Problem 96 - Precalculus Essentials

Question

Let \(f\) and \(g\) be defined by the following table: $$\begin{array}{ccc} x & f(x) & g(x) \\ \hline-2 & 6 & 0 \\ -1 & 3 & 4 \\ 0 & -1 & 1 \\ 1 & -4 & -3 \\ 2 & 0 & -6 \end{array}$$ $$\text { Find } \sqrt{f(-1)-f(0)}-[g(2)]^{2}+f(-2) \div g(2) \cdot g(-1)$$ $$\text { Find }|f(1)-f(0)|-[g(1)]^{2}+g(1) \div f(-1) \cdot g(2)$$

Step-by-Step Solution

Answer

The first expression is undefined and the second expression evaluates to \(-8\).

Step 1: Substitute the function values for the first expression

Use the value of \(f\) and \(g\) at various points as needed in the equation \(\sqrt{f(-1)-f(0)}-[g(2)]^{2}+f(-2) \div g(2) \cdot g(-1)\). After substituting, the expression becomes \(\sqrt{3 - (-1)} - (0)^{2} + 6 \div 0 * 4\)

Step 2: Simplifying the expression further

This exercise involves undefined calculation as we're making a division by zero in the last operation of the expression. Since division by zero is undefined in mathematics, we cannot provide a numerical value for this expression.

Step 3: Substitute the function values for the second expression

Similarly, substitute the values of \(f\) and \(g\) in the second expression \(|f(1)-f(0)|-[g(1)]^{2}+g(1) \div f(-1) \cdot g(2)\), which becomes \(|-4 - (-1)| - (-3)^{2} + -3 \div 3 * -6\) after substitution.

Step 4: Solving the expression

After the necessary substitutions, the expression becomes \(|3| - 9 - 2). Then, solve the modulus, which gives us (3 - 9 - 2)\) to evaluate.

Step 5: Final Calculation

Evaluate the expression \((3 - 9 - 2)\) gives the answer \(-8\).

Key Concepts

Function Values Substitution

Understanding the concept of function values substitution is essential in precalculus and foundational for tackling complex problems. To put it simply, function values substitution involves replacing the variable in a function with a specific value, which essentially gives you the output for that input. In the exercise provided, values for the functions f(x) and g(x) are determined by substituting x with the numbers given in the table. This is a crucial step for evaluating the expressions that combine these functions.

For example, when calculating \(\sqrt{f(-1)-f(0)}\), you look up the values of \(f(-1)\) and \(f(0)\) in the table and substitute them directly, which in this case are 3 and -1, respectively. After substitution, it turns into a calculation involving only numbers: \(\sqrt{3 - (-1)}\), which simplifies to \(\sqrt{4}\) or 2. The process gets trickier when operations involve two functions, such as \(f(-2) \div g(2) \cdot g(-1)\), but the approach remains the same: find the corresponding values in the table, and substitute them into the expression. Remember, the key is to always replace the variables methodically and ensure you are substituting the correct values.

Piecewise Functions

Although not directly demonstrated in the provided exercise, the concept of piecewise functions is another common thread in precalculus studies. A piecewise function is simply a function that has different expressions for different intervals of the input variable x. Imagine a function that behaves one way for negative values of x and another for positive values. To deal with such functions, you need to know which 'piece' of the function to use when evaluating it for any given value of x.

In real-world scenarios, such functions can represent situations like tax brackets, where a different rate applies depending on income, or shipping costs that vary with the weight of the package. It's important to recognize which part of the function applies to the situation at hand before proceeding to solve it, much like choosing the correct tool for a job. If you encounter a problem involving a piecewise function, make sure to understand each 'piece' and its corresponding interval fully before substituting values or making any calculations.

Undefined Mathematical Expressions

Encountering undefined mathematical expressions can be a puzzling experience. One common example is division by zero. In mathematics, dividing a number by zero does not yield a meaningful result; thus, the expression is considered undefined. This is visible in the initial exercise where we have the term f(-2) \div g(2). As per the given table, \(g(2) = 0\), and attempting to divide by zero, as in 6 \div 0, leads to an undefined expression.

Undefined expressions don't just stop at division by zero; they can also arise in cases of taking square roots of negative numbers (when not considering complex numbers) and logarithms of non-positive numbers. It is essential for students to recognize these expressions because they signal that either the problem has no solution, or there might be an error in the calculation or assumptions leading up to that point. Always keep a lookout for such conditions in the problem-solving process to avoid unnecessary confusion.