Chapter 3: Problem 11
In Exercises \(7-12,\) find a linearization at a suitably chosen integer near \(a\) at which the given function and its derivative are easy to evaluate. $$ f(x)=\sqrt[3]{x}, \quad a=8.5 $$
Short Answer
Expert verified
Linearization at \( x = 8 \) is \( L(x) = 2 + \frac{1}{12}(x - 8) \).
Step by step solution
01
Identify the Function and the Point
The given function is \( f(x) = \sqrt[3]{x} \). We need to find the linearization near \( a = 8.5 \). However, evaluating the function and its derivative is simpler at integer points close to \( 8.5 \), like \( x = 8 \) or \( x = 9 \). We choose \( a = 8 \) because it has a neat cube root.
02
Calculate the Function Value at the Chosen Point
Evaluate \( f(x) \) at \( x = 8 \).\[ f(8) = \sqrt[3]{8} = 2 \]
03
Calculate the Derivative of the Function
Find the derivative of \( f(x) = x^{1/3} \) using the power rule:\[ f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} \]
04
Evaluate the Derivative at the Chosen Point
Substitute \( x = 8 \) into \( f'(x) \):\[ f'(8) = \frac{1}{3\sqrt[3]{8^2}} = \frac{1}{3\times 4} = \frac{1}{12} \]
05
Write the Linearization Formula
The linearization at a point \( a \) can be expressed as:\[ L(x) = f(a) + f'(a)(x - a) \]Substitute \( a = 8 \), \( f(8) = 2 \), and \( f'(8) = \frac{1}{12} \) into the formula:\[ L(x) = 2 + \frac{1}{12}(x - 8) \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linearization
Linearization is a powerful tool in calculus that provides an approximation of a function close to a point by using a linear function. The basic idea is to use the tangent line at a particular point on the curve to approximate the function's behavior nearby. This is especially useful when dealing with complex functions that are difficult to evaluate precisely.
- Linearization helps simplify calculations by replacing a complex curve with a straight line approximation.
- It's often used when a quick estimation is needed or when the exact computation might be resource-intensive.
Derivative
In calculus, a derivative is a measure of how a function changes as its input changes. It is a fundamental tool and concept, often used to find the slope of a tangent line to a curve at a given point. For a given function, the derivative indicates the rate of change of the function with respect to one of its variables.
- The derivative of a function is typically denoted by \( f'(x) \) or \( \frac{df}{dx} \).
- It provides the slope of the tangent line to the curve of the function at any point \( x \).
Power Rule
The power rule is a basic differentiation rule used to find the derivative of a power of \( x \). It is a simple yet powerful tool in calculus that simplifies the process of differentiation when dealing with polynomials or expressions involving powers. The power rule states: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- The power rule makes calculations quick and straightforward for polynomial functions.
- It is applicable to any real number power, including fractions and negative numbers.
Cube Root
A cube root is a number that, when multiplied by itself twice, gives the original number. The cube root function \( f(x) = \sqrt[3]{x} \) is the inverse of the cube function \( x^3 \). This means if \( x = y^3 \), then \( y = \sqrt[3]{x} \).
- Cube roots are useful when working with volumes or dimensions involving three dimensions.
- The function \( \sqrt[3]{x} \) is defined for all real numbers, as cubes can be either positive or negative.
Function Evaluation
Function evaluation involves calculating the output value of a function for a given input. This is an essential step in many calculus problems, particularly when dealing with linearization, as both the function and its derivative need to be evaluated at a specific point.
- Function evaluation is straightforward with basic functions but can be more complex with advanced ones.
- It's a fundamental part of checking the accuracy of approximations or linearizations.