Problem 35 - A Graphical Approach to Precalculus with Limits

Question

Use the table to evaluate each expression, if possible. (a) \((f+g)(2)\)(b) \((f-g)(4)\)(c) \((f g)(-2)\)(d) \(\left(\frac{f}{g}\right)(0)\) $$\begin{array}{|r|r|r|}\hline x & f(x) & g(x) \\\\\hline-2 & 0 & 6 \\\\\hline 0 & 5 & 0 \\\\\hline 2 & 7 & -2 \\\\\hline 4 & 10 & 5 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

(a) 5, (b) 5, (c) 0, (d) Undefined

Step 1: Evaluate (f+g)(2)

To find \(f+g\)(2), we need to add \(f(2)\) and \(g(2)\). From the table, \(f(2) = 7\) and \(g(2) = -2\). Therefore, \(f(2) + g(2) = 7 + (-2) = 5\).

Step 2: Evaluate (f-g)(4)

To find \(f-g\)(4), we need to subtract \(g(4)\) from \(f(4)\). From the table, \(f(4) = 10\) and \(g(4) = 5\). Therefore, \(f(4) - g(4) = 10 - 5 = 5\).

Step 3: Evaluate (f g)(-2)

To find \(f g\)(-2), we need to multiply \(f(-2)\) and \(g(-2)\). From the table, \(f(-2) = 0\) and \(g(-2) = 6\). Therefore, \(f(-2) \, \cdot \, g(-2) = 0 \, \cdot \, 6 = 0\).

Step 4: Evaluate (f/g)(0)

To find \left(\frac{f}{g}\right)(0), we need to divide \(f(0)\) by \(g(0)\). From the table, \(f(0) = 5\) and \(g(0) = 0\). Division by zero is undefined, hence this expression cannot be evaluated.

Key Concepts

Arithmetic Operations

Understanding arithmetic operations is key when solving function evaluation problems. These tasks often involve adding, subtracting, multiplying, or dividing the values of functions. These operations are straightforward but follow specific rules that mirror everyday arithmetic.

When adding functions, as in the expression \(f+g\)(2), you simply add the outputs of the functions at a given input value. For instance, if \(f(2) = 7\) and \(g(2) = -2\) from the table provided, then \(f(2) + g(2) = 7 + (-2) = 5\). This operation emphasizes the importance of paying attention to signs, especially when integrating negative numbers.

• Be careful with negative numbers - they affect the outcome significantly.
• Always substitute values directly from the table for accuracy.

For subtraction and multiplication, similar principles apply. Subtracting requires attention to detail in preserving the order of operations. For \(f-g\)(4), you subtract \(g(4) = 5\) from \(f(4) = 10\), resulting in 5.

Multiplication, as shown in \(f\cdot g\)(-2), requires multiplying the outputs directly. With \(f(-2) = 0\) and \(g(-2) = 6\), the result is simply 0, showing how zero can dominate a multiplication operation.

• Remember multiplication and addition are commutative; subtraction is not.
• In multiplication, any term that is zero nullifies the product.

Undefined Expressions

Undefined expressions crop up mostly in division operations. It's a crucial point to grasp that you cannot divide any number by zero, as this renders the expression undefined.

In the expression \left(\frac{f}{g}\right)(0)\, you are instructed to divide \(f(0) = 5\) by \(g(0) = 0\). Because dividing by zero is mathematically undefined, the expression cannot proceed to a solution. This is a common occurrence in calculus and algebra.

• Dividing by zero is undefined - always check the denominator first.
• An undefined result means it cannot be calculated with set rules.

Spotting undefined expressions is vital for function evaluation. It saves time as you already know that the operation can't be computed. Instead of wasting time trying to solve an impossible fraction, recognize it, and note it as undefined.

Table Interpretation

Tables are fundamental tools in evaluating functions. They provide all necessary data, like input-output pairs, for accurate function computation.

Understanding how to read and use tables can greatly simplify complex calculations. Each row in a table corresponds to a specific input value and its respective function outputs. Look at the table from the problem statement; each row offers the exact values needed for evaluation.

• Find the \(x\)-value in the first column and use the corresponding \(f(x)\) and \(g(x)\) values.
• Double-check that you select the correct row to avoid mistakes.

Using the table correctly allows you to swiftly compute operations such as addition or multiplication of function outputs. The goal is to effectively read off values to substitute into expressions, all while maintaining precision.