Problem 27 - Algebra and Trigonometry

Question

Evaluate the piecewise defined function at the indicated values. $$ \begin{array}{l}{f(x)=\left\\{\begin{array}{ll}{x^{2}} & {\text { if } x<0} \\\ {x+1} & {\text { if } x \geq 0}\end{array}\right.} \\ {f(-2), f(-1), f(0), f(1), f(2)}\end{array} $$

Step-by-Step Solution

Answer

f(-2)=4, f(-1)=1, f(0)=1, f(1)=2, f(2)=3.

Step 1: Evaluate f(-2)

Since {-2 < 0} , we use the first piece of the function, which is {f(x) = x^2} . Substitute {x = -2} : {f(-2) = (-2)^2 = 4} .

Step 2: Evaluate f(-1)

Since {-1 < 0} , we again use the first piece of the function, {f(x) = x^2} . Substitute {x = -1} : {f(-1) = (-1)^2 = 1} .

Step 3: Evaluate f(0)

Since {0 ot< 0} (0 is not less than 0), we use the second piece of the function, {f(x) = x + 1} , because {0 ext{ is } ext{greater than or equal to 0}} . Substitute {x = 0} : {f(0) = 0 + 1 = 1} .

Step 4: Evaluate f(1)

Since {1 ext{ is } ext{greater than 0}} , we use the second piece of the function, {f(x) = x + 1} . Substitute {x = 1} : {f(1) = 1 + 1 = 2} .

Step 5: Evaluate f(2)

Since {2 ext{ is } ext{greater than 0}} , we use the second piece of the function, {f(x) = x + 1} . Substitute {x = 2} : {f(2) = 2 + 1 = 3} .

Key Concepts

Function Evaluation

Function evaluation is the process of determining the output of a function for a given input. For the function \( f(x) = \begin{cases} x^2 & \text{if } x < 0 \ x+1 & \text{if } x \geq 0 \end{cases} \), we evaluate it by plugging specific values of \( x \) into the pieces of the function. In this example, the inputs are \( f(-2), f(-1), f(0), f(1), \text{and } f(2) \). For each input, consider its condition and choose the corresponding function piece to evaluate:

• If the input is negative, use the function \( f(x) = x^2 \).
• If the input is zero or positive, use the function \( f(x) = x + 1 \).

For instance, evaluating \( f(-2) \) involves using \( f(x) = x^2 \) because \( -2 < 0 \), resulting in \( f(-2) = (-2)^2 = 4 \). For \( f(0) \), since zero is not less than zero, we use the second piece, obtaining \( f(0) = 0 + 1 = 1 \). By understanding which part of the function to use, you can accurately find each function value.

Mathematical Reasoning

Mathematical reasoning involves logical thinking to select the correct piece of a piecewise defined function based on the input. It helps us decide which part of the function applies. For our function example, logical steps guide us on how to handle different inputs:

• For \( x < 0 \), logical reasoning shows that the function is equal to \( x^2 \). Evaluating inputs like \( -2 \) and \( -1 \) meets this condition, hence, the calculations \( (-2)^2 = 4 \) and \( (-1)^2 = 1 \) are straightforward.
• For \( x \geq 0 \), logical reasoning leads us to use \( x + 1 \). With inputs like \( 0, 1, \) and \( 2 \), we substitute and compute the results as \( 0 + 1 = 1, 1 + 1 = 2, \) and \( 2 + 1 = 3 \), respectively.

By practicing these logical steps, you build a strong foundation in mathematical reasoning. It ensures that you consistently apply the right operations, leading to correct evaluations of piecewise functions.

Piecewise Defined Function

Piecewise defined functions are mathematical expressions specified by multiple sub-functions. Each sub-function applies to a specific interval of the function's domain. In our example, \( f(x) = \begin{cases} x^2 & \text{if } x < 0 \ x+1 & \text{if } x \geq 0 \end{cases} \), there are two parts:

• \( x^2 \) is used when \( x < 0 \), meaning it describes the function's behavior for negative inputs.
• \( x + 1 \) applies to inputs where \( x \geq 0 \), including zero and any positive values.

The piecewise approach models situations where a rule changes based on different conditions, like physics problems with different forces at different times or tax calculations that apply varied rates depending on income levels. By understanding how piecewise defined functions switch rules for distinct input sets, we can effectively model and solve real-world problems that require separate logical cases.