Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 6
Let \(f(x, y)=x y .\) Show that \(f(2,3+k)-f(2,3)=2 k\).
Short Answer
Expert verified
\( f(2, 3+k) - f(2, 3) = 2k \)
Step by step solution
01
Understand the given function
The function provided is \( f(x, y) = xy \). This means that when you substitute values for \( x \) and \( y \), you multiply them together to get the result.
02
Substitute for 饾惐 and 饾惒 in 饾拠(饾煇, 饾煈+饾拰)
To find \( f(2, 3+k) \), substitute \( x = 2 \) and \( y = 3 + k \) into the function \( f(x, y) \). Hence, \( f(2, 3+k) = 2(3+ k) = 6 + 2k \).
03
Substitute for 饾惐 and 饾惒 in 饾拠(饾煇, 饾煈)
For \( f(2, 3) \), substitute \( x = 2 \) and \( y = 3 \) into the function \( f(x, y) \). Thus, \( f(2, 3) = 2(3) = 6 \).
04
Find the difference 饾拠(饾煇, 饾煈+饾拰)鈭掟潚�(饾煇, 饾煈)
Subtract \( f(2, 3) \) from \( f(2, 3+k) \). Therefore, \( f(2, 3+k) - f(2, 3) = (6 + 2k) - 6 = 2k \).
05
Conclusion
We have now shown that \( f(2, 3+k) - f(2, 3) = 2k \).
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Multivariable Functions
A multivariable function is a function that depends on two or more variables. In our exercise, the function is given by 饾憮(饾懃, 饾懄) = 饾懃饾懄, which means it takes two inputs, 饾懃 and 饾懄, and outputs their product.
Understanding multivariable functions is crucial because many real-world phenomena depend on more than one variable. For instance, the area of a rectangle depends both on its length and width.
When dealing with these functions, it's important to understand how changing one variable while keeping the others constant affects the overall outcome. This will help you get comfortable with the concept of partial derivatives, which are like regular derivatives but applied to multivariable functions.
In our exercise, we are particularly interested in how changing 饾懄 (specifically transitioning from 3 to 3+k) while keeping 饾懃 constant (at 2) affects the function's value.
Understanding multivariable functions is crucial because many real-world phenomena depend on more than one variable. For instance, the area of a rectangle depends both on its length and width.
When dealing with these functions, it's important to understand how changing one variable while keeping the others constant affects the overall outcome. This will help you get comfortable with the concept of partial derivatives, which are like regular derivatives but applied to multivariable functions.
In our exercise, we are particularly interested in how changing 饾懄 (specifically transitioning from 3 to 3+k) while keeping 饾懃 constant (at 2) affects the function's value.
Differentiation
Differentiation in calculus is a process that helps us find how a function changes as its input changes. When dealing with multivariable functions, we use partial differentiation.
Partial differentiation allows us to look at how the function changes with respect to one variable while holding the other variables constant. For the function 饾憮(饾懃, 饾懄) = 饾懃饾懄, if we want to see how it changes with respect to 饾懄, we differentiate it partially with respect to 饾懄.
This results in:
\(\frac{\text{鈭倉f}{\text{鈭倉y} = x \).
This tells us that for every unit increase in 饾懄, the function increases by 饾懃 times that unit. In our specific exercise, since 饾懃 = 2, it means for every unit change in 饾懄, the function changes by 2 times that unit (or simply 2k if 饾懄 changes by k).
This forms the core understanding of how partial derivatives are applied. Differentiation is vital because it lays the groundwork for analyzing optimization problems, understanding rates of change, and much more.
Partial differentiation allows us to look at how the function changes with respect to one variable while holding the other variables constant. For the function 饾憮(饾懃, 饾懄) = 饾懃饾懄, if we want to see how it changes with respect to 饾懄, we differentiate it partially with respect to 饾懄.
This results in:
\(\frac{\text{鈭倉f}{\text{鈭倉y} = x \).
This tells us that for every unit increase in 饾懄, the function increases by 饾懃 times that unit. In our specific exercise, since 饾懃 = 2, it means for every unit change in 饾懄, the function changes by 2 times that unit (or simply 2k if 饾懄 changes by k).
This forms the core understanding of how partial derivatives are applied. Differentiation is vital because it lays the groundwork for analyzing optimization problems, understanding rates of change, and much more.
Function Evaluation
Function evaluation involves substituting given values into the function to calculate an output.
For the exercise, the function 饾憮(饾懃, 饾懄) = 饾懃饾懄 needs to be evaluated at specific points. First, we substitute x = 2 and y = 3+k into the function to get 饾憮(2, 3+k). This gives us 2(3+k) = 6 + 2k.
Next, we evaluate the function at another specific point, x = 2 and y = 3, to get 饾憮(2, 3), which results in 2(3) = 6.
Finally, to find the difference between these two results, we subtract one from the other: 6 + 2k - 6, simplifying to 2k.
This illustrates that proper function evaluation using substitution is essential for solving most calculus problems involving specific inputs.
By mastering function evaluation, you set a solid foundation for tackling more complex calculus problems with confidence.
For the exercise, the function 饾憮(饾懃, 饾懄) = 饾懃饾懄 needs to be evaluated at specific points. First, we substitute x = 2 and y = 3+k into the function to get 饾憮(2, 3+k). This gives us 2(3+k) = 6 + 2k.
Next, we evaluate the function at another specific point, x = 2 and y = 3, to get 饾憮(2, 3), which results in 2(3) = 6.
Finally, to find the difference between these two results, we subtract one from the other: 6 + 2k - 6, simplifying to 2k.
This illustrates that proper function evaluation using substitution is essential for solving most calculus problems involving specific inputs.
By mastering function evaluation, you set a solid foundation for tackling more complex calculus problems with confidence.
