Problem 4 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(h(x, y)=(x+y) /(x-y) .\) Compute \(h(0,1), h(-1,1)\), \(h(2,1)\), and \(h(\pi,-\pi) .\)

Step-by-Step Solution

Answer

The short answer is: h(0, 1) = \(-1\) h(-1, 1) = \(0\) h(2, 1) = \(3\) h(π, -π) = \(0\)

Step 1: 1. Plug in values for h(0,1)

We plug x=0 and y=1 into the function.

Step 2: 2. Compute h(0,1)

When x=0 and y=1, we have: h(0, 1) = \(\frac{0+1}{0-1}\) = \(-1\)

Step 3: 3. Plug in values for h(-1,1)

We plug x=-1 and y=1 into the function.

Step 4: 4. Compute h(-1,1)

When x=-1 and y=1, we have: h(-1, 1) = \(\frac{-1+1}{-1-1}\) Since the numerator is 0 and the denominator is -2, we have: h(-1, 1) = \(\frac{0}{-2}\) = \(0\)

Step 5: 5. Plug in values for h(2,1)

We plug x=2 and y=1 into the function.

Step 6: 6. Compute h(2,1)

When x=2 and y=1, we have: h(2, 1) = \(\frac{2+1}{2-1}\) = \(\frac{3}{1}\) = \(3\)

Step 7: 7. Plug in values for h(π,-π)

We plug x=π and y=-π into the function.

Step 8: 8. Compute h(π,-π)

When x=π and y=-π, we have: h(π, -π) = \(\frac{\pi-\pi}{\pi+\pi}\) = \(\frac{0}{2\pi}\) = \(0\) In conclusion, we have computed the following output values for h(x, y): h(0, 1) = \(-1\) h(-1, 1) = \(0\) h(2, 1) = \(3\) h(π, -π) = \(0\)

Key Concepts

Applied Mathematics

Applied mathematics encompasses a vast range of mathematical tools that are used to solve real-world problems across various fields such as science, engineering, business, computer science, and industry.

The computation of a rational function like the one in our example, where we evaluate the function
\(h(x, y) = \frac{x+y}{x-y}\), is integral in applying mathematical concepts to practical situations. By calculating the values for \(h(0,1), h(-1,1), h(2,1)\), and \(h(\pi,-\pi)\), we're seeing how mathematical analysis can help us understand relationships and predict outcomes in systems where variables can change, like studying economics, predicting population growth, or even in more abstract areas such as game theory.

Function Evaluation

Function evaluation is a fundamental practice in algebra that involves substituting variables within a function with specific values to determine the function's output.

When evaluating \( h(x, y) = \frac{x+y}{x-y} \), such as in the provided exercise, we took particular inputs, applied them to the function, and calculated the outputs. This process can be applied to any function to understand how a change in the inputs affects the results, which is essential in fields like physics or economics where functional relationships are key to modeling real-world situations.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operators (like addition, subtraction, multiplication, and division) that define a particular mathematical statement or function.

In the problem at hand, \(x+y\) and \(x-y\) are algebraic expressions which together form a rational function. Knowing how to manipulate these expressions is crucial for simplifying and solving problems. In the real world, the ability to understand and reformulate expressions allows us to turn complex problems into more manageable ones, paving the way for breakthroughs in science and engineering.

Mathematical Computations

Mathematical computations involve the process of performing algebraic, arithmetic, or logical operations to reach a conclusion. This is the actual 'work' part of math, where you perform the required calculations to find answers.

The exercise showed a straightforward example, with steps like substituting values into the function and then carrying out the divisions to find the values of \(h(x,y)\). Performing these computations correctly is crucial, as even small errors can significantly impact the outcome in fields where precision is necessary, such as in engineering or data analysis.