Chapter 4: Problem 24
Use the formal definition to find the derivative of \(y=\frac{1}{x}\) at \(x=2\). (b) Show that the point \(\left(2, \frac{1}{2}\right)\) is on the graph of \(y=\frac{1}{x}\), and find the equation of the normal line at the point \(\left(2, \frac{1}{2}\right)\). (c) Graph \(y=\frac{1}{x}\) and the tangent line at the point \(\left(2, \frac{1}{2}\right)\) in the same coordinate system.
Short Answer
Step by step solution
Define the Derivative Formally
Simplify the Difference Quotient
Evaluate the Limit
Calculate the Derivative at \(x=2\)
Confirm the Point is on the Graph
Find the Normal Line Equation
Graph the Function and Lines
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Formal Definition of Derivative
The formal definition of a derivative is given by:
- \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]
- Find \(f(x + h)\) by substituting \(x + h\) into the function.
- Calculate the difference quotient \(\frac{f(x+h) - f(x)}{h}\).
- Simplify the expression to make taking the limit as \(h\) approaches 0 easier.
- Evaluate the limit to find the derivative \(f'(x)\).
Tangent Line
To find the equation of the tangent line at a point:
- First, compute the derivative of the function at that point. For instance, if we have the derivative \(f'(2) = -\frac{1}{4}\), this value is the slope of the tangent line.
- Use the point-slope form: \[y - y_1 = m(x - x_1)\] where \(m\) is the slope (\(-\frac{1}{4}\)) and \((x_1, y_1)\) is the point \((2, \frac{1}{2})\).
- \[y - \frac{1}{2} = -\frac{1}{4}(x - 2)\]
Normal Line
If the slope of the tangent line is known, say \(-\frac{1}{4}\), the normal line's slope is the negative reciprocal of this slope. This is calculated as follows:
- Slope of normal line = \(-\frac{1}{\left(-\frac{1}{4}\right)} = 4\)
- \[y - \frac{1}{2} = 4(x - 2)\]
- \[y = 4x - 7.5\]