Problem 49 - Precalculus with Limits

Question

In Exercises 37-52, evaluate the function at each specified value of the independent variable and simplify. \( f(x) = \left\\{ \begin{array}{ll} 2x + 1, & \mbox{ \) x < 0 \(} \\ 2x + 2, & \mbox{ \) x \ge 0 \(} \end{array} \right.\) (a) \(f(-1)\) (b) \(f(0)\) (c) \(f(2)\)

Step-by-Step Solution

Answer

The given function \(f(x)\) evaluates to \(f(-1) = -1\), \(f(0) = 2\), and \(f(2) = 6\).

Step 1: Solve for \(f(-1)\)

Since -1 is less than 0, we use the rule \(2x + 1\) of the piecewise function. We insert -1 in place of \(x\) to yield \(2(-1) + 1 = -2 + 1 = -1\)

Step 2: Solve for \(f(0)\)

Now, we are dealing with 0 which is not less than 0, so we use the second rule of the piecewise function, \(2x + 2\). Substituting 0 in place of \(x\), we get \(2(0) + 2 = 2\)

Step 3: Solve for \(f(2)\)

Lastly, we substitute 2, which is greater than 0, so we use the second rule \(2x+2\). Hence we have \(2(2)+2=6\)

Key Concepts

Function Evaluation

Function evaluation is an essential skill when dealing with piecewise functions. It involves determining the output of a function for specific input values. For our given function, the piecewise rule has two parts. Depending on the input, the function uses either one rule or the other. It works like a decision-maker.
For instance, if we evaluate the function at \( x = -1 \), we check which condition \( x \) meets. Since \( -1 < 0 \), we apply the formula on the left of our piecewise definition:

• Use \( 2x + 1 \) for values less than 0.
• Use \( 2x + 2 \) for values equal to or greater than 0.

Plug \(-1\) into \(2x + 1\) and compute the result: \[ f(-1) = 2(-1) + 1 = -2 + 1 = -1 \] This shows how piecewise functions calculate different outputs based on where \( x \) falls on the number line.

Independent Variable

The independent variable is a vital component of functions, often denoted as \( x \). It represents the input value that dictates which rule of the piecewise function to apply. In our exercise, \( x \) can take on any value, and this value directs us to choose either the condition for \( x < 0 \) or \( x \ge 0 \).
Let's explore using our exercise's example:

• For \( x = 0 \) and \( x = 2 \), both cases apply the rule for \( x \ge 0 \).
• Hence, both are calculated using \( 2x + 2 \).

When \( x = 0 \), this results in: \[ f(0) = 2(0) + 2 = 2 \]When \( x = 2 \), this gives the evaluation: \[ f(2) = 2(2) + 2 = 6 \]Understanding the independent variable helps determine the correct section of the piecewise function to use, ensuring accurate calculations.

Simplification

Simplification in mathematical expressions is about reducing them to their simplest form. After evaluating a function based on the piecewise definition, it's crucial to simplify the result to make it clearer and more concise.
Consider the solution for each part of the exercise. By substituting the independent variable into the expression, you often end up with more immediate results:

• For \( f(-1) = 2(-1) + 1 \), simplifying gives us \(-2 + 1\).
• The simplified result is \(-1\).
• For \( f(0) = 2(0) + 2 \) and \( f(2) = 2(2) + 2 \), they simplify directly to the summations themselves.

This process eliminates any unnecessary operations, making it manageable and easier to interpret. In practice, simplification ensures that the results from function evaluations are accessible and less prone to error when referenced later. By practicing these simplifications, one becomes adept at reading and reducing expressions efficiently.