Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(\mathrm{F}\) be the collection of ordered number pairs to which an ordered pair \((x, y)\) belongs if and only if \(x\) is a number and \(y=x^{2}+x\). a. Which of the ordered number pairs belong to F? (0,1),(0,0),(1,1),(1,3),(1,-1),(-1,1) , (-1,0),(-1,-1) b. Is there any uncertainty as to the members of \(\mathrm{F} ?\) c. What is the domain of \(\mathrm{F} ?\) d. What is the range of \(\mathrm{F} ?\)

Step-by-Step Solution

Answer

a. (0,0) and (-1,0) belong to F. b. No uncertainty about members. c. The domain is all real numbers. d. The range is [-1/4, ∞).

Step 1: Understanding the Definition of F

The collection \( \mathrm{F} \) consists of ordered pairs \((x, y)\) such that \( y = x^2 + x \). This means for each \( x \), \( y \) is determined by substituting \( x \) into the expression \( x^2 + x \).

Step 2: Evaluate Ordered Pairs

Evaluate each provided pair to determine if it satisfies the rule \( y = x^2 + x \):- For \((0, 1)\), \(y = 0^2 + 0 = 0\); not in \( F \).- For \((0, 0)\), \(y = 0^2 + 0 = 0\); belongs to \( F \).- For \((1, 1)\), \(y = 1^2 + 1 = 2\); not in \( F \).- For \((1, 3)\), \(y = 1^2 + 1 = 2\); not in \( F \).- For \((1, -1)\), \(y = 1^2 + 1 = 2\); not in \( F \).- For \((-1, 1)\), \(y = (-1)^2 + (-1) = 0\); not in \( F \).- For \((-1, 0)\), \(y = (-1)^2 + (-1) = 0\); belongs to \( F \).- For \((-1, -1)\), \(y = (-1)^2 + (-1) = 0\); not in \( F \).

Step 3: Addressing Uncertainty About Members of F

There is no uncertainty when identifying members of \( F \). An ordered pair either satisfies \( y = x^2 + x \) or it does not.

Step 4: Finding the Domain of F

The domain of \( F \) is the set of all possible \( x \) values for which the function \( y = x^2 + x \) is defined. Since \( x \) is a real number, the domain is all real numbers, \( \mathbb{R} \).

Step 5: Determining the Range of F

The range of \( F \) is the set of all possible \( y \) values that result from \( y = x^2 + x \) for \( x \) in the domain of real numbers. Completing the square for \( x^2 + x + \frac{1}{4} - \frac{1}{4} = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \), we see that \( y \geq -\frac{1}{4} \). Thus, the range is \( [-\frac{1}{4}, \infty) \).

Key Concepts

Ordered Pairs

In mathematics, an ordered pair is simply a pair of elements grouped together in a specific sequence. The notation used for an ordered pair is \((x, y)\). Here, \(x\) is typically the first component, and \(y\) is the second. This sequence matters because in ordered pairs, \((x, y)\) is not the same as \((y, x)\) unless \(x = y\).These pairs are crucial in describing points on a coordinate plane, where \(x\) is the horizontal axis (or abscissa) and \(y\) is the vertical axis (or ordinate). In the context of functions like the one presented in the exercise, we verify each ordered pair by using the given function equation, \(y = x^2 + x\). When checking if a pair \((x, y)\) belongs to the set \(F\), substitute \(x\) into the equation, and check if the resulting \(y\) matches the second element of the pair.For example:

• Check \((0, 0)\): Substitute \(x = 0\), which gives \(y = 0^2 + 0 = 0\). So \((0, 0)\) is in the set \(F\).
• Check \((1, 1)\): Substitute \(x = 1\), which gives \(y = 1^2 + 1 = 2\). So \((1, 1)\) is not in \(F\).

This process clarifies whether an ordered pair satisfies the condition set by the function.

Domain and Range

The domain and range are fundamental concepts in understanding functions. The **domain** of a function is the set of all possible input values (or \(x\)-values) which the function can accept without causing any undefined behavior. For the function \(y = x^2 + x\), the domain encompasses all real numbers, \(\mathbb{R}\). This means that no matter what real number we choose for \(x\), \(y\) will take a definite value.On the other hand, the **range** of a function denotes all possible output values (or \(y\)-values) the function can produce, given the inputs from its domain. Determining the range often requires more depth of understanding about the function itself. For our function, after some algebraic manipulation like completing the square, we find the expression can be rewritten as \(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\). The minimum value of \(\left(x + \frac{1}{2}\right)^2\) is 0, leading to a minimum \(y\)-value of \(-\frac{1}{4}\). Thus, the range of the function is the interval \([-\frac{1}{4}, \infty)\), which means \(y\) can be any value from \(-\frac{1}{4}\) to positive infinity.

Completing the Square

Completing the square is a technique used to simplify calculations with quadratic expressions. It makes it easier to analyze the properties of a quadratic function, such as finding the vertex or determining the range.To complete the square for the function \(y = x^2 + x\), we manipulate it to take the form \((x + a)^2 + b\). Begin by identifying the coefficient of the \(x\) term, dividing it by 2, and squaring the result:

• The coefficient of \(x\) in \(x^2 + x\) is 1.
• Divide by 2 and square it: \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

Add and subtract \(\frac{1}{4}\) within the expression: \[y = x^2 + x + \frac{1}{4} - \frac{1}{4}\]. This simplifies to \[\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\], which is the vertex form.By completing the square, we now see that the vertex of the parabola described by \(y = x^2 + x\) is at \((-\frac{1}{2}, -\frac{1}{4})\), giving us insights into the minimum values the function can achieve.