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Chapter 3: Problem 1

In Exercises \(1-5,\) find the linearization \(L(x)\) of \(f(x)\) at \(x=a\) $$ f(x)=x^{3}-2 x+3, \quad a=2 $$

Short Answer

Expert verified
The linearization is \( L(x) = 10x - 13 \).

Step by step solution

01

Understand the Formula for Linearization

The linearization of a function \( f(x) \) at \( x = a \) is given by the formula \( L(x) = f(a) + f'(a)(x-a) \). This means we need to find the value of \( f(a) \) and the derivative \( f'(x) \) at \( x = a \).
02

Compute the Derivative of f(x)

Start by finding the derivative of \( f(x) = x^3 - 2x + 3 \). Using basic differentiation rules, the derivative \( f'(x) \) is \( 3x^2 - 2 \).
03

Evaluate the Function and Its Derivative at x = a

Evaluate \( f(x) \) and \( f'(x) \) at \( x = 2 \):- \( f(2) = 2^3 - 2 \cdot 2 + 3 = 8 - 4 + 3 = 7 \).- \( f'(2) = 3(2)^2 - 2 = 12 - 2 = 10 \).
04

Construct the Linearization L(x)

Substitute the values of \( f(2) \) and \( f'(2) \) into the linearization formula: \[ L(x) = f(2) + f'(2)(x-2) \]\[ L(x) = 7 + 10(x - 2) \]
05

Simplify the Expression for L(x)

Simplify the expression for \( L(x) \): \[ L(x) = 7 + 10x - 20 \]\[ L(x) = 10x - 13 \]Thus, the linearization of \( f(x) \) at \( x = 2 \) is \( L(x) = 10x - 13 \).

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Key Concepts

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

Derivative
A derivative essentially tells us how a function changes as its input changes. Mathematically, it is the slope of the tangent line to the graph of the function at any given point. In simpler terms: if you imagine hiking up a hill, the derivative tells you how steep the hill is at any point.

For the function given in the problem, which is \[ f(x) = x^3 - 2x + 3 \]we need to find its derivative. Using differentiation rules, we compute:
  • The derivative of \( x^3 \) is \( 3x^2 \). This is because we bring down the exponent, 3, and multiply it by \( x \) raised to the power of one less, which is 2.
  • The derivative of \(-2x \) is \(-2\). Here, the coefficient \(-2\) remains constant as the exponent of \( x \), which is 1, reduces to 0, making \( x^0 = 1 \).
  • The derivative of a constant, like 3, is 0, because constants do not change as \( x \) changes.
So, the derivative of the function, \( f'(x) \), will be: \[ f'(x) = 3x^2 - 2 \]This derivative provides us the tools to understand how \( f(x) \) increases or decreases as we move along the \( x \)-axis.
Function Evaluation
Function evaluation involves computing the value of a function at a certain point, designated here as \( x = 2 \). For our function, let's evaluate \( f(x) \) at this point to find \( f(2) \).

By substituting 2 into the function \( f(x) = x^3 - 2x + 3 \), we perform the calculations as follows:
  • First, calculate \( 2^3 = 8 \).
  • Then compute \( - 2 \times 2 = -4 \).
  • Finally, add the results: \( 8 - 4 + 3 = 7 \).
Therefore, \( f(2) = 7 \).

Evaluating the function provides us with a specific value, which in combination with the derivative evaluation forms the basis for finding the linear approximation, or linearization, near \( x = a \).
Differentiation Rules
Differentiation rules are the guidelines that help us take the derivatives of different types of functions. These rules make the process of finding derivatives systematic and manageable.

Some basic rules include:
  • Power Rule: For a term \( x^n \), the derivative is \( nx^{n-1} \).
  • Constant Rule: The derivative of a constant is 0.
  • Constant Factor Rule: The derivative of a constant multiplied by a function is the constant multiplied by the derivative of the function.
  • Sum Rule: The derivative of a sum of functions is the sum of their derivatives. For example, for \( f(x) = g(x) + h(x) \), \( f'(x) = g'(x) + h'(x) \).
Applying these rules to our function \( f(x) = x^3 - 2x + 3 \) ensures we find its derivative accurately.

Knowing and using these differentiation rules is crucial for students to grasp how to compute derivatives quickly and correctly, as seen in deriving the function in our exercise.

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