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Chapter 1: Problem 34

Find \(f^{\prime}(x)\). $$ f(x)=\frac{2 x}{3} $$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{2}{3} \).

Step by step solution

01

Identify the Function Type

The given function is \( f(x) = \frac{2x}{3} \). This is a linear function because it can be rewritten as \( f(x) = \frac{2}{3}x \) which is in the form \( ax + b \), where \( a = \frac{2}{3} \) and \( b = 0 \).
02

Recall the Derivative of a Linear Function

The derivative of a linear function \( f(x) = ax + b \) is simply \( a \). This is because the derivative measures the rate of change, and for a linear function, this rate of change is constant and equals the coefficient of \( x \).
03

Apply the Derivative Rule

Using the rule from the previous step, the derivative of \( f(x) = \frac{2}{3}x \) is simply the coefficient of \( x \), which is \( \frac{2}{3} \). Thus, \( f'(x) = \frac{2}{3} \).

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Key Concepts

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

Understanding Linear Functions
A linear function is one of the simplest types of functions in mathematics. It is commonly expressed in the form \( f(x) = ax + b \), where \( a \) and \( b \) are constants. Linear functions graph as straight lines and have a constant slope.

To visualize it:
  • \( a \) determines the slope or steepness of the line.
  • \( b \) represents the y-intercept, which is where the line crosses the y-axis.
Linear functions are straightforward yet foundational as they form the basis for more complicated functions in calculus and algebra. A key characteristic is their constant rate of change, which stays the same regardless of which two points you pick on the line.
Exploring the Rate of Change
The rate of change of a function tells us how much the output value changes for a given change in the input value. For linear functions, this rate of change is constant and directly equals the slope \( a \) in the expression \( f(x) = ax + b \).

Here's what you need to know:
  • In the context of a linear function, the rate of change is uniform, meaning the function increases or decreases steadily.
  • This is why the derivative of a linear function, which measures this rate, remains constant.
Understanding the rate of change is crucial because it simplifies the process of finding derivatives in linear functions. It gives insight into the behavior of functions and prepares you for more complex topics like slopes of tangent lines in calculus.
Importance of the Coefficient of x
The coefficient of \( x \) in a linear function \( f(x) = ax + b \) plays a vital role in determining the function's behavior. This coefficient influences both the slope of the line on a graph and the derivative of the function.

When talking about derivatives:
  • The derivative of a linear function \( f(x) = ax + b \) is \( f'(x) = a \). This derivative represents the function's slope or rate of change.
  • The derivative is constant because the coefficient \( a \) remains the same across any interval of \( x \).
By understanding the coefficient of \( x \), you can easily determine key attributes of linear functions, helping to quickly solve problems and interpret graphs.

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Most popular questions from this chapter

On Earth, an object travels \(4.905 \mathrm{~m}\) after 1 sec of free fall. Thus, by symmetry, an athlete would require 1 sec to jump \(4.905 \mathrm{~m}\) high, and another second to come back down. Is it possible for a person to stay in the air for (have a "hang time" of) 2 sec? Can a person have a hang time of 1.5 sec? 1 sec? What do you think is the longest possible hang time achievable by humans jumping from level ground?

On the moon, all free-fall distance functions are of the form \(s(t)=0.81 t^{2},\) where \(t\) is in seconds and \(s(t)\) is in meters. An object is dropped from a height of 200 meters above the moon. After \(t=2 \mathrm{sec}\) a) How far has the object fallen? b) How fast is it traveling? c) What is its acceleration? d) Explain the meaning of the second derivative of this free-fall function.

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=-0.025 x^{2}+4 x $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=x^{3}-6 x+1 $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow-2}\left(\frac{x^{2}-4}{x+2}\right) $$
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