Problem 1 - Precalculus

Question

Use the pair of functions \(f\) and \(g\) to find the following values if they exist. \- \((f+g)(2)\) \- \((f g)\left(\frac{1}{2}\right)\) -\((f-g)(-1)\) -\(\left(\frac{f}{g}\right)(0)\) -\((g-f)(1)\) -\(\left(\frac{g}{f}\right)(-2)\) $$f(x)=3 x+1 \text { and } g(x)=4-x$$

Step-by-Step Solution

Answer

(f+g)(2)=9; (fg)(\frac{1}{2})=\frac{35}{4}; (f-g)(-1)=-7; \left(\frac{f}{g}\right)(0)=\frac{1}{4}; (g-f)(1)=-1; \left(\frac{g}{f}\right)(-2)=-\frac{6}{5}.

Step 1: Understanding (f+g)(2)

To find \((f+g)(2)\), we need to calculate \(f(2) + g(2)\). Start by finding \(f(2)\) and \(g(2)\). Use the given functions: \(f(x) = 3x + 1\) and \(g(x) = 4 - x\).

Step 2: Calculation for f(2)

Substitute \(x = 2\) in \(f(x) = 3x + 1\): \[ f(2) = 3(2) + 1 = 6 + 1 = 7 \]

Step 3: Calculation for g(2)

Substitute \(x = 2\) in \(g(x) = 4-x\): \[ g(2) = 4 - 2 = 2 \]

Step 4: Result for (f+g)(2)

Add the results from Step 2 and Step 3: \[ (f+g)(2) = f(2) + g(2) = 7 + 2 = 9 \]

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

Find \((fg)\left(\frac{1}{2}\right)\) by calculating \(f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right)\).

Step 6: Calculation for f(\frac{1}{2})

Substitute \(x = \frac{1}{2}\) in \(f(x) = 3x + 1\): \[ f\left(\frac{1}{2}\right) = 3\left(\frac{1}{2}\right) + 1 = \frac{3}{2} + 1 = \frac{5}{2} \]

Step 7: Calculation for g(\frac{1}{2})

Substitute \(x = \frac{1}{2}\) in \(g(x) = 4-x\): \[ g\left(\frac{1}{2}\right) = 4 - \frac{1}{2} = \frac{8}{2} - \frac{1}{2} = \frac{7}{2} \]

Step 8: Result for (fg)(\frac{1}{2})

Multiply the results from Step 6 and Step 7: \[(fg)\left(\frac{1}{2}\right) = \frac{5}{2} \cdot \frac{7}{2} = \frac{35}{4}\]

Step 9: Understanding (f-g)(-1)

Find \((f-g)(-1)\) by calculating \(f(-1) - g(-1)\).

Step 10: Calculation for f(-1)

Substitute \(x = -1\) in \(f(x) = 3x + 1\): \[ f(-1) = 3(-1) + 1 = -3 + 1 = -2 \]

Step 11: Calculation for g(-1)

Substitute \(x = -1\) in \(g(x) = 4-x\): \[ g(-1) = 4 - (-1) = 4 + 1 = 5 \]

Step 12: Result for (f-g)(-1)

Subtract the results from Step 11 from Step 10: \[(f-g)(-1) = f(-1) - g(-1) = -2 - 5 = -7\]

Step 13: Understanding \(\left(\frac{f}{g}\right)(0)\)

Find \(\left(\frac{f}{g}\right)(0)\) by calculating \(\frac{f(0)}{g(0)}\), ensuring \(g(0) eq 0\).

Step 14: Calculation for f(0)

Substitute \(x = 0\) in \(f(x) = 3x + 1\): \[ f(0) = 3(0) + 1 = 1 \]

Step 15: Calculation for g(0)

Substitute \(x = 0\) in \(g(x) = 4-x\): \[ g(0) = 4 - 0 = 4 \]

Step 16: Result for \(\left(\frac{f}{g}\right)(0)\)

Since \(g(0) eq 0\), calculate: \[\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)} = \frac{1}{4}\]

Step 17: Understanding (g-f)(1)

Find \((g-f)(1)\) by calculating \(g(1) - f(1)\).

Step 18: Calculation for f(1)

Substitute \(x = 1\) in \(f(x) = 3x + 1\): \[ f(1) = 3(1) + 1 = 3 + 1 = 4 \]

Step 19: Calculation for g(1)

Substitute \(x = 1\) in \(g(x) = 4-x\): \[ g(1) = 4 - 1 = 3 \]

Step 20: Result for (g-f)(1)

Subtract the result of Step 18 from Step 19: \[(g-f)(1) = g(1) - f(1) = 3 - 4 = -1\]

Step 21: Understanding \(\left(\frac{g}{f}\right)(-2)\)

Find \(\left(\frac{g}{f}\right)(-2)\) by calculating \(\frac{g(-2)}{f(-2)}\), ensuring \(f(-2) eq 0\).

Step 22: Calculation for f(-2)

Substitute \(x = -2\) in \(f(x) = 3x + 1\): \[ f(-2) = 3(-2) + 1 = -6 + 1 = -5 \]

Step 23: Calculation for g(-2)

Substitute \(x = -2\) in \(g(x) = 4-x\): \[ g(-2) = 4 - (-2) = 4 + 2 = 6 \]

Step 24: Result for \(\left(\frac{g}{f}\right)(-2)\)

Since \(f(-2) eq 0\), calculate: \[\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)} = \frac{6}{-5} = -\frac{6}{5}\]

Key Concepts

Composite Functions

Composite functions combine two or more functions into a single function through operations such as addition, subtraction, multiplication, division, or composition itself.
Consider the functions provided: \(f(x) = 3x + 1\) and \(g(x) = 4 - x\). Using these functions, we can form new functions like \((f+g)(x)\) or \((f \cdot g)(x)\).
For example, the function \((f+g)(x)\) will result in the sum: \((f+g)(x) = f(x) + g(x) = (3x + 1) + (4 - x) = 2x + 5\). Each composite operation requires individually evaluating the contributing functions and then applying the operation accordingly, just like in arithmetic.
It's essential to note that composite functions can transform the inputs through various operations, which makes them powerful for mathematical analysis and problem-solving. When dealing with composite functions, keep an eye on domains since some operations might introduce restrictions.

Polynomial Functions

Polynomial functions are among the most commonly used types of functions in mathematics. A polynomial function is defined as a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients.
In our exercise, the polynomial function \(f(x) = 3x + 1\) is a first-degree polynomial, also known as a linear function. It takes the form \(ax + b\), where \(a\) and \(b\) are constants.
Key characteristics of polynomial functions include:

• Degree: The highest power of the variable in the function determines its degree. For example, if the highest power of \(x\) is 1, the polynomial is linear; if 2, it is quadratic, and so on.
• Continuous and smooth graphs: Polynomial functions are continuous and differentiable everywhere, which results in smooth graphs without breaks or sharp corners.
• Varied behavior based on degree: Linear functions produce straight lines, quadratic functions form parabolic curves, and higher-degree polynomials can produce more complex shapes.

Understanding polynomial functions and their properties is crucial for solving equations in algebra and calculus, making them a cornerstone of mathematical studies.

Function Evaluation

Function evaluation involves finding the output of a function given a specific input. It is a fundamental concept that allows us to understand how functions behave across different inputs.
To evaluate a function like \(f(x) = 3x + 1\) at a particular value, such as \(x = 2\), we simply substitute the input into the function: \(f(2) = 3(2) + 1 = 7\). This process is repeated for different values to determine how the function responds at those points.
When performing operations on functions, individuals often need to evaluate each function at the specified input before executing the operation. For instance, in the exercise question \((f-g)(-1)\), both \(f(-1)\) and \(g(-1)\) must be calculated before subtracting to find \((f-g)(-1) = -7\).
This concept is not only important in mathematical calculations but also gives insights into real-world scenarios modeled by functions. It's a core skill that supports problem-solving and analytical thinking in various academic disciplines.