Problem 4 - Precalculus

Question

Use the pair of functions \(f\) and \(g\) to find the following values if they exist. $$ \begin{array}{lll} \bullet(f+g)(2) & \bullet(f-g)(-1) & \bullet(g-f)(1) \\ \bullet(f g)\left(\frac{1}{2}\right) & \bullet\left(\frac{f}{g}\right)(0) & \bullet\left(\frac{g}{f}\right)(-2) \end{array} $$ $$ f(x)=2 x^{3} \text { and } g(x)=-x^{2}-2 x-3 $$

Step-by-Step Solution

Answer

5, 0, -8, -\frac{13}{16}, 0, \frac{3}{16}

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

Calculate \(f(2)\) and \(g(2)\) first. \[ f(2) = 2(2)^3 = 16 \] \[ g(2) = -(2)^2 - 2(2) - 3 = -4 - 4 - 3 = -11 \] Now, substitute these back into \((f+g)(2) = f(2) + g(2)\): \[ (f+g)(2) = 16 + (-11) = 5 \]

Step 2: Evaluate \((f-g)(-1)\)

Calculate \(f(-1)\) and \(g(-1)\) first. \[ f(-1) = 2(-1)^3 = -2 \] \[ g(-1) = -(-1)^2 - 2(-1) - 3 = -1 + 2 - 3 = -2 \] Now, substitute these into \((f-g)(-1) = f(-1) - g(-1)\): \[ (f-g)(-1) = -2 - (-2) = 0 \]

Step 3: Evaluate \((g-f)(1)\)

Calculate \(f(1)\) and \(g(1)\) first. \[ f(1) = 2(1)^3 = 2 \] \[ g(1) = -(1)^2 - 2(1) - 3 = -1 - 2 - 3 = -6 \] Now, substitute these into \((g-f)(1) = g(1) - f(1)\): \[ (g-f)(1) = -6 - 2 = -8 \]

Step 4: Evaluate \((fg)(\frac{1}{2})\)

Calculate \(f(\frac{1}{2})\) and \(g(\frac{1}{2})\) first. \[ f(\frac{1}{2}) = 2(\frac{1}{2})^3 = 2(\frac{1}{8}) = \frac{1}{4} \] \[ g(\frac{1}{2}) = -\left(\frac{1}{2}\right)^2 - 2(\frac{1}{2}) - 3 = -\left(\frac{1}{4}\right) - 1 - 3 = -\frac{1}{4} - 1 - 3 = -\frac{13}{4} \] Now, substitute into \((fg)(\frac{1}{2}) = f(\frac{1}{2}) \cdot g(\frac{1}{2})\): \[ (fg)(\frac{1}{2}) = \frac{1}{4} \cdot \left(-\frac{13}{4}\right) = -\frac{13}{16} \]

Step 5: Evaluate \(\left(\frac{f}{g}\right)(0)\)

Calculate \(f(0)\) and \(g(0)\) first. \[ f(0) = 2(0)^3 = 0 \] \[ g(0) = -(0)^2 - 2(0) - 3 = -3 \] Now, substitute these into \(\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)}\): \[ \left(\frac{f}{g}\right)(0) = \frac{0}{-3} = 0 \]

Step 6: Evaluate \(\left(\frac{g}{f}\right)(-2)\)

Calculate \(f(-2)\) and \(g(-2)\) first. \[ f(-2) = 2(-2)^3 = 2(-8) = -16 \] \[ g(-2) = -(-2)^2 - 2(-2) - 3 = -4 + 4 - 3 = -3 \] Now, substitute these into \(\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)}\): \[ \left(\frac{g}{f}\right)(-2) = \frac{-3}{-16} = \frac{3}{16} \]

Key Concepts

Polynomial Functions

Polynomial functions are expressions that involve a sum of powers of a variable, each multiplied by a coefficient. Polynomials are characterized by their degree, which is determined by the highest exponential power of the variable. In the exercise provided, the polynomial functions are:

• \( f(x) = 2x^3 \), a cubic polynomial of degree 3.
• \( g(x) = -x^2 - 2x - 3 \), a quadratic polynomial of degree 2.

These functions are made up of monomials (individual terms) that are added or subtracted from each other.

Polynomial functions can significantly vary in shape based on the degree and coefficients of the terms. Higher degree polynomials will have more turning points and can grow or shrink more rapidly than lower degree polynomials.
Understanding these basic properties will help in grasping further operations involving polynomial functions.

Function Evaluation

Function evaluation involves computing the value of a function at a particular point. It’s essentially plugging in a number into the function's formula and performing the required arithmetic operations.

For instance, in the exercise:

• To evaluate \( f(2) \), we calculate \( 2(2)^3 = 16 \).
• For \( g(2) \), substitute 2 into the expression: \( -2^2 - 2(2) - 3 = -11 \).

This process is straight-forward but essential, as it is the first step in solving many problems involving function operations.

Function evaluation provides the groundwork by allowing one to determine specific points on the curve described by the function, which can be useful for plotting graphs or performing further operations.

Arithmetic Operations on Functions

Arithmetic operations such as addition, subtraction, and multiplication can be performed on functions similarly to arithmetic done with numbers. These operations are done by combining functions’ outputs for the same input.

Consider these examples from the exercise:

• Addition: \((f+g)(x) = f(x) + g(x)\), consequently for \( x = 2 \), \( 16 + (-11) = 5 \).
• Subtraction: \((f-g)(x) = f(x) - g(x)\), for \( x = -1 \), \(-2 - (-2) = 0 \).
• Multiplication: \((fg)(x) = f(x) \cdot g(x)\), such as \((fg)(\frac{1}{2}) = \frac{1}{4} \cdot \left(-\frac{13}{4}\right) = -\frac{13}{16}\).

These operations allow for the creation of new functions with combined characteristics and can simplify complex function manipulations.
Understanding how these operations affect functions’ step-by-step behavior is crucial for higher-level mathematics.

Division of Functions

Division of functions is similar to dividing numbers but requires attention to the denominator since division by zero is undefined. When dividing two functions, \(\left(\frac{f}{g}\right)(x)\) is computed similar to arithmetic division:

• For \(\left(\frac{f}{g}\right)(0)\), we find \(f(0) = 0\) and \(g(0) = -3\), yielding \(\frac{0}{-3} = 0\).
• For \(\left(\frac{g}{f}\right)(-2)\), calculations give \(\frac{-3}{-16} = \frac{3}{16}\).

The critical aspect is ensuring that the denominator is not zero; otherwise, the operation is undefined. Dividing functions can also yield rational expressions characterized by their domain restrictions, specifically where the denominator equals zero.
Being comfortable with this operation is essential for calculus and solving fractional functions in advanced algebra.