Problem 10 - Precalculus

Question

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

Step-by-Step Solution

Answer

(f+g)(2) = \frac{26}{5}, (fg)(\frac{1}{2}) = 1, (f-g)(-1) = \frac{3}{2}, (\frac{f}{g})(0) = 1, (g-f)(1) = -\frac{3}{2}, (\frac{g}{f})(-2) = \frac{1}{25}.

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

To find \((f+g)(2)\), we add the values of \(f(2)\) and \(g(2)\). First, calculate \(f(2) = 2^2 + 1 = 4 + 1 = 5\). Then, calculate \(g(2) = \frac{1}{2^2 + 1} = \frac{1}{4 + 1} = \frac{1}{5}\). Therefore, \((f+g)(2) = f(2) + g(2) = 5 + \frac{1}{5} = \frac{25}{5} + \frac{1}{5} = \frac{26}{5}\).

Step 2: Evaluate (fg)(1/2)

To find \((fg)\left(\frac{1}{2}\right)\), we multiply the values \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\).First, \(f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 1 = \frac{1}{4} + 1 = \frac{5}{4}\).Then, \(g\left(\frac{1}{2}\right) = \frac{1}{\left(\frac{1}{2}\right)^2 + 1} = \frac{1}{\frac{1}{4} + 1} = \frac{1}{\frac{5}{4}} = \frac{4}{5}\).Therefore, \((fg)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right) = \frac{5}{4} \cdot \frac{4}{5} = 1\).

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

To find \((f-g)(-1)\), subtract \(g(-1)\) from \(f(-1)\).First, \(f(-1) = (-1)^2 + 1 = 1 + 1 = 2\).Then, \(g(-1) = \frac{1}{(-1)^2 + 1} = \frac{1}{1 + 1} = \frac{1}{2}\).Therefore, \((f-g)(-1) = f(-1) - g(-1) = 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}\).

Step 4: Evaluate (f/g)(0)

To find \(\left(\frac{f}{g}\right)(0)\), divide \(f(0)\) by \(g(0)\).First, \(f(0) = 0^2 + 1 = 1\).Then, \(g(0) = \frac{1}{0^2 + 1} = \frac{1}{1} = 1\).Therefore, \(\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)} = \frac{1}{1} = 1\).

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

To find \((g-f)(1)\), subtract \(f(1)\) from \(g(1)\).First, \(f(1) = 1^2 + 1 = 1 + 1 = 2\).Then, \(g(1) = \frac{1}{1^2 + 1} = \frac{1}{1 + 1} = \frac{1}{2}\).Therefore, \((g-f)(1) = g(1) - f(1) = \frac{1}{2} - 2 = \frac{1}{2} - \frac{4}{2} = -\frac{3}{2}\).

Step 6: Evaluate (g/f)(-2)

To find \(\left(\frac{g}{f}\right)(-2)\), divide \(g(-2)\) by \(f(-2)\).First, \(f(-2) = (-2)^2 + 1 = 4 + 1 = 5\).Then, \(g(-2) = \frac{1}{(-2)^2 + 1} = \frac{1}{4 + 1} = \frac{1}{5}\).Therefore, \(\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)} = \frac{1/5}{5} = \frac{1}{25}\).

Key Concepts

Addition of Functions

Combining two functions by adding them is a fundamental operation in function operations. When we talk about the addition of functions, it simply means summing up the outputs of two functions for the same input value. Let's see how it works with a simple example.

Suppose we have two functions, say, \( f(x) \) and \( g(x) \). To find the sum \((f+g)(x)\), we compute \( f(x) + g(x) \). It's all about taking the result of each function and adding them together.

• For example, if \( f(x) = x^2 + 1 \) and \( g(x) = \frac{1}{x^2 + 1} \), to add these, just sum their outputs:
• \( (f+g)(x) = (x^2 + 1) + \left(\frac{1}{x^2 + 1}\right) \)

When substituting \( x = 2 \), the result would be \( (f+g)(2) = 5 + \frac{1}{5} \). Remember, here we use basic algebraic addition rules to add the fractions correctly.

Multiplication of Functions

When functions are multiplied, we look at a different operation to combine values from two functions. The multiplication of functions involves multiplying the results of those functions for the same input.

• Take two functions, \( f(x) \) and \( g(x) \). Their multiplication \((f \cdot g)(x) \) is basically multiplying \( f(x) \) by \( g(x) \).
• So, \((f \cdot g)(x) = f(x) \cdot g(x) \).

Consider our example functions, \( f(x) = x^2 + 1 \) and \( g(x) = \frac{1}{x^2 + 1} \). To multiply these at \( x = \frac{1}{2} \), compute:

\[(f \cdot g)\left(\frac{1}{2}\right) = \left(\frac{5}{4}\right) \cdot \left(\frac{4}{5}\right) = 1\]
Here, multiplying fractions means multiplying the numerators together and the denominators together, then simplifying the result.

Subtraction of Functions

The subtraction of functions is much like addition but involves a slight twist. With subtraction, we are not combining but rather finding the difference between the values produced by each function.

• Given functions \( f(x) \) and \( g(x) \), the subtraction \((f-g)(x)\) calculates \( f(x) - g(x) \).

Let's look at this process using our example functions, where \( f(x) = x^2 + 1 \) and \( g(x) = \frac{1}{x^2 + 1} \).

To subtract these when \( x = -1 \), do as follows:

\[(f-g)(-1) = 2 - \frac{1}{2} = \frac{3}{2}\]
This involves combining like terms and ensuring you handle fractions accurately. You subtract \( g(x) \) from \( f(x) \), considering correct fraction subtraction rules.

Division of Functions

Division of functions is slightly more complex because it involves fractions. In this operation, you divide one function's value by another for a specific input, provided the divisor function's result is not zero.

• With functions \( f(x) \) and \( g(x) \), the division \( \left(\frac{f}{g}\right)(x) \) is expressed as \( \frac{f(x)}{g(x)} \).

Consider \( f(x) = x^2 + 1 \) and \( g(x) = \frac{1}{x^2 + 1} \) at \( x = 0 \). Here’s how you proceed:

• First calculate \( f(0) \) and \( g(0) \).
• Then divide: \( \left(\frac{f}{g}\right)(0) = \frac{1}{1} = 1\)

Remember that division is only defined when the denominator is non-zero. Always ensure \( g(x) eq 0 \) before dividing. Division requires careful attention to the possibilities of undefined values when the denominator equals zero for certain inputs.