Problem 43 - Calculus with Concepts in Calculus

Question

Determine which of the following define a function. Explain your reason for any that do not define a function. a. The domain consists of the number \(-2\), which is assigned the number \(\pi\). b. The domain consists of the number \(-2\), which is assigned the numbers \(-2\) and \(\pi\). c. \(f(x)=\pm \sqrt{x}\) d. \(f(x)=\pm \sqrt{x^{2}+1}\) e. \(g(x)=\left\\{\begin{array}{l}x-1 \text { for } x<0 \\ 12 x-6 \text { for } x>0\end{array}\right.\) f. \(g(x)=\left\\{\begin{array}{l}2-4 x \text { for } x<0 \\ x^{2} \text { for } x>1\end{array}\right.\) g. \(g(x)=\left\\{\begin{array}{l}4 x+1 \text { for } x \leq 2 \\ 2 x^{3}-7 \text { for } x \geq 2\end{array}\right.\) h. \(g(x)=\left\\{\begin{array}{l}2-3 x^{3} \text { for } x \leq 1 \\ 3 x^{4}-3 \text { for } x \geq 1\end{array}\right.\) i. \(f(t)=\left\\{\begin{array}{l}t^{2} \text { for } t \text { rational } \\\ t \text { for } t \text { irrational }\end{array}\right.\) j. \(f(t)=\left\\{\begin{array}{l}t^{2} \text { for } t^{2} \text { rational } \\\ t \text { for } t \text { irrational }\end{array}\right.\)

Step-by-Step Solution

Answer

Functional: a, e, g, and i; non-functional: b, c, d, f, h, and j.

Step 1: Analyze Option (a)

A function assigns each element from the domain to exactly one element in the range. In the given option (a), the domain consists of a single number, \(-2\), which is assigned to \(\pi\). This satisfies the condition of a function because each element in the domain maps to just one element in the range.

Step 2: Analyze Option (b)

Option (b) states that the domain consists of the number \(-2\), which is assigned the numbers \(-2\) and \(\pi\). This violates the definition of a function as each input in the domain must map to exactly one output in the range. Therefore, option (b) does not define a function.

Step 3: Analyze Option (c)

The expression \((x) = \pm \sqrt{x}\) implies two possible outputs (positive and negative) for a single input value of \(x\). This does not meet the criteria for a function because it assigns two different outputs to the same input value.

Step 4: Analyze Option (d)

For \((x) = \pm \sqrt{x^{2}+1}\), similar to option (c), this expression assigns two values (positive and negative) for a given \(x\). As such, it does not meet the definition of a function.

Step 5: Analyze Option (e)

The piecewise function \((x)=\begin{cases}x-1, & x<0 \ 12x-6, & x>0 \ \end{cases}\) defines a function because there is no overlap of conditions; each input \(x\) leads to exactly one output and there is no break or discontinuity in the domain.

Step 6: Analyze Option (f)

In \(g(x)=\begin{cases}2-4 x, & x<0 \ x^{2}, & x>1 \end{cases}\), it lacks definition in the interval \(0 \leq x \leq 1\) which means not every input has an associated output. However, within its defined segments, it behaves as a function assigning exactly one output per input.

Step 7: Analyze Option (g)

The function \(g(x)=\begin{cases}4x+1, & x \leq 2 \ 2x^{3}-7, & x \geq 2 \end{cases}\) at \(x=2\) needs careful consideration since two expressions could overlap. If the computations at \(x=2\) give inconsistent outputs, it fails to be a function. But calculating both yields \(4(2)+1 = 9\) and \(2(2)^{3}-7 = 9\). Therefore, it assigns exactly one output per input including at the overlap.

Step 8: Analyze Option (h)

In \(g(x)=\begin{cases}2-3x^{3}, & x \leq 1 \ 3x^{4}-3, & x \geq 1 \end{cases}\), different equations provide outputs for \(x=1\). Evaluating both conditions results in \(2-3(1)^{3} = -1\) and \(3(1)^{4}-3 = 0\), suggesting conflicting values hence it fails to define a function.

Step 9: Analyze Option (i)

For \(f(t)=\begin{cases}t^{2}, & t \text{ rational } \ t, & t \text{ irrational } \end{cases}\), a function definition is violated if a single value of \(t\) being both rational and irrational leads to different outputs. However, no number can be both rational and irrational, thus each value of \(t\) provides a well-defined unique output.

Step 10: Analyze Option (j)

Here, \(f(t)=\begin{cases}t^{2}, & t^{2} \text{ rational } \ t, & t \text{ irrational } \end{cases}\) involves checking rationality of \(t^{2}\), complicating single output guarantee as numbers like \(\sqrt{2}\) offer different outcomes under this definition. Thus, it does not universally satisfy function definitions.

Key Concepts

Piecewise Functions

Piecewise functions are a type of function where different expressions are used to compute outputs based on specific intervals or conditions of the input. Imagine breaking down a road into sections, where each section has a different speed limit. Similarly, in a piecewise function:

• There are distinct rules or parts for different intervals of the domain.
• Each interval employs its own expression to produce outputs.

For example, in the step-by-step solution, option (e) is defined as a piecewise function:\[ g(x)=\begin{cases} x-1, & x<0 \ 12x-6, & x>0 \end{cases} \]In this function, when \(x\) is less than zero, the expression \(x - 1\) is used, and when \(x\) is greater than zero, the expression \(12x - 6\) takes over. The beauty of piecewise functions lies in their ability to describe complex behaviors concisely.
When working with piecewise functions, ensure each piece or rule applies only within its specified conditions, and look out for points where conditions meet, like in options (g) and (h) in the exercise.

Domain and Range

The domain and range are fundamental concepts in understanding functions.

• The domain of a function is the complete set of possible input values (the x-values), for which the function is defined.
• The range, on the other hand, is the full set of possible output values (the y-values) that result from using the domain.

Think about it like this: if a vending machine only accepts coins, the coins are its domain. The snacks you can get out of it represent the range. For instance, in option (e) of the exercise, the domain is all real numbers except zero, due to the conditions \(x < 0\) and \(x > 0\). But for option (f), the interval between zero and one is missing in the domain, meaning it lacks definition here.
Understanding the domain and range helps ensure all inputs correspond to a specific and meaningful output. Always check that the domain is appropriately handled in a function definition to avoid errors or undefined outputs.

Criteria for a Function

To qualify as a function, a mathematical relation has certain criteria it must meet. Let's explore what it means for something to be a function:

• Every element in the domain is associated with exactly one element in the range.
• If you input the same value, you always get the same output.

In the exercise, option (a) clearly defines a function because \(-2\) maps to exactly one number, \(\pi\). However, with option (b), \(-2\) is assigned to both \(-2\) and \(\pi\), violating the one-to-one rule, and thus, it does not define a function.
In options (c) and (d), expressions like \(\pm \sqrt{x}\) introduce ambiguity due to the positive and negative outputs possible for each \(x\), meaning these are not functions. Always ensure there's a unique and consistent outcome for each input before declaring a relation a function. This uniqueness in output is the bedrock of function theory.