Problem 6 - Precalculus

Question

Use the pair of functions \(f\) and \(g\) to find the following values if they exist. \- \((f+g)(2)\) \- \((f g)\left(\frac{1}{2}\right)\) -\((f-g)(-1)\) -\(\left(\frac{f}{g}\right)(0)\) -\((g-f)(1)\) -\(\left(\frac{g}{f}\right)(-2)\) $$f(x)=\sqrt{4-x} \text { and } g(x)=\sqrt{x+2}$$

Step-by-Step Solution

Answer

\((f+g)(2) = \sqrt{2} + 2\); \((f g)(\frac{1}{2}) = \sqrt{\frac{35}{4}}\); \((f-g)(-1) = \sqrt{5} - 1\); \((\frac{f}{g})(0) = \sqrt{2}\); \((g-f)(1) = 0\); \((\frac{g}{f})(-2) = 0\).

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

Calculate the value of \(f(x)\) and \(g(x)\) at \(x=2\). Then add the results.\[f(2) = \sqrt{4-2} = \sqrt{2}\]\[g(2) = \sqrt{2+2} = \sqrt{4} = 2\]Therefore, \[(f+g)(2) = f(2) + g(2) = \sqrt{2} + 2\]

Step 2: Evaluate \((f g)\left(\frac{1}{2}\right)\)

Calculate the product of \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\).\[f\left(\frac{1}{2}\right) = \sqrt{4-\frac{1}{2}} = \sqrt{\frac{7}{2}}\]\[g\left(\frac{1}{2}\right) = \sqrt{\frac{1}{2} + 2} = \sqrt{\frac{5}{2}}\]The product is \[(f g)\left(\frac{1}{2}\right) = \sqrt{\frac{7}{2}} \cdot \sqrt{\frac{5}{2}} = \sqrt{\frac{35}{4}}\]

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

Calculate the value of \(f(-1)\) and \(g(-1)\), and then subtract \(g(-1)\) from \(f(-1)\).\[f(-1) = \sqrt{4 - (-1)} = \sqrt{5}\]\[g(-1) = \sqrt{-1 + 2} = \sqrt{1} = 1\]So, \[(f-g)(-1) = f(-1) - g(-1) = \sqrt{5} - 1\]

Step 4: Evaluate \(\left(\frac{f}{g}\right)(0)\)

Calculate the quotient \(\frac{f(0)}{g(0)}\) after computing\(f(0)\) and \(g(0)\).\[f(0) = \sqrt{4 - 0} = \sqrt{4} = 2\]\[g(0) = \sqrt{0 + 2} = \sqrt{2}\]Then, \[\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)} = \frac{2}{\sqrt{2}} = \sqrt{2}\]

Step 5: Evaluate \((g-f)(1)\)

Calculate the values \(g(1)\) and \(f(1)\), then subtract \(f(1)\) from \(g(1)\).\[g(1) = \sqrt{1 + 2} = \sqrt{3}\]\[f(1) = \sqrt{4 - 1} = \sqrt{3}\]Thus,\[(g-f)(1) = g(1) - f(1) = \sqrt{3} - \sqrt{3} = 0\]

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

Determine \(g(-2)\) and \(f(-2)\), then compute the quotient \(\frac{g(-2)}{f(-2)}\).\[g(-2) = \sqrt{-2 + 2} = \sqrt{0} = 0\]\[f(-2) = \sqrt{4 - (-2)} = \sqrt{6}\]Since \(g(-2) = 0\),\[\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)} = \frac{0}{\sqrt{6}} = 0\]

Key Concepts

Function Addition

Adding two functions, denoted by \((f + g)(x)\), means you're adding the outputs that each function provides when given the same input. It's like asking two friends what they want to eat for lunch and **combining** their wishlists!

To add two functions:

• First, compute the value of each function at a particular input. For example, if we want to find \((f + g)(2)\), we need to find both \(f(2)\) and \(g(2)\).
• Next, simply add the results together to get the final value.

For instance, in the given problem, we calculated:

• \(f(2) = \sqrt{4-2} = \sqrt{2}\)
• \(g(2) = \sqrt{2+2} = \sqrt{4} = 2\)
• Thus, \((f + g)(2) = \sqrt{2} + 2\)

Adding functions gives you a new function that includes aspects of both original functions. It’s a powerful way to analyze combined effects on a single input.

Function Multiplication

When we talk about **function multiplication**, written as \((f \, g)(x)\), we mean multiplying the results of the functions for the same input value.

Think of it as two factories producing parts, and you're curious about the total number of parts produced by both factories combined!

Here's how to multiply two functions:

• First, find the output of each function for a particular input. For \((f \, g)\left(\frac{1}{2}\right)\), find both \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\).
• Then, multiply these results together.

In our specific example, we have:

• \(f\left(\frac{1}{2}\right) = \sqrt{\frac{7}{2}}\)
• \(g\left(\frac{1}{2}\right) = \sqrt{\frac{5}{2}}\)
• So, \((f \, g)\left(\frac{1}{2}\right) = \sqrt{\frac{7}{2}} \, \cdot \, \sqrt{\frac{5}{2}} = \sqrt{\frac{35}{4}}\)

Function multiplication helps to model scenarios where the interaction of two processes or values at the same input is examined.

Function Subtraction

Function subtraction, indicated by \((f - g)(x)\), is similar to everyday subtraction. You're measuring the difference between the two functions' outputs. Imagine you're comparing scores of two games to see who won more!

To subtract two functions:

• First, determine the function values at the given input for both functions. For example, calculate \(f(-1)\) and \(g(-1)\).
• Next, subtract the output of \(g(x)\) from \(f(x)\).

For this exercise, the steps were:

• \(f(-1) = \sqrt{5}\)
• \(g(-1) = 1\)
• Therefore, \((f - g)(-1) = \sqrt{5} - 1\)

This subtraction operation is quite useful for spotting the difference in value changes or analyzing how much one function surpasses or lacks compared to the other at any point.

Function Division

Performing function division, denoted by \(\left(\frac{f}{g}\right)(x)\), is a bit like dividing one friend's earnings by another's to find who earns more per hour!

The goal is to compare how one function's output relates to another’s for the same input.

Here's how you can divide two functions:

• First, evaluate both functions at the given input value. Get both \(f(x)\) and \(g(x)\).
• Next, divide the result of \(f(x)\) by \(g(x)\).

In our example, we computed:

• \(f(0) = 2\)
• \(g(0) = \sqrt{2}\)
• Thus, \(\left(\frac{f}{g}\right)(0) = \frac{2}{\sqrt{2}} = \sqrt{2}\)

Remember that you cannot divide by zero, as it makes the operation undefined!