Problem 10 - Precalculus

Question

Use the given pair of functions to find the following values if they exist. $$\bullet (g \circ f)(0)$$ $$\bullet (f \circ g)(-1)$$ $$\bullet (f \circ f)(2)$$ $$\bullet (g \circ f)(-3)$$ $$\bullet (f \circ g)\left(\frac{1}{2}\right)$$ $$\bullet (f \circ f)(-2)$$ $$f(x)=\frac{x}{x+5}, g(x)=\frac{2}{7-x^{2}}$$

Step-by-Step Solution

Answer

\((g \circ f)(0) = \frac{2}{7}, (f \circ g)(-1) = \frac{1}{16}, (f \circ f)(2) = \frac{2}{37}, (g \circ f)(-3) = \frac{8}{19}, (f \circ g)(\frac{1}{2}) = \frac{8}{143}, (f \circ f)(-2) = \frac{-2}{13}\).

Step 1: Compute \((g \circ f)(0)\)

Start by using the function \(f(x)\) first. Substitute 0 into \(f\): \[ f(0) = \frac{0}{0+5} = 0 \]. Now use this as the input for \(g\): \[ g(0) = \frac{2}{7 - 0^2} = \frac{2}{7} \]. Therefore, \((g \circ f)(0) = \frac{2}{7} \).

Step 2: Compute \((f \circ g)(-1)\)

First, use \(g(x)\) with \(-1\) as the input: \[ g(-1) = \frac{2}{7 - (-1)^2} = \frac{2}{6} = \frac{1}{3} \]. Use this result as the input for \(f\): \[ f\left(\frac{1}{3}\right) = \frac{\frac{1}{3}}{\frac{1}{3} + 5} = \frac{\frac{1}{3}}{\frac{16}{3}} = \frac{1}{16}\]. Thus, \((f \circ g)(-1) = \frac{1}{16} \).

Step 3: Compute \((f \circ f)(2)\)

First, apply \(f(2)\): \[ f(2) = \frac{2}{2+5} = \frac{2}{7} \]. Use this result as the new input for \(f\): \[ f\left(\frac{2}{7}\right) = \frac{\frac{2}{7}}{\frac{2}{7} + 5} = \frac{\frac{2}{7}}{\frac{37}{7}} = \frac{2}{37} \]. Thus, \((f \circ f)(2) = \frac{2}{37} \).

Step 4: Compute \((g \circ f)(-3)\)

Start by evaluating \(f(-3)\): \[ f(-3) = \frac{-3}{-3+5} = \frac{-3}{2} \]. Take this result as the input for \(g\): \[ g\left(\frac{-3}{2}\right) = \frac{2}{7 - \left(\frac{-3}{2}\right)^2} = \frac{2}{7 - \frac{9}{4}} = \frac{2}{\frac{19}{4}} = \frac{8}{19} \]. Thus, \((g \circ f)(-3) = \frac{8}{19} \).

Step 5: Compute \((f \circ g)(\frac{1}{2})\)

First, apply \(g\left(\frac{1}{2}\right)\): \[ g\left(\frac{1}{2}\right) = \frac{2}{7 - \left(\frac{1}{2}\right)^2} = \frac{2}{7 - \frac{1}{4}} = \frac{2}{\frac{27}{4}} = \frac{8}{27} \]. Now use this result with \(f\): \[ f\left(\frac{8}{27}\right) = \frac{\frac{8}{27}}{\frac{8}{27} + 5} = \frac{\frac{8}{27}}{\frac{143}{27}} = \frac{8}{143} \]. Thus, \((f \circ g)\left(\frac{1}{2}\right) = \frac{8}{143} \).

Step 6: Compute \((f \circ f)(-2)\)

First, evaluate \(f(-2)\): \[ f(-2) = \frac{-2}{-2+5} = \frac{-2}{3} \]. Use the result as the input for \(f\): \[ f\left(\frac{-2}{3}\right) = \frac{\frac{-2}{3}}{\frac{-2}{3} + 5} = \frac{\frac{-2}{3}}{\frac{13}{3}} = \frac{-2}{13} \]. Thus, \((f \circ f)(-2) = \frac{-2}{13} \).

Key Concepts

Function Evaluation

When we talk about function evaluation, we're focusing on how to substitute specific values into a function to obtain a result. To evaluate a function like \( f(x) = \frac{x}{x+5} \), simply replace \( x \) with the given number and solve the expression. For example, if you need to find \( f(0) \), plug in \( x = 0 \) and simplify:

• \[ f(0) = \frac{0}{0 + 5} = 0 \]

For composition of functions, function evaluation becomes a two-step process. You start by evaluating the inner function first and use the result as the input for the outer function. Suppose you need to find \( (g \circ f)(0) \). First, evaluate \( f(0) \) and then use the result to find \( g(0) \):

• \[ f(0) = 0, \quad g(0) = \frac{2}{7 - 0^2} = \frac{2}{7} \]

By practicing these steps, you can become comfortable evaluating functions and understanding their outcomes.

Inverse Functions

Inverse functions essentially reverse the effect of the original function. If you have a function \( f(x) \) and its inverse \( f^{-1}(x) \), applying \( f \) and then \( f^{-1} \) (or vice versa) will bring you back to your original value:

• \( f(f^{-1}(x)) = x \)
• \( f^{-1}(f(x)) = x \)

To find the inverse of a function, switch \( x \) and \( y \) in the equation, and then solve for \( y \). For example, if \( f(x) = \frac{x}{x+5} \), to find the inverse, you'll:

• Set \( y = \frac{x}{x+5} \).
• Switch \( x \) and \( y \), which gives \( x = \frac{y}{y+5} \).
• Solve for \( y \), which might involve algebraic manipulation to rewrite the function.

Keep in mind that not all functions have inverses that are also functions.

Function Domain

The domain of a function includes all the possible input values (usually \( x \)-values) where the function can be evaluated without any issues such as division by zero or taking the square root of a negative number. For example, the domain of \( f(x) \) where \( f(x) = \frac{x}{x+5} \) is all real numbers except \( x = -5 \), because at \( x = -5 \), the function becomes undefined:

• \[ x + 5 = 0 \Rightarrow x = -5 \]

Similarly, for \( g(x) = \frac{2}{7 - x^2} \), the domain is all real numbers except where \( 7 - x^2 = 0 \) (i.e., \( x = \pm\sqrt{7} \)):

• Solve \( 7 - x^2 = 0 \), yielding \( x = \pm \sqrt{7} \).

Understanding the domain is essential for correctly using and composing functions.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying expressions to make calculations easier, especially when working with complex functions. For instance, when solving \( (f \circ g)(-1) \), algebraic manipulation is key:1. Evaluate \( g(-1) \):

• \[ g(-1) = \frac{2}{7 - (-1)^2} = \frac{2}{6} = \frac{1}{3} \]

2. Use this value to find \( f \left( \frac{1}{3} \right) \):

• \[ f \left( \frac{1}{3} \right) = \frac{\frac{1}{3}}{rac{1}{3} + 5} = \frac{\frac{1}{3}}{\frac{16}{3}} = \frac{1}{16} \]

Effective algebraic manipulation can make finding the result much easier. Practice simplifies the process, turning involved expressions into manageable pieces one step at a time.