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Chapter 3: Problem 2

Let \(f(x)=17 x+11 .\) Use the definition of the derivative to calculate \(f^{\prime}(x)\).

Short Answer

Expert verified
The derivative \( f'(x) \) is 17.

Step by step solution

01

Recall the Definition of the Derivative

The derivative of a function \( f(x) \) with respect to \( x \), denoted as \( f'(x) \), is defined as the limit: \[ f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \] Our task is to use this definition to find the derivative of the function \( f(x) = 17x + 11 \).
02

Substitute into the Derivative Definition

Substitute the given function \( f(x) = 17x + 11 \) into the formula: \[ f'(x) = \lim_{{h \to 0}} \frac{{(17(x+h) + 11) - (17x + 11)}}{h} \] Simplify the expression inside the limit.
03

Simplify the Expression

First distribute \(17\) in the term \(17(x+h)\): \[ 17(x+h) = 17x + 17h \]Now substitute and simplify the expression:\[ f'(x) = \lim_{{h \to 0}} \frac{{(17x + 17h + 11) - (17x + 11)}}{h} \]Cancel like terms:\[ f'(x) = \lim_{{h \to 0}} \frac{17h}{h} \]
04

Simplify and Evaluate the Limit

Cancel \( h \) in the fraction:\[ f'(x) = \lim_{{h \to 0}} 17 \] Since \( 17 \) is a constant, the limit simplifies to:\[ f'(x) = 17 \]
05

Conclusion

The derivative of the function \( f(x) = 17x + 11 \) is \( f'(x) = 17 \). For a linear function \( f(x) = mx + b \), the derivative is simply the slope \( m \), which confirms our calculation was correct.

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Key Concepts

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

Definition of the Derivative
In calculus, the concept of a derivative is a cornerstone for understanding how functions change. The derivative of a function at a point gives you the rate at which the function is changing at that point. Mathematically, the derivative of a function \( f(x) \) is written as \( f'(x) \) and is defined using a limit:
  • \( f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h} \)
This formula tells us that we consider the difference in the function value at \( x \) and \( x + h \) as \( h \) approaches zero. Calculating derivatives can help you understand the behavior of functions across different values of \( x \). In the exercise, the derivative of a linear function \( f(x) = 17x + 11 \) was easily calculated using this definition.
Limit Process
The limit process is a fundamental aspect of calculus and is crucial for defining derivatives. The limit, \( \lim_{{h \to 0}} \), in the derivative formula means that we want to see what happens to our expression as \( h \) gets incredibly close to zero. This is because we're interested in the instantaneous rate of change at a single point, not over an interval.
  • In the step-by-step solution, after substituting \( f(x) = 17x + 11 \) into the derivative definition, simplifying shows that many terms cancel.
  • The expression eventually simplifies to \( \lim_{{h \to 0}} 17 \) because the change in \( x \) vanishes as \( h \) approaches zero.
The beauty of limits lies in understanding how a function behaves as it approaches certain values. When dealing with derivatives, limits allow us to explore the micro-behavior of functions in an infinitely small neighborhood around a point.
Linear Functions
Linear functions are perhaps the simplest type of function in mathematics. They take the form \( f(x) = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In geometry, the slope \( m \) describes the steepness of the line. The function \( f(x) = 17x + 11 \) is a linear function with a slope of 17.
  • The derivative of a linear function is always the slope \( m \).
  • This is because as \( h \to 0 \), the change in the constant \( b \) becomes irrelevant, and only the rate \( m \) affects the steepness.
Therefore, finding the derivative of a linear function becomes straightforward: it simply equals the coefficient of \( x \). In the given exercise, this understanding was affirmed when we calculated the derivative to be 17.
Calculus
Calculus is a branch of mathematics that studies continuous change. It is divided into two main parts: differentiation and integration. Differentiation, which includes finding derivatives, is concerned with understanding rates of change and slopes of curves at given points.
  • The exercise you encountered is an example of differentiation, focusing on computing the derivative of a linear function.
  • Calculus employs the limit process among other tools to handle tasks involving continuous change smoothly.
Through differentiation, one can discern how functions behave at any specific point along their curve. While derivatives for linear functions provide constant rates of change like 17 for the function \( f(x) = 17x + 11 \), calculus enables this powerful analysis across various types of functions, providing valuable insights in fields like physics, engineering, and economics.

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Most popular questions from this chapter

Explain what is wrong with the statement. Given \(f(2)=6, f^{\prime}(2)=3,\) and \(f^{-1}(3)=4,\) we have $$\left(f^{-1}\right)^{\prime}(2)=\frac{1}{f^{-1}\left(f^{\prime}(2)\right)}=\frac{1}{f^{-1}(3)}=\frac{1}{4}$$.

Find the limit of the function as \(x \rightarrow \infty\) $$\frac{e^{2 x}}{\sinh (2 x)}$$

(a) Find tanh 0 (b) For what values of \(x\) is tanh \(x\) positive? Negative? Explain your answer algebraically. (c) On what intervals is tanh \(x\) increasing? Decreasing? Use derivatives to explain your answer. (d) Find \(\lim _{x \rightarrow \infty} \tanh x\) and \(\lim _{x \rightarrow-\infty} \tanh x .\) Show this information on a graph. (e) Does tanh \(x\) have an inverse? Justify your answer using derivatives.

Imagine you are zooming in on the graph of each of the following functions near the origin: $$\begin{aligned} &y=x \quad y=\sqrt{x}\\\ &y=x^{2} \quad y=\sin x\\\ &y=x \sin x \quad y=\tan x\\\ &y=\sqrt{x /(x+1)} y=x^{3}\\\ &y=\ln (x+1) \quad y=\frac{1}{2} \ln \left(x^{2}+1\right)\\\ &y=1-\cos x \quad y=\sqrt{2 x-x^{2}} \end{aligned}$$ Which of them look the same? Group together those functions which become indistinguishable, and give the equations of the lines they look like.

Sketch the circles \(y^{2}+x^{2}=1\) and \(y^{2}+(x-3)^{2}=4\) There is a line with positive slope that is tangent to both circles. Determine the points at which this tangent line touches each circle.
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