Problem 5 - Precalculus

Question

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

Step-by-Step Solution

Answer

(f+g)(2) = \(\sqrt{5} + 3\); (fg)(1/2) = 0; (f-g)(-1) = \(\sqrt{2} + 3\); (f/g)(0) = \(-\sqrt{3}\); (g-f)(1) = -1; (g/f)(-2) = -5.

Step 1: Understanding the Function Operations

For any two functions \(f(x)\) and \(g(x)\), we can define several operations: - \((f+g)(x) = f(x) + g(x)\)- \((f \cdot g)(x) = f(x) \cdot g(x)\)- \((f-g)(x) = f(x) - g(x)\)- \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\) provided \(g(x) eq 0\).Apply these operations to the given functions \(f(x) = \sqrt{x+3}\) and \(g(x) = 2x-1\).

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

Plug the value \(x = 2\) into both functions. Calculate:\[ f(2) = \sqrt{2+3} = \sqrt{5} \]\[ g(2) = 2 \times 2 -1 = 3 \]Combine the results:\[ (f+g)(2) = f(2) + g(2) = \sqrt{5} + 3 \]

Step 3: Calculate \((fg)\left(\frac{1}{2}\right)\)

Evaluate both functions at \(x = \frac{1}{2}\):\[ f\left(\frac{1}{2}\right) = \sqrt{\frac{1}{2} + 3} = \sqrt{\frac{7}{2}} \]\[ g\left(\frac{1}{2}\right) = 2 \times \frac{1}{2} - 1 = 0 \]Multiply the results:\[ (fg)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right) = \sqrt{\frac{7}{2}} \cdot 0 = 0 \]

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

Find the values at \(x = -1\):\[ f(-1) = \sqrt{-1 + 3} = \sqrt{2} \]\[ g(-1) = 2 \times (-1) - 1 = -3 \]Subtract these results:\[ (f-g)(-1) = f(-1) - g(-1) = \sqrt{2} - (-3) = \sqrt{2} + 3 \]

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

Evaluate both functions at \(x = 0\):\[ f(0) = \sqrt{0 + 3} = \sqrt{3} \]\[ g(0) = 2 \times 0 - 1 = -1 \]Since \(g(0) eq 0\), divide:\[ \left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)} = \frac{\sqrt{3}}{-1} = -\sqrt{3} \]

Step 6: Calculate \((g-f)(1)\)

Find the values at \(x = 1\):\[ f(1) = \sqrt{1 + 3} = 2 \]\[ g(1) = 2 \times 1 - 1 = 1 \]Calculate the difference:\[ (g-f)(1) = g(1) - f(1) = 1 - 2 = -1 \]

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

Evaluate both functions at \(x = -2\):\[ f(-2) = \sqrt{-2 + 3} = 1 \]\[ g(-2) = 2 \times (-2) - 1 = -5 \]Since \(f(-2) eq 0\), divide:\[ \left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)} = \frac{-5}{1} = -5 \]

Key Concepts

Addition of Functions

When adding two functions, we combine their values by adding the outputs for the same input. Imagine having two separate machines: function "\(f\)" and function "\(g\)". Each machine takes an "\(x\)" value, performs its special operation, and gives back a number. For addition, you give the same "\(x\)" to both, collect the two numbers they return, and sum them together. The rule is simple:

• \((f+g)(x) = f(x) + g(x)\)

This rule means you run each function separately and then add their results.
Say you have \(f(x) = \sqrt{x+3}\) and \(g(x) = 2x-1\). If you want \((f+g)(2)\), you compute \(f(2)\) and \(g(2)\), which are \(\sqrt{5}\) and \(3\) respectively, and then add them to get \(\sqrt{5} + 3\).
Remember, this just means you're getting results from both functions independently and then adding those results together.

Multiplication of Functions

The multiplication of functions follows a method similar to their addition, but now instead of adding, we multiply the outputs. Think of it as taking the result from function "\(f\)" and multiplying it by the result from function "\(g\)" for a given "\(x\)" value. The formula for this operation is straightforward:

• \((f \cdot g)(x) = f(x) \cdot g(x)\)

This means you apply both functions to the same "\(x\)" and multiply the outputs.
For instance, with \(f(x) = \sqrt{x+3}\) and \(g(x) = 2x-1\), to find \((fg)\left(\frac{1}{2}\right)\), you calculate \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\), which are \(\sqrt{\frac{7}{2}}\) and \(0\) respectively, and then multiply them.
The result turns out to be \(0\), because multiplying any number by zero is zero. This concept is a reminder that multiplication impacts function results differently than addition does.

Subtraction of Functions

Subtraction of functions is the reverse process of adding. Instead of adding the outputs of two functions, here you subtract the output of "\(g\)" from "\(f\)". This operation is defined as:

• \((f-g)(x) = f(x) - g(x)\)

This operation involves evaluating each function separately at the same "\(x\)" value and then performing a subtraction.
Take, for example, the functions \(f(x) = \sqrt{x+3}\) and \(g(x) = 2x-1\). To find \((f-g)(-1)\), you first calculate \(f(-1)\) and \(g(-1)\), obtaining \(\sqrt{2}\) and \(-3\). Subtract these results: \(\sqrt{2} - (-3) = \sqrt{2} + 3\).
Notice how subtracting a negative is equivalent to adding a positive, hence the change in the sign from subtraction to addition. Understanding this can clarify the function subtraction process for various scenarios.

Division of Functions

Division of functions requires special attention, as you need to ensure the divisor isn't zero to avoid undefined results. This is written as:

• \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}\)

Provided that \(g(x) eq 0\). This checks the output of "\(g\)" before proceeding with the division.
Consider the functions \(f(x) = \sqrt{x+3}\) and \(g(x) = 2x-1\). To calculate \(\left(\frac{f}{g}\right)(0)\), first ensure that \(g(0) eq 0\).
Calculate \(f(0) = \sqrt{3}\) and \(g(0) = -1\). Then \(\left(\frac{f}{g}\right)(0)\) becomes \(\frac{\sqrt{3}}{-1} = -\sqrt{3}\).
When dividing such functions, be cautious about values where "\(g(x)\)" becomes zero, as division by zero is undefined. Always check the denominator to assure the operation is valid.