Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Let \(f(x)=x^{2}+3 x\) and \(g(x)=2 x-1 .\) Perform the composition or operation indicated. $$(f-g)(-2)$$

Step-by-Step Solution

Answer

The result of \((f-g)(-2)\) is 3.

Step 1: Find Individual Functions at x = -2

First, we need to evaluate each function at \(x = -2\). Start with \(f(x)\). The function \(f(x) = x^2 + 3x\). Plug in \(-2\) for \(x\):\[f(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2\]Now evaluate \(g(x) = 2x - 1\):\[g(-2) = 2(-2) - 1 = -4 - 1 = -5\].

Step 2: Compute the Difference of Functions at x = -2

Now that we have \(f(-2)\) and \(g(-2)\), compute the difference \((f-g)(-2)\):\[(f-g)(-2) = f(-2) - g(-2)\]Substitute the values we found:\[ (f-g)(-2) = (-2) - (-5) = -2 + 5 = 3\].

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving terms that are powers of a variable, usually denoted as \( x \). The term "polynomial" refers to the sum of more than one monomial. Each monomial is a product of a constant (known as the coefficient) and a non-negative integer power of \( x \). The general form of a polynomial can be expressed as:
\[a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\]

• Here, \( a_n, a_{n-1}, \ldots, a_0 \) are constants, and \( n \) is a non-negative integer known as the degree of the polynomial.
• The degree of the polynomial is determined by the highest power of \( x \) in the polynomial.
• A polynomial with a degree of 0 is a constant function, a degree of 1 is linear, a degree of 2 is quadratic, and so on.

In our problem, \( f(x) = x^2 + 3x \) is a quadratic polynomial because the highest power of \( x \) is 2. Understanding different types of polynomial functions allows you to recognize patterns and solve equations more efficiently.

Difference of Functions

The difference of functions involves subtracting one function from another. When given two functions, \( f(x) \) and \( g(x) \), the difference \((f-g)(x)\) represents a new function.
This operation combines the individual values of \( f(x) \) and \( g(x) \) to produce new results:
\[(f-g)(x) = f(x) - g(x)\]When calculating in practice, it's important to handle signs carefully:

• Subtracting a negative, like in \( -(-5) \), actually turns into addition: \(-(-5) = +5\).
• The result of this operation is dependent on the values of \( f(x) \) and \( g(x) \) at specific points.

In the exercise, after evaluating the functions at \( x = -2 \), we found:

• \( f(-2) = -2 \)
• \( g(-2) = -5 \)
• Thus, the difference is \((f-g)(-2) = -2 - (-5) = 3\).

Evaluate Functions

Evaluating functions is determining the output value of a function for a specific input value of \( x \). This involves substituting the input value directly into the function and performing any necessary arithmetic operations.

• For each function, you replace \( x \) with the given number and simplify.
• This process reveals how the function behaves at particular points.
• Careful attention to arithmetic will ensure accurate results.

In our exercise, we evaluated two functions:
\( f(x) = x^2 + 3x \)
and
\( g(x) = 2x - 1 \).
For \( x = -2 \):

• \( f(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2 \)
• \( g(-2) = 2(-2) - 1 = -4 - 1 = -5 \)

By evaluating each function, we can then perform operations such as finding the difference, as demonstrated in the exercise.