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Chapter 7: Problem 13

For each function, evaluate the given expression. $$ w(u, v)=\frac{1+2 u+3 v}{u v}, \text { find } w(-1,1) $$

Short Answer

Expert verified
The value of \( w(-1, 1) \) is \(-2\).

Step by step solution

01

Understand the Function

The function given is \( w(u, v) = \frac{1 + 2u + 3v}{uv} \). This means for any value of \( u \) and \( v \), you substitute these into the formula to calculate \( w(u, v) \).
02

Substitute the Values

We need to find \( w(-1, 1) \), which means substituting \( u = -1 \) and \( v = 1 \) into the function. So,\[w(-1, 1) = \frac{1 + 2(-1) + 3(1)}{(-1)(1)}.\]
03

Simplify the Numerator

Calculate the numerator:\[1 + 2(-1) + 3(1) = 1 - 2 + 3 = 2.\]
04

Simplify the Denominator

Calculate the denominator:\[(-1)(1) = -1.\]
05

Evaluate the Expression

Now substitute the simplified numerator and denominator into the expression:\[w(-1, 1) = \frac{2}{-1} = -2.\]
06

Conclude the Calculation

The final value of the function for \( w(-1, 1) \) is \(-2\).

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Key Concepts

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

Substitution in Functions
When dealing with functions, understanding substitution is essential. The process involves replacing the variables in a function with specific values. In essence, it's about plugging numbers into an algebraic expression to find a specific output.
If you're working with a function like \( w(u, v) = \frac{1 + 2u + 3v}{uv} \), and you're asked to evaluate \( w(-1, 1) \), you substitute \( u = -1 \) and \( v = 1 \). This transforms the expression as follows:
  • Everywhere you see 'u', insert -1.
  • Everywhere you see 'v', insert 1.
This process is often the first step in evaluating a function and is crucial for understanding how changes in input affect the output.
Numerator Simplification
Once you've substituted the values into the function, the next step is to simplify the numerator of the fraction. The numerator refers to the top part of a fraction, and simplifying it helps clarify the expression.
In our exercise, after substitution, the numerator became \(1 + 2(-1) + 3(1)\). Simplifying involves performing the arithmetic within this part:
  • Multiply: Calculate \(2(-1) = -2\).
  • Add: Combine to find \(1 - 2 + 3 = 2\).
Through careful simplification, we arrive at a much simpler expression, which makes it easier to move on to the denominator. Remember, working step-by-step helps prevent errors and ensures precision.
Denominator Simplification
Simplifying the denominator is just as important as the numerator. The denominator is the bottom part of a fraction that shows how many equal parts the numerator is divided by.
In this example, simplifying involves finding the product of \((-1)(1)\). This is straightforward:
  • Multiply: The expression \((-1)(1)\) simplifies directly to \(-1\).
With a simplified denominator, it's easier to proceed to the final step of evaluating the function expression. Keep in mind the sign of your result, as this can affect the final outcome of your calculation.
Algebraic Expressions
Algebraic expressions are the backbone of mathematical functions. They're made up of variables, numbers, and operators working together systematically.
In dealing with expressions like \(w(u, v) = \frac{1 + 2u + 3v}{uv}\), you encounter a blend of multiplication, addition, and division. Understanding how each part functions together is crucial:
  • Variables: \(u\) and \(v\) are placeholders that can represent any number.
  • Coefficients: Numbers like 2 and 3 that scale variables.
  • Operators: Symbols like +, -, and / dictate the operations performed.
By mastering these components, evaluating complex algebraic expressions becomes less daunting. Remember that simplifying expressions piece by piece, as done here, leads to accurate outcomes.

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Most popular questions from this chapter

The following table shows how a driver's blood-alcohol level (\% grams per dekaliter) affects the probability of being in a collision. A collision factor of 3 means that the probability of a collision is 3 times as large as normal. Fit an exponential curve to the data. Then use your curve to estimate the collision factor for a blood-alcohol level of \(15 .\) $$ \begin{array}{cc} \hline \text { Blood-Alcohol } & \\ \text { Level } & \text { Collision Factor } \\ 0 & 1 \\ 6 & 1.1 \\ 8 & 3 \\ 10 & 6 \end{array} $$

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Use least squares to find the exponential curve \(y=B e^{A x}\) for the following tables of points. $$ \begin{array}{c|c} x & y \\ \hline-2 & 20 \\ 0 & 12 \\ 2 & 9 \\ 4 & 6 \\ 5 & 5 \end{array} $$

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