Problem 2 - Precalculus

Question

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

Step-by-Step Solution

Answer

1, -2, -2, 0, 0, \(\frac{5}{4}\).

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

First, determine \(f(2)\) and \(g(2)\). \(f(2) = 2^2 = 4\) and \(g(2) = -2(2) + 1 = -4 + 1 = -3\). Then, \((f+g)(2) = f(2) + g(2) = 4 - 3 = 1\).

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

Find \(f(-1)\) and \(g(-1)\). \(f(-1) = (-1)^2 = 1\) and \(g(-1) = -2(-1) + 1 = 2 + 1 = 3\). Therefore, \((f-g)(-1) = f(-1) - g(-1) = 1 - 3 = -2\).

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

Determine \(f(1)\) and \(g(1)\). \(f(1) = 1^2 = 1\) and \(g(1) = -2(1) + 1 = -2 + 1 = -1\). Then, \((g-f)(1) = g(1) - f(1) = -1 - 1 = -2\).

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

Find \(f(\frac{1}{2})\) and \(g(\frac{1}{2})\). \(f(\frac{1}{2}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and \(g(\frac{1}{2}) = -2\left(\frac{1}{2}\right) + 1 = -1 + 1 = 0\). \((fg)(\frac{1}{2}) = f(\frac{1}{2}) \times g(\frac{1}{2}) = \frac{1}{4} \times 0 = 0\).

Step 5: Calculate \((\frac{f}{g})(0)\)

Evaluate \(f(0)\) and \(g(0)\). \(f(0) = 0^2 = 0\) and \(g(0) = -2(0) + 1 = 1\). Therefore, \((\frac{f}{g})(0) = \frac{f(0)}{g(0)} = \frac{0}{1} = 0\).

Step 6: Calculate \((\frac{g}{f})(-2)\)

Calculate \(f(-2)\) and \(g(-2)\). \(f(-2) = (-2)^2 = 4\) and \(g(-2) = -2(-2) + 1 = 4 + 1 = 5\). Hence, \((\frac{g}{f})(-2) = \frac{g(-2)}{f(-2)} = \frac{5}{4}\).

Key Concepts

Adding Functions

When adding two functions, you're essentially combining the output of each function by summing them together. To visualize this, think of it as adding the values from two separate tables.

If given two functions, say, \( f(x) \) and \( g(x) \), the sum \( (f+g)(x) \) is simply \( f(x) + g(x) \). For instance, if \( f(x) = x^2 \) and \( g(x) = -2x + 1 \), you add them together for a new function: \( f(x) + g(x) = x^2 - 2x + 1 \).

This means for any input \( x \), you substitute \( x \) into both \( f(x) \) and \( g(x) \), calculate the results, then add those results together. It’s a straightforward yet powerful operation that helps produce a blended perspective from two separate insights.

Subtracting Functions

Subtracting functions involves determining the difference between two functions for any given input. If you have \( f(x) \) and \( g(x) \), then \( (f-g)(x) \) equates to \( f(x) - g(x) \).

The act of subtraction is fundamentally a means of comparing outputs. With subtraction, you can discern which function's value is greater at any point or see how closely aligned the functions are.

For example, with \( f(x) = x^2 \) and \( g(x) = -2x+1 \), calculate them separately and subtract: \( x^2 - (-2x + 1) \) transforms into \( x^2 + 2x - 1 \). Always remember to carefully manage signs, especially when dealing with negative values, to avoid common errors.

Multiplying Functions

Multiplying functions involves taking the product of two functions at a specific input. This operation is not as intuitive as simple addition or subtraction because the resulting magnitude can change dramatically.

Given functions \( f(x) \) and \( g(x) \), their product \( (f \cdot g)(x) \) is \( f(x) \times g(x) \). For \( f(x) = x^2 \) and \( g(x) = -2x + 1 \), the process involves multiplying the outputs: \( x^2 \cdot (-2x + 1) \). Here, you'll need to employ distribution: \( x^2 \times -2x = -2x^3 \) and \( x^2 \times 1 = x^2 \), resulting in \( -2x^3 + x^2 \).

Such multiplication blends both functions in a new way, producing outputs that may explore entirely different behaviors of \( f(x) \) and \( g(x) \), which can be particularly useful for composite system modeling.

Dividing Functions

Dividing functions is an operation that involves dividing the outputs of one function by another. This creates a ratio, showing how much one function is in comparison to another at any given input.

If you have \( f(x) \) and \( g(x) \), the quotient \( (\frac{f}{g})(x) \) is \( \frac{f(x)}{g(x)} \). For \( f(x) = x^2 \) and \( g(x) = -2x + 1 \), you'll find \( \frac{x^2}{-2x + 1} \).

When dealing with division operations, it is critical to be wary of points where \( g(x) = 0 \). Such points make the expression undefined since division by zero is impossible. This operation requires close attention to potential values where disallowed division may occur, further illustrating the importance of checking the function's domain.