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Chapter 3: Problem 14

Find the linear approximation to each function and evaluate it at the given values of \(x\) and \(d x\). \(\sqrt{x}\) at \(x=16\) and \(d x=-2\)

Short Answer

Expert verified
The linear approximation is \( \frac{15}{4} \).

Step by step solution

01

Identify the Function and Point of Tangency

The function we need to find the linear approximation for is \( f(x) = \sqrt{x} \). We will perform the approximation around the point where \( x = 16 \).
02

Find the Derivative of the Function

Calculate the derivative of \( f(x) = \sqrt{x} \). Using the power rule, the derivative is \( f'(x) = \frac{1}{2\sqrt{x}} \).
03

Evaluate the Derivative at the Given Point

Evaluate the derivative at \( x = 16 \). Thus, \( f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8} \).
04

Write the Linear Approximation Formula

The linear approximation (or tangent line) at \( x = a \) is given by the formula: \( L(x) = f(a) + f'(a)(x - a) \). Here, \( a = 16 \).
05

Substitute Values into the Linear Approximation

Substitute \( a = 16 \), \( f(16) = \sqrt{16} = 4 \), and \( f'(16) = \frac{1}{8} \) into the formula: \[ L(x) = 4 + \frac{1}{8}(x - 16) \].
06

Evaluate the Linear Approximation at \( x = 16 + dx \)

Substitute \( x = 16 + (-2) = 14 \) into the approximation: \[ L(14) = 4 + \frac{1}{8}(14 - 16) = 4 + \frac{1}{8}(-2) = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \].
07

Final Answer

The linear approximation of \( \sqrt{x} \) at \( x = 16 \) for \( dx = -2 \) is \( \frac{15}{4} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a fundamental branch of mathematics that studies rates of change and accumulation. It's like having a mathematical toolkit that helps us understand how things change and accumulate over time. Calculus is divided into two main branches:
  • Differential Calculus: Focuses on the concept of the derivative, which represents rates of change.
  • Integral Calculus: Concerns itself with accumulation and areas under curves.
In the context of linear approximation, we primarily deal with differential calculus. Linear approximation allows us to estimate the value of a function at a point by using the tangent at a specific point, which involves finding the derivative.
Derivative
The derivative is a central concept in differential calculus. It measures how a function changes as its input changes. Think of it as the function's rate of change at a particular point.
  • If a function describes a road's slope, the derivative will tell you how steep the road is at any given point.
  • A positive derivative indicates an increasing function, while a negative derivative signifies a decreasing function.
To find the derivative of a function like \(f(x) = \sqrt{x}\), we apply the power rule, resulting in \(f'(x) = \frac{1}{2\sqrt{x}}\). This derivative helps us construct a linear approximation around a specific point.
Tangent Line
A tangent line to a curve at a given point is the straight line that touches the curve at that point. The line doesn鈥檛 cross the curve at the immediate vicinity of the point; it just touches it.
  • The slope of the tangent line is given by the derivative of the function at that point.
  • For example, at \(x = 16\) for \(f(x) = \sqrt{x}\), the slope is \(f'(16) = \frac{1}{8}\).
We use the tangent line to approximate the curve nearby. This straight line serves as a simple way to estimate the function's value at points close to the point of tangency.
Function Evaluation
Function evaluation is the process of finding the output of a function given a particular input. It involves substituting the input value into the function to determine its output.
  • For instance, evaluating \(f(x) = \sqrt{x}\) at \(x = 16\) gives \(f(16) = \sqrt{16} = 4\).
  • This output is used when calculating the linear approximation.
When performing linear approximation, you use function evaluation at the point of tangency and nearby values to understand how the function behaves. This approach provides a simplified model of the function around a specific point, assisting in predictions and understanding of the function's behavior.

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