Problem 7 - Calculus for Biology and Medicine

Question

Evaluate each function at the given point. (a) \(f_{1}(x, y)=2 x-3 y^{2}\) at \((-1,2)\) (b) \(f_{2}(y, x)=2 x-3 y^{2}\) at \((-1,2)\)

Step-by-Step Solution

Answer

(a) \(f_{1}(-1, 2) = -14\); (b) \(f_{2}(-1, 2) = 1\).

Step 1: Identify the given points for (a)

The function is \(f_{1}(x, y)=2x-3y^{2}\) and the given point is \((-1, 2)\). This means \(x = -1\) and \(y = 2\).

Step 2: Substitute the values into f1

Substitute \(x = -1\) and \(y = 2\) into the function \(f_{1}(x, y) = 2x - 3y^{2}\).\[ f_{1}(-1, 2) = 2(-1) - 3(2)^{2} \]

Step 3: Calculate f1

Calculate the expression from \[ f_{1}(-1, 2) = -2 - 3(4) = -2 - 12 = -14 \]

Step 4: Identify the given points for (b)

The function is \(f_{2}(y, x)=2x-3y^{2}\) at the given point \((-1, 2)\). This means \(y = -1\) and \(x = 2\).

Step 5: Substitute the values into f2

Substitute \(x = 2\) and \(y = -1\) into the function \(f_{2}(y, x) = 2x - 3y^{2}\).\[ f_{2}(-1, 2) = 2(2) - 3(-1)^{2} \]

Step 6: Calculate f2

Calculate the expression from the previous step:\[ f_{2}(-1, 2) = 4 - 3(1) = 4 - 3 = 1 \]

Key Concepts

Function Evaluation

Function evaluation in calculus is all about finding the value of a function at a specific point. Imagine you have a formula that connects two or more variables, like a recipe that tells you how much sugar and flour to mix for a cake. When we evaluate a function, we substitute the given values of the variables into that formula.

This process lets us find the result for particular conditions. For example, considering the function \( f(x, y) = 2x - 3y^2 \), if we want to know what this formula gives us when \( x = -1 \) and \( y = 2 \), we directly plug in these values.

Function evaluation involves these clear steps:

• Identify each variable and its corresponding value from the given point.
• Replace these in the expression.
• Simplify the resulting expression to get the function's value at the desired point.

This makes function evaluation a practical tool for solving real-world problems by breaking them down into mathematical models.

Substitution Method

The substitution method is our go-to technique for solving equations that involve multiple variables. It's like swapping out the letters in a formula for numbers. If you've ever replaced ingredients in a recipe with their alternatives, you've already used a similar concept!

Here's how to master substitution in calculus:

• Read the problem carefully to identify which values need substituting.
• Take each given variable and substitute it with its corresponding value in your equation.
• Solve the resulting equation to find the value of the function at that point.

For instance, with \( f_1(x, y) = 2x - 3y^2 \) and the point \((-1, 2)\):
1. Identify \( x = -1 \) and \( y = 2 \).
2. Substitute: \( f_1(-1, 2) = 2(-1) - 3(2)^2 \).
The method of substitution helps simplify complex equations and is essential for making sense of mathematical functions.

Coordinate Points

Coordinate points let us pinpoint locations in a plane, much like using a map to find your route. Each coordinate is a pair of numbers that tell us exactly where to look both horizontally and vertically.

An ordered pair is typically written as \((x, y)\), denoting the horizontal and vertical positions, respectively. In the context of our function problems, these points determine the inputs for our functions. The first number represents one variable's value, and the second number represents another's.

Understanding coordinate points involves:

• Recognizing the format as \((x, y)\) or even \((y, x)\) depending on how the function is expressed.
• Using these to substitute directly into the function.
• Ensuring each part of the point matches its corresponding variable in the function.

Such a method is critical for pinpointing the exact inputs needed for function evaluation, plot graphs accurately, and solving complex math problems with precision.