Problem 66 - Elementary and Intermediate Algebra

Question

Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=\frac{x^{2}+x-2}{x^{2}-5 x} $$

Step-by-Step Solution

Answer

\(h(5)\) is undefined and \(h(-2) = 0\).

Step 1: Substitute 5 into the Function

To find \(h(5)\), we substitute \(x = 5\) into the function. This gives us \[ h(5) = \frac{5^2 + 5 - 2}{5^2 - 5 \cdot 5} \].

Step 2: Simplify the Numerator for h(5)

Calculate the numerator: \(5^2 + 5 - 2 = 25 + 5 - 2 = 28\).

Step 3: Simplify the Denominator for h(5)

Calculate the denominator: \(5^2 - 5 \cdot 5 = 25 - 25 = 0\).

Step 4: Conclusion of h(5)

Since the denominator is zero, \(h(5)\) is undefined as division by zero is not possible.

Step 5: Substitute -2 into the Function

To find \(h(-2)\), substitute \(x = -2\) into the function. This leads to \[ h(-2) = \frac{(-2)^2 + (-2) - 2}{(-2)^2 - 5(-2)} \].

Step 6: Simplify the Numerator for h(-2)

Calculate the numerator: \((-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0\).

Step 7: Simplify the Denominator for h(-2)

Calculate the denominator: \((-2)^2 - 5(-2) = 4 + 10 = 14\).

Step 8: Conclusion of h(-2)

The fraction simplifies to \( \frac{0}{14} = 0 \). Therefore, \(h(-2) = 0\).

Key Concepts

Algebraic Expressions

Algebraic expressions form the core of many mathematical concepts, acting as the building blocks behind functions like the one in our exercise: \( h(x) = \frac{x^{2}+x-2}{x^{2}-5x} \). An algebraic expression combines numbers, variables, and arithmetic operations, such as addition, subtraction, multiplication, and division. In this function:

• The numerator is \(x^2 + x - 2\), which is a quadratic expression.
• The denominator is \(x^2 - 5x\), another quadratic expression.

These expressions dictate the behavior and characteristics of the function itself. Simplifying algebraic expressions can often help in easier computation and understanding of functions. Here, simplifying involves substituting values for \(x\) and reducing the operations' results accordingly. Remember, changes in the algebraic expression directly affect the calculations and the interpretations of the functions based on those expressions.

Division by Zero

In mathematics, division by zero is an undefined operation. This concept is crucial when evaluating rational functions like in our problem. The function \( h(x) = \frac{x^{2}+x-2}{x^{2}-5x} \) highlights the rule: never divide by zero.

• The operation becomes undefined if the denominator equals zero, which disrupts the function's ability to produce a result.
• For example, when evaluating \( h(5) \), the denominator \(5^2 - 5 \cdot 5\) becomes zero. This means, at \(x = 5\), the function is undefined due to division by zero.

Understanding where a function is undefined involves finding values of \(x\) that lead the denominator to zero, marking them as restrictions on the function's domain. Avoid division by zero to maintain mathematically valid expressions.

Function Evaluation

Function evaluation involves substituting specific values for a variable within a function expression and simplifying to obtain a numerical result. In the given function \( h(x) = \frac{x^{2}+x-2}{x^{2}-5x} \), evaluating \( h(5) \) and \( h(-2) \) involves replacing \(x\) with these values.Evaluating involves:

• Substituting the value directly into both the numerator and denominator of the rational function for \(x\).
• Simplifying both the resulting expressions separately.

For \( h(5) \), substitution leads to a denominator of zero, hence it's undefined. But for \( h(-2) \), the function simplifies to \( \frac{0}{14} = 0 \), showing a defined value of the function at \(x = -2\).It's crucial to proceed systematically, following logical steps, so evaluation accurately reflects the function's inherent behavior at specific points.