Problem 3 - Precalculus

Question

Use the pair of functions \(f\) and \(g\) to find the following values if they exist. $$ \begin{array}{lll} \bullet(f+g)(2) & \bullet(f-g)(-1) & \bullet(g-f)(1) \\ \bullet(f g)\left(\frac{1}{2}\right) & \bullet\left(\frac{f}{g}\right)(0) & \bullet\left(\frac{g}{f}\right)(-2) \end{array} $$ $$ f(x)=x^{2}-x \text { and } g(x)=12-x^{2} $$

Step-by-Step Solution

Answer

(f+g)(2) = 10, (f-g)(-1) = -9, (g-f)(1) = 11, (fg)(1/2) = -47/16, (f/g)(0) = 0, (g/f)(-2) = 4/3.

Step 1: Understand the Problem

We have two functions, \( f(x) = x^2 - x \) and \( g(x) = 12 - x^2 \). We need to evaluate combinations of these functions at given points: \((f+g)(2)\), \((f-g)(-1)\), \((g-f)(1)\), \((fg)(\frac{1}{2})\), \((\frac{f}{g})(0)\), and \((\frac{g}{f})(-2)\).

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

Add the functions and evaluate at \( x = 2 \):\[(f+g)(2) = f(2) + g(2)\]\[f(2) = 2^2 - 2 = 2\] \[g(2) = 12 - 2^2 = 8\] \[(f+g)(2) = 2 + 8 = 10\]

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

Subtract the functions and evaluate at \( x = -1 \):\[(f-g)(-1) = f(-1) - g(-1)\] \[f(-1) = (-1)^2 - (-1) = 2\] \[g(-1) = 12 - (-1)^2 = 11\] \[(f-g)(-1) = 2 - 11 = -9\]

Step 4: Calculate (g-f)(1)

Subtract the functions in reverse order and evaluate at \( x = 1 \):\[(g-f)(1) = g(1) - f(1)\] \[f(1) = 1^2 - 1 = 0\] \[g(1) = 12 - 1^2 = 11\] \[(g-f)(1) = 11 - 0 = 11\]

Step 5: Calculate (fg)(\( \frac{1}{2} \))

Multiply the functions and evaluate at \( x = \frac{1}{2} \):\[(fg)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right)\] \[f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\] \[g\left(\frac{1}{2}\right) = 12 - \left(\frac{1}{2}\right)^2 = 12 - \frac{1}{4} = \frac{47}{4}\] \[(fg)\left(\frac{1}{2}\right) = -\frac{1}{4} \cdot \frac{47}{4} = -\frac{47}{16}\]

Step 6: Calculate (\( \frac{f}{g} \))(0)

Divide the functions and evaluate at \( x = 0 \):\[\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)}\] \[f(0) = 0^2 - 0 = 0\] \[g(0) = 12 - 0^2 = 12\] \[\left(\frac{f}{g}\right)(0) = \frac{0}{12} = 0\]

Step 7: Calculate (\( \frac{g}{f} \))(-2)

Divide the functions and evaluate at \( x = -2 \): Check if \( f(-2) eq 0 \) before dividing. \[\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)}\] \[f(-2) = (-2)^2 - (-2) = 4 + 2 = 6\] \[g(-2) = 12 - (-2)^2 = 12 - 4 = 8\] \[\left(\frac{g}{f}\right)(-2) = \frac{8}{6} = \frac{4}{3}\]

Key Concepts

Addition of Functions

In the realm of mathematics, adding functions is a simple yet powerful concept. The process involves combining two functions into a single function by summing their outputs for a given input. Suppose we have two functions, \( f(x) \) and \( g(x) \). The addition of these functions is denoted by \((f+g)(x)\), which means computing \( f(x) + g(x) \).

For example, if \( f(x) = x^2 - x \) and \( g(x) = 12 - x^2 \), to find \((f+g)(2)\), we evaluate both functions at \(x = 2\) and add the results:
\[ f(2) = 2^2 - 2 = 2 \]
\[ g(2) = 12 - 2^2 = 8 \]
Thus,
\[ (f+g)(2) = f(2) + g(2) = 2 + 8 = 10 \]
This simple addition can be useful in various applications such as physics, where you might want to combine different forces or energies.

Subtraction of Functions

Subtracting functions is just as straightforward as adding them. You take two functions and compute the difference between their outputs for a given input. This is represented by \((f-g)(x)\), meaning \(f(x) - g(x)\).

Let's illustrate this with our functions \( f(x) = x^2 - x \) and \( g(x) = 12 - x^2 \). To evaluate \((f-g)(-1)\), we find:
\[ f(-1) = (-1)^2 - (-1) = 2 \]
\[ g(-1) = 12 - (-1)^2 = 11 \]
Then subtract:
\[ (f-g)(-1) = f(-1) - g(-1) = 2 - 11 = -9 \]
This operation helps in scenarios where you might need to determine differences, such as calculating net values in economics.

Multiplication of Functions

Multiplying functions involves creating a new function where each input yields the product of the outputs of the original functions. When multiplying, the notation \((fg)(x)\) or \(f(x) \cdot g(x)\) is used.

Using the same example functions \( f(x) = x^2 - x \) and \( g(x) = 12 - x^2 \), let's find \((fg)(\frac{1}{2})\):

• Find \( f(\frac{1}{2}) = (\frac{1}{2})^2 - \frac{1}{2} = -\frac{1}{4} \)
• Find \( g(\frac{1}{2}) = 12 - (\frac{1}{2})^2 = \frac{47}{4} \)
• Multiply: \( (fg)(\frac{1}{2}) = -\frac{1}{4} \cdot \frac{47}{4} = -\frac{47}{16} \)

Multiplying functions is useful in areas such as statistics, where you may need to compute joint probabilities.

Division of Functions

Division of functions requires caution, as it involves dividing their outputs, which can result in undefined values if the denominator is zero. This operation is expressed as \(\left( \frac{f}{g} \right)(x)\) for \(f(x) \div g(x)\).

For our functions \( f(x) = x^2 - x \) and \( g(x) = 12 - x^2 \), let's compute \(\left( \frac{f}{g} \right)(0)\):

• Find \( f(0) = 0^2 - 0 = 0 \)
• Find \( g(0) = 12 - 0^2 = 12 \)
• Compute: \( \left( \frac{f}{g} \right)(0) = \frac{0}{12} = 0 \)

Similarly, evaluate \(\left( \frac{g}{f} \right)(-2)\):

• Find \( f(-2) = 6 \) and \( g(-2) = 8 \)
• Compute: \( \left( \frac{g}{f} \right)(-2) = \frac{8}{6} = \frac{4}{3} \)

It's important in real-world applications to check the division validity, especially in fields like engineering where precise calculations matter.