Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the function at each specified value of the independent variable and simplify. \(f(x)=\left\\{\begin{array}{ll}x^{2}+1, & x \leq 1 \\ 2 x-3, & x>1\end{array}\right.\) (a) \(f(-2)\) (b) \(f(1)\) (c) \(f\left(\frac{3}{2}\right)\) (d) \(f(0)\)

Step-by-Step Solution

Answer

The function \(f(x)\) evaluated at the specified values is as follows: \(f(-2)=5\), \(f(1)=2\), \(f\left(\frac{3}{2}\right)=0\), and \(f(0)=1\).

Step 1: Evaluate \(f(-2)\)

Since \(-2 \leq 1\), we will use the first part of the function, which is \(x^{2}+1\). Plugging in -2, we get \((-2)^{2}+1 = 4 + 1 = 5\). So, \(f(-2)=5\).

Step 2: Evaluate \(f(1)\)

While 1 is a boundary value, the function specifies that we should use \(x^{2}+1\) when \(x \leq 1\). Therefore, when we substitute 1 into that part of the function, we get \(1^{2}+1 = 1 + 1 = 2\). So, \(f(1) = 2\).

Step 3: Evaluate \(f\left(\frac{3}{2}\right)\)

Since \(\frac{3}{2} > 1\), we should use the second part of the function, which is \(2x-3\). Substituting \(x = \frac{3}{2}\) into this function gives us \(2*\frac{3}{2}-3 = 3 - 3 = 0\). So, \(f\left(\frac{3}{2}\right) = 0\).

Step 4: Evaluate \(f(0)\)

0 is less than or equal to 1, so we will use the first part of the function, \(x^{2}+1\). When we plug 0 into this part of the function, we get \(0^{2}+1 = 0 + 1 = 1\). So, \(f(0) = 1\).

Key Concepts

Function Evaluation

The first step in working with piecewise functions is function evaluation. This involves determining which part of the given function applies to the specific value of the independent variable (commonly "x"). In this case, our piecewise function is given as:

• For values of \(x\) less than or equal to 1, use \(f(x) = x^2 + 1\).
• For values of \(x\) greater than 1, use \(f(x) = 2x - 3\).

To evaluate at a specific point, simply identify which part of the function to use based on the value of \(x\). For example, when evaluating \(f(-2)\), since \(-2 \leq 1\), we use the first expression \((-2)^2 + 1\), resulting in \(f(-2) = 5\). This process involves substituting \(x\) into the appropriate equation and performing the arithmetic steps.

Domain and Range

Understanding the domain and range in the context of piecewise functions is crucial. The domain refers to all possible input values ("x" values), while the range is all potential outputs ("f(x)" values). With piecewise functions, the domain is often partitioned, meaning different rules apply to different intervals of \(x\).For instance, in this function:

• The domain for \(x^2 + 1\) is \(x \leq 1\), capturing inputs from negative infinity up to and including 1.
• The domain for \(2x - 3\) is \(x > 1\), capturing values from just above 1 to infinity.

The range would be calculated by examining each piece separately, determining the output values corresponding to the given domain sections. For example:

• For \(x^2 + 1\), as \(x\) covers its domain, the function outcome varies from 1 upwards.
• For \(2x - 3\), outputs include those greater than -1 (since as \(x\) increases, \(f(x)\) increases without bound).

Mathematical Notation

Mathematical notation is the language used to express mathematical ideas succinctly. Piecewise functions are a specific form of notation that combines multiple functions into one overarching rule. Formalizing this in practice:The piecewise function here is written as:\[ f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq 1 \ 2x - 3 & \text{if } x > 1 \end{cases} \]This notation tells us clearly which mathematical expression to use depending on the input \(x\). The structure uses curly braces to encapsulate the different expressions, alongside conditional statements that segment the domain (e.g., \(x \leq 1\) and \(x > 1\)). This ensures clarity when translating the function into computations.When presented with a problem, understanding the notation helps to effortlessly interpret the matching rules and conditions. This makes it easier to proceed accurately with evaluations and calculations.