Problem 9 - Precalculus

Question

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

Step-by-Step Solution

Answer

(f+g)(2)=\frac{17}{4}, (f-g)(-1)=0, (g-f)(1)=0, (f \cdot g)(\frac{1}{2})=1, \left(\frac{f}{g}\right)(0) is undefined, \left(\frac{g}{f}\right)(-2)=\frac{1}{16}.

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

To find \((f+g)(2)\), we evaluate both functions at \(x=2\) and add the results. First, find \(f(2): f(2) = 2^2 = 4\).Next, find \(g(2): g(2) = \frac{1}{2^2} = \frac{1}{4}\). Add them together: \((f+g)(2) = f(2) + g(2) = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}\).

Step 2: Calculate \((f-g)(-1)\)

To find \((f-g)(-1)\), evaluate both functions at \(x=-1\) and subtract the outputs. Find \(f(-1): f(-1) = (-1)^2 = 1\).Find \(g(-1): g(-1) = \frac{1}{(-1)^2} = 1\).Subtract the results: \((f-g)(-1) = f(-1) - g(-1) = 1 - 1 = 0\).

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

To find \((g-f)(1)\), evaluate both functions at \(x=1\) and subtract \(f\) from \(g\).Find \(f(1): f(1) = 1^2 = 1\).Find \(g(1): g(1) = \frac{1}{1^2} = 1\).Subtract: \((g-f)(1) = g(1) - f(1) = 1 - 1 = 0\).

Step 4: Calculate \((f \cdot g)(\frac{1}{2})\)

To find \((f \cdot g)(\frac{1}{2})\), evaluate both functions at \(x=\frac{1}{2}\) and multiply the results. Find \(f\left(\frac{1}{2}\right): f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).Find \(g\left(\frac{1}{2}\right): g\left(\frac{1}{2}\right) = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{4}} = 4\).Multiply them: \((f \cdot g)(\frac{1}{2}) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right) = \frac{1}{4} \cdot 4 = 1\).

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

To find \(\left(\frac{f}{g}\right)(0)\), we need to divide \(f(0)\) by \(g(0)\). Find \(f(0): f(0) = 0^2 = 0\).Find \(g(0): g(0) = \frac{1}{0^2}\), which is undefined since division by zero is not possible. Thus, \(\left(\frac{f}{g}\right)(0)\) is undefined.

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

To find \(\left(\frac{g}{f}\right)(-2)\), divide \(g(-2)\) by \(f(-2)\). Find \(f(-2): f(-2) = (-2)^2 = 4\).Find \(g(-2): g(-2) = \frac{1}{(-2)^2} = \frac{1}{4}\).Divide to get: \(\left(\frac{g}{f}\right)(-2) = \frac{g(-2)}{f(-2)} = \frac{\frac{1}{4}}{4} = \frac{1}{16}\).

Key Concepts

Function Operations

Function operations work just like arithmetic operations but with algebraic functions. Here's how you can understand each operation:

• **Addition**: When we add two functions, say \(f(x) + g(x)\), we evaluate each function at a specific value of \(x\) and then add the outcomes. For example, using the functions from our exercise, to find \((f+g)(2)\), you calculate both \(f(2)\) and \(g(2)\) separately and then add them together: \(f(2) + g(2) = 4 + \frac{1}{4} = \frac{17}{4}\).


• **Subtraction**: Subtraction between functions is similar to addition but instead of adding, you subtract the results: if \(f(x) - g(x)\), you find the difference between the values of each function. For example, \((f-g)(-1)\) simply means you subtract \(g(-1)\) from \(f(-1)\), which gives zero here.


• **Multiplication**: Multiply the function values directly. For \((f \cdot g)(x)\), you evaluate both and multiply the results. So, \((f \cdot g)(\frac{1}{2})\) results from multiplying their outcomes at \(x=\frac{1}{2}\), resulting in one because \(f(\frac{1}{2}) \cdot g(\frac{1}{2}) = \frac{1}{4} \times 4 = 1\).


• **Division**: Dividing functions means first ensuring values are defined, especially avoiding division by zero scenarios. As shown, \(\frac{f(0)}{g(0)}\) becomes undefined at 0 because \(g(x) = \frac{1}{x^2}\) returns an undefined result when \(x=0\).

Rational Functions

Rational functions are expressions constructed by dividing two polynomials. In our specific case, function \(g(x) = \frac{1}{x^2}\) is a classic example of a rational function. Here's what you should consider:

• **Structure**: A rational function is in the form \(\frac{p(x)}{q(x)}\), where both \(p(x)\) and \(q(x)\) are polynomials. For instance, with \(g(x)\), the numerator is 1 (a constant polynomial), and \(q(x) = x^2\).


• **Domain Considerations**: It's crucial to ensure the denominator doesn't equal zero as this makes the function undefined. The exercise highlights this at \(x=0\) for \(g(x)\), causing \(g(x)\) to be undefined.


• **Behavior**: Rational functions often have interesting asymptotic behaviors as \(x\) approaches certain values, which means their values can become very large or very small. Understanding these properties helps us predict how such functions behave.

Function Evaluation

Function evaluation involves finding the output of a function given a specific input. This is a nifty skill when dealing with algebraic functions:

• **Direct Substitution**: Just plug the value of \(x\) into the function's expression. For \(f(x) = x^2\), if you evaluate it at \(x = 2\), substitute 2 for \(x\), resulting in \(f(2) = 4\).


• **Identify Undefined Points**: Some functions may not work for all values due to operations like division by zero. For instance, when evaluating \(g(x)\) at zero, it becomes undefined. Always check your denominators!


• **Finding Outcomes for Combined Operations**: After evaluating initial values of each function, such as \(f(-1)\) or \(g(\frac{1}{2})\), perform the necessary operations as instructed in the problem — whether adding, subtracting, multiplying, or dividing the outcomes. Like with calculations for \((f \cdot g)(\frac{1}{2})\), you need both function evaluations before multiplying their results together.

Remember, consistent practice with these evaluations helps build a solid foundation in solving more complex problems! Learn to love these little evaluations—they simplify big challenges in math.