/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Find \(f^{\prime}(x)\) for \(f(x... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find \(f^{\prime}(x)\) for \(f(x)=m x+b\).

Short Answer

Expert verified
The derivative \(f^{ ext{'}}(x)\) is \(m\).

Step by step solution

01

Understanding the Function

The function given is a linear function in the form of a standard equation for a line, which is represented as \(f(x) = mx + b\). Here, \(m\) is the slope of the line, and \(b\) is the y-intercept.
02

Recall the Derivative of a Linear Function

For the standard linear function \(f(x) = mx + b\), recall that the derivative of a linear function \(f(x) = ax + c\) is simply \(a\). This is because the derivative measures the rate of change of the function, and in a line, the rate of change is constant and is equal to the slope \(m\).
03

Apply the Derivative Rule

Apply the derivative rule to our function \( f(x) = mx + b \). According to the derivative rules, the derivative of \(mx\) is \(m\), and since the derivative of a constant \(b\) is 0, we have \( f^{ ext{'}}(x) = m \).
04

Conclusion

Thus, the derivative \(f^{ ext{'}}(x)\) of the linear function \(f(x)=mx+b\) is \(m\), which represents the slope of the line.

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Key Concepts

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

Derivative of a Linear Function
The concept of differentiation can initially seem intimidating, but with a linear function, it’s quite straightforward. A linear function, in the form of \(f(x) = mx + b\), represents a straight line. The derivative of a function tells us how the function's output value changes as its input value changes, essentially measuring the function's rate of change.
The derivative of a linear function is particularly simple to calculate because these functions have a constant rate of change, denoted by the slope \(m\).
When we differentiate, we find \(f'(x)\), which in this case, directly gives us the slope of the line.
  • The derivative of \(mx\) is \(m\), as you multiply \(m\) by the derivative of \(x\) which is 1.
  • The derivative of the constant \(b\) is 0 because constants do not change.
Putting these together, the derivative \(f'(x) = m\). This result reinforces the idea that the derivative of a linear function is a constant, reflecting its uniform rate of change.
Slope of a Line
Understanding the slope of a line is crucial in calculus and algebra. The slope is denoted by \(m\) in the equation \(f(x) = mx + b\). It describes how steep or flat a line is. In simple terms, the slope tells you how much \(y\) changes for a change in \(x\).

For example, if the slope \(m\) is 2, every increase of 1 in \(x\) results in an increase of 2 in \(f(x)\). This "rise over run" concept is fundamental for understanding graph lines:
  • A positive slope means the line ascends as you move from left to right.
  • A negative slope indicates the line descends as you move from left to right.
  • A zero slope corresponds to a horizontal line.
Thus, the slope \(m\) helps determine the line's direction and steepness on a graph.
Rate of Change
The rate of change in mathematics is a crucial concept, representing how a quantity changes with respect to another. In a linear function, this concept is captured by its slope \(m\), which is consistent across the function.

In practical terms, for the linear function \(f(x) = mx + b\), the rate of change is uniform. Every step along the horizontal axis \(x\) changes the output \(y\) by exactly \(m\) units.
This type of consistent change is what makes linear functions so straightforward. Their rate of change is simple to understand and apply:
  • It's constant, meaning it never changes.
  • It allows for easy prediction of future values of the function.
In real-world applications, this concept of rate of change helps model relationships where one variable affects another at a steady rate, such as traveling at constant speed.

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Most popular questions from this chapter

Below are the steps in the simplification of the difference quotient for \(f(x)=\sqrt{x}\) (see Example 8). Provide a brief justification for each step. \(\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\) a) \(=\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\) b) \(=\frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})}\) c) \(=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\) d) \(=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\) e) \(=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

At the beginning of a trip, the odometer on a car reads \(30,680,\) and the car has a full tank of gas. At the end of the trip, the odometer reads \(31,077 .\) It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

Fill in each blank so that \(\lim _{x \rightarrow 2} f(x)\) exists. $$ f(x)=\left\\{\begin{array}{ll} x^{2}-9, & \text { for } x<2, \\ -x^{2}+\ldots, & \text { for } x>2 \end{array}\right. $$

The function given by \(R(x)=11.74 x^{1 / 4}\) can be used to approximate the maximum range \(\mathrm{R}(x)\) in miles, of an ARSR-3 surveillance radar with a peak power of \(x\) watts (W). (Source: Introduction to RADAR Techniques, Federal Aviation Administration, \(1988 .\). a) Find the rate at which the maximum radar range changes as peak power increases from \(40,000 \mathrm{~W}\) to \(60,000 \mathrm{~W}\) b) Find \(\frac{R(60,000)-R(50,000)}{60,000-50,000}\). What does this rate represent?

Graph fand \(f^{\prime}\) and then determine \(f^{\prime}(1) .\) $$ f(x)=x^{4}-x^{3} $$

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