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Chapter 3: Problem 4

Find \(f^{\prime}(x)\) using the limit definition of \(f^{\prime}(x)\). $$ f(x)=2 x^{2}-x $$

Short Answer

Expert verified
The derivative is \( f'(x) = 4x - 1 \).

Step by step solution

01

Write the Limit Definition of the Derivative

The derivative of a function \(f(x)\) at a point \(x\) is given by the limit definition: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. \]
02

Substitute the Function into the Limit Definition

Given the function \( f(x) = 2x^2 - x \), find \( f(x+h) \): \[ f(x+h) = 2(x+h)^2 - (x+h). \]Simplify \( f(x+h) \):- \((x+h)^2 = x^2 + 2xh + h^2\)- So, \( f(x+h) = 2(x^2 + 2xh + h^2) - (x+h) \) becomes \( 2x^2 + 4xh + 2h^2 - x - h \).
03

Formulate the Difference Quotient

Plug \( f(x+h) \) and \( f(x) \) into the difference quotient:\[ \frac{f(x+h) - f(x)}{h} = \frac{(2x^2 + 4xh + 2h^2 - x - h) - (2x^2 - x)}{h}. \]
04

Simplify the Expression

Simplify the expression inside the numerator:\[ = 2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x \]Which simplifies to:\[ = 4xh + 2h^2 - h. \]Take this over \(h\):\[ \frac{4xh + 2h^2 - h}{h} = 4x + 2h - 1. \]
05

Take the Limit as \(h\) Approaches 0

Now, apply the limit as \( h \to 0 \):\[ f'(x) = \lim_{h \to 0} (4x + 2h - 1) = 4x - 1. \]
06

Conclusion: The Derivative of the Function

Therefore, the derivative of the function \( f(x) = 2x^2 - x \) using the limit definition is:\[ f'(x) = 4x - 1. \]

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Key Concepts

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

Limit Definition
The limit definition is the basis for calculating derivatives and understanding how a function behaves at specific points. The mathematical expression for the derivative using the limit is:
  • \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \)
This represents the instantaneous rate of change of the function, essentially the slope of the tangent line at any point \(x\) on the curve. Breaking it down:
  • \( f(x+h) \) is the function evaluated at a small increment \(h\) from \(x\).
  • \( f(x) \) is the original function value.
  • The difference \( f(x+h) - f(x) \) over \(h\) gives the average rate of change.
  • Taking the limit as \(h\) approaches zero pinpoints the exact rate at that point.
This method underpins the concept of differentiation and is fundamental in calculus.
Differentiation
Differentiation is the process of finding the derivative of a function, which expresses how the function changes at any given point. This is critical for understanding behaviors such as:
  • Finding slopes of lines tangent to a curve.
  • Identifying local maxima and minima.
  • Understanding increasing or decreasing functions.
In our exercise, differentiation was accomplished by applying the limit definition:
  • Set \( f(x + h) = 2(x+h)^2 - (x+h) \).
  • Simplify and calculate: \( 2(x^2 + 2xh + h^2) - x - h = 2x^2 + 4xh + 2h^2 - x - h \).
After constructing the difference quotient, simplifying, and taking the limit, we find:
  • \( f'(x) = 4x - 1 \)
This gives the gradient of the function's curve at any point \(x\).
Polynomial Function
Polymer function refers to mathematical expressions that include terms summed together, each with a coefficient and an exponential component. In our case, the function \( f(x) = 2x^2 - x \) is a basic quadratic polynomial, characterized by:
  • The highest power of \(x\) being 2, i.e., a quadratic term \( 2x^2 \).
  • A linear term \(-x\).
These functions are differentiable using straightforward algebraic manipulations. For such functions:
  • The derivative of a polynomial \( ax^n \) is \( anx^{n-1} \).
By applying this rule, you can manually verify the limit-derived result as well. An understanding of polynomial differentiation helps in obtaining derivatives quickly, such as using their power rule properties, making them essential in calculus and analytical computations.

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Most popular questions from this chapter

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