Problem 65 - Elementary and Intermediate Algebra

Question

Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=\frac{x^{2}+2 x-35}{x^{2}+5 x+6} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Function

The function given is a rational function: \[ h(x) = \frac{x^2 + 2x - 35}{x^2 + 5x + 6} \] Our goal is to find the values of \(h(5)\) and \(h(-2)\).

Step 2: Calculate \(h(5)\)

Substitute \(x = 5\) into the function: \[ h(5) = \frac{5^2 + 2(5) - 35}{5^2 + 5(5) + 6} \] Calculate the numerator: \(5^2 = 25\), \(2 \times 5 = 10\), so the numerator is \(25 + 10 - 35 = 0\).Calculate the denominator: \(5^2 = 25\), \(5 \times 5 = 25\), so the denominator is \(25 + 25 + 6 = 56\). Thus, \(h(5) = \frac{0}{56} = 0\).

Step 3: Calculate \(h(-2)\)

Substitute \(x = -2\) into the function: \[ h(-2) = \frac{(-2)^2 + 2(-2) - 35}{(-2)^2 + 5(-2) + 6} \] Calculate the numerator: \((-2)^2 = 4\), \(2 \times (-2) = -4\), so the numerator is \(4 - 4 - 35 = -35\).Calculate the denominator: \((-2)^2 = 4\), \(5 \times (-2) = -10\), so the denominator is \(4 - 10 + 6 = 0\). Thus, \(h(-2)\) is undefined because the denominator is 0.

Key Concepts

Substitution in Functions

When it comes to mathematical functions, specifically rational functions like the one we have here, substitution is a straightforward technique for finding specific values. In essence, substitution involves replacing the variable, in this case, \(x\), with a specific number to evaluate the function. For the rational function provided, our task is to find \(h(5)\) and \(h(-2)\).
This step involves substituting the given values directly into the formula of the function:

• For \(h(5)\), replace every \(x\) in the equation with 5.
• For \(h(-2)\), replace every \(x\) with -2.

Substitution is crucial because it allows us to identify specific outputs of the function based on the given inputs.

Numerator and Denominator

Understanding the concept of numerator and denominator is key when dealing with rational functions. A rational function is essentially a fraction where both the numerator and the denominator are polynomials. To evaluate such functions after substituting a value, you calculate both:

• The numerator, which is the top part of the fraction.
• The denominator, the bottom part of the fraction.

In our problem, after substituting \(x = 5\), we compute:

• The numerator: \(5^2 + 2(5) - 35 = 0\).
• The denominator: \(5^2 + 5(5) + 6 = 56\).

Meanwhile, after substituting \(x = -2\), we get:

• The numerator: \((-2)^2 + 2(-2) - 35 = -35\).
• The denominator: \((-2)^2 + 5(-2) + 6 = 0\).

These calculations will help determine the actual value of the function or indicate if it's undefined.

Undefined Expressions in Mathematics

In mathematics, expressions become undefined when certain conditions are met, causing the operation to be invalid. For rational functions, an expression becomes undefined if the denominator equals zero. This is because division by zero does not yield a finite or meaningful result.
In our example, when we substitute \(x = -2\) into the denominator, the calculation results in zero:

• \((-2)^2 + 5(-2) + 6 = 0\)

This renders the whole function undefined at \(h(-2)\).
Recognizing undefined expressions in rational functions is crucial. It ensures accurate graphing, calculation, and understanding of where discontinuities or breaks happen in a mathematical model.