \frac{1}{2} \). Thus, the linearization is \( L_1(x) = 1 + \frac{1}{2}(x-0) = 1 + \frac{1}{2}x \).

Step 3: Linearization of \( \sin x \) at \( x=0 \)

Evaluate \( \sin x \) at \( x = 0 \), which is \( \sin 0 = 0 \). The derivative of \( \sin x \) is \( \cos x \), and at \( x = 0 \), \( \cos 0 = 1 \). Therefore, the linearization is \( L_2(x) = 0 + 1(x-0) = x \).

Step 4: Linearization of \( f(x) = \sqrt{x+1} + \sin x \) at \( x = 0 \)

Since \( f(x) = \sqrt{x+1} + \sin x \), its value at \( x=0 \) is \( f(0) = 1 + 0 = 1 \). The derivative \( f'(x) = \frac{1}{2\sqrt{x+1}} + \cos x \). At \( x=0 \), this becomes \( f'(0) = \frac{1}{2} + 1 = \frac{3}{2} \). Thus, the linearization is \( L(x) = 1 + \frac{3}{2}(x-0) = 1 + \frac{3}{2}x \).

Step 5: Comparing Individual and Combined Linearizations

The combined linearization \( L(x) = 1 + \frac{3}{2}x \) results from linearly adding the individual linearizations \( L_1(x) = 1 + \frac{1}{2}x \) and \( L_2(x) = x \). This reflects how the derivative of the sum of functions is the sum of derivatives.

Key Concepts

Linearization

In calculus, linearization provides us with a way to approximate a function near a given point. This method essentially involves replacing the function with a straight line—a tangent line—at the specified point. At the heart of linearization is the formula:

• \( L(x) = f(a) + f'(a)(x-a) \)

Here, \( L(x) \) is the linear approximation of the function \( f(x) \) near \( x = a \). The value \( f(a) \) is the function value at that point, and \( f'(a) \) is the derivative which gives us the slope of the tangent line. This formula shows how closely the line can model the function for small intervals around \( a \). Linearization is especially useful in situations where computing the actual value of the function might be complex. By simplifying functions into lines, we can deal with easier calculations while maintaining reasonable accuracy.

Derivative

The derivative is a fundamental concept in calculus that helps us understand the behavior of a function. Specifically, it measures how a function changes as its input changes. Think of it as a way to quantify the slope of the tangent line at any given point on the function's curve.

• The derivative of \( f(x) \), denoted as \( f'(x) \), represents this rate of change.
• At a point \( x = a \), the derivative \( f'(a) \) gives us the slope of the tangent line to the curve at that point.

For the function \( f(x) = \sqrt{x+1} + \sin x \), we calculate the derivative using basic derivative rules such as the chain rule and the derivatives of standard functions.

• The derivative of \( \sqrt{x+1} \) is \( \frac{1}{2\sqrt{x+1}} \).
• The derivative of \( \sin x \) is \( \cos x \).

These derivatives allow us to calculate the linearization of the function by providing the rate of change needed to approximate the function with a tangent line.

Tangent line

The tangent line is a straight line that lightly "touches" a curve at a single point. In calculus, this contact helps us approximate the behavior of the curve near that point. By using the derivative, we find the slope of the tangent line, which plays a key role in linearization.

• The equation of the tangent line at \( x = a \) is given by:
  • \( y = f(a) + f'(a)(x-a) \)
• This line shares the slope \( f'(a) \) with the curve at the point \( a \), ensuring it aligns perfectly up to the first order.

In the example of \( f(x) = \sqrt{x+1} + \sin x \):

• At \( x = 0 \), the function has a value \( f(0) = 1 \).
• The combined derivative \( f'(0) = \frac{3}{2} \) gives the slope.
• Therefore, the tangent line equation is \( y = 1 + \frac{3}{2}x \).

This tangent line offers a simplified model of the more complex function "close" to \( x = 0 \).

Approximation

Approximation in calculus allows us to replace complicated function behavior with simpler expressions, leading to easier computations. Through linearization, we approximate functions using their tangent lines.

• This is particularly useful when dealing with values close to the point of tangency since the linear model is most accurate there.
• For \( f(x) = \sqrt{x+1} + \sin x \) at \( x = 0 \), we approximate it using the tangent line \( 1 + \frac{3}{2}x \).
• Individually, \( \sqrt{x+1} \) and \( \sin x \) are approximated by their own tangent lines: \( 1 + \frac{1}{2}x \) and \( x \), respectively.

By combining these approximations, we arrive at the function's overall linearization. This process not only simplifies calculations but also provides a way to understand how the function behaves in close proximity to certain points, offering an insightful look into the function’s local properties.