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Chapter 2: Problem 57

Learning Theory In a psychology experiment, a person could memorize \(x\) words in \(f(x)=2 x^{2}-x\) seconds (for \(0 \leq x \leq 10)\). a. Find \(f^{\prime}(x)\) by using the definition of the derivative. b. Find \(f^{\prime}(5)\) and interpret it as an instantaneous rate of change in the proper units.

Short Answer

Expert verified
\( f'(x) = 4x - 1 \); \( f'(5) = 19 \) seconds per word.

Step by step solution

01

Understanding the function

The function given is \( f(x) = 2x^2 - x \). This function models the time it takes for a person to memorize \( x \) words.
02

Definition of the Derivative

By the definition, the derivative of \( f(x) \) at any point \( x \) is given by \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \). This represents the rate of change of the function.
03

Calculate \( f(x+h) \)

First, calculate \( f(x+h) \). Substitute \( x+h \) into the original function: \( f(x+h) = 2(x+h)^2 - (x+h) \). Expand this expression to simplify it. \( f(x+h) = 2(x^2 + 2xh + h^2) - x - h = 2x^2 + 4xh + 2h^2 - x - h \).
04

Calculate \( f(x+h) - f(x) \)

Subtract \( f(x) \) from \( f(x + h) \): \( f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - x - h) - (2x^2 - x) = 4xh + 2h^2 - h \). Simplify this to: \( f(x+h) - f(x) = h(4x + 2h - 1) \).
05

Limit of the Difference Quotient

Now compute the derivative as: \( f'(x) = \lim_{{h \to 0}} \frac{h (4x + 2h - 1)}{h} = \lim_{{h \to 0}} (4x + 2h - 1) \). As \( h \to 0 \), this simplifies to: \( f'(x) = 4x - 1 \).
06

Calculate \( f'(5) \)

Substitute \( x = 5 \) into the derivative \( f'(x) = 4x - 1 \): \( f'(5) = 4(5) - 1 = 20 - 1 = 19 \).
07

Interpretation of \( f'(5) \)

The value of \( f'(5) = 19 \) represents the instantaneous rate of change of the time with respect to the number of words memorized. In proper units, it means that at \( x = 5 \) words, the time required to memorize is increasing at a rate of 19 seconds per additional word.

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Key Concepts

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

Instantaneous Rate of Change
The idea of instantaneous rate of change is central to understanding derivatives in calculus. It tells us how fast a function is changing at a specific point.
For example, in the context of our psychology experiment, it shows how quickly the time to memorize words changes as more words are added.

Think of it as the speedometer on a car, showing how fast you are going at an exact moment.
  • When you calculate the derivative at a particular point, you find the instantaneous rate of change.
  • In the original exercise, we calculated this rate at the point where 5 words are memorized.
  • This gives the rate of change in seconds per additional word, indicating how the process includes more words to memorize.
It's essential in numerous applications, helping us understand systems that change over time.
Limit of a Function
The limit is a foundational concept in calculus that allows us to define a derivative.
Limits help in understanding how a function behaves as it approaches a certain value or point.
In our derivative definition, the limit was used to determine the slope of the tangent line to the curve at any given point.

Let's explore how limits work:
  • We use limits to understand what happens when we make very tiny changes (like approaching zero).
  • In the solution, the expression \( \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \) represents the limit of the difference quotient.
  • Calculating this limit gives us the derivative \( f'(x) \), revealing the rate of change.
The limit helps smooth out errors and fixes undefined expressions, offering precision in calculus problems.
Differentiation Techniques
Differentiation techniques are crucial in solving calculus problems involving rate of change.
They provide us with methods to find derivatives of different types of functions.
This process involves using the limit of a function as well as specific rules to calculate derivatives efficiently.

Some basic differentiation techniques include:
  • Power Rule: Useful for polynomial functions, where you multiply the power by the coefficient and reduce the power by one.
  • Chain Rule: Applies when you have a function inside another function, helping to differentiate complex compositions.
  • Product and Quotient Rules: Help differentiate products or quotients of functions.
In our problem, the derivative was calculated directly using the definition, but these techniques often simplify derivative calculations in practice.

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Most popular questions from this chapter

Determine whether each function is continuous or discontinuous. If discontinuous, state where it is discontinuous. $$ \begin{array}{l} f(x)=\left\\{\begin{array}{ll} 3-x & \text { if } x \leq 4 \\ 10-2 x & \text { if } x>4 \end{array}\right.\\\ \text { [Hint: See Exercise 37.] } \end{array} $$

Based on a recent study, the "happiness" of people who live in a country whose average temperature is \(t\) degrees Fahrenheit is given by \(h(t)=8.2-(0.01 t-2.8)^{2}\), for \(35 \leq t \leq 72\). ("Happiness" was rated from \(1=\) "not at all happy" to \(4=\) "very happy".) Find \(h(40)\) and \(h^{\prime}(40)\). Interpret your answers.

A student once said: "The limit is where a function is going, even if it never gets there." Explain what the student meant.

Use the Generalized Power Rule to find the derivative of each function. $$ f(x)=\left(\frac{x+1}{x-1}\right)^{3} $$

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