Problem 6 - Thomas Calculus

Question

If \(f(x)=x-1\) and \(g(x)=1 /(x+1),\) find the following. $$ \begin{array}{ll}{\text { a. } f(g(1 / 2))} & {\text { b. } g(f(1 / 2))} \\\ {\text { c. } f(g(x))} & {\text { d. } g(f(x))}\end{array} $$ $$ \begin{array}{ll}{\text { e. } f(f(2))} & {\text { f. } g(g(2))} \\ {\text { g. } f(f(x))} & {\text { h. } g(g(x))}\end{array} $$

Step-by-Step Solution

Answer

a. \(-\frac{1}{3}\); b. 2; c. \(-\frac{x}{x+1}\); d. \(\frac{1}{x}\); e. 0; f. \(\frac{3}{4}\); g. \(x-2\); h. \(\frac{x+1}{x+2}\).

Step 1: Calculate g(1/2)

First, we need to find the value of \( g(1/2) \). The function \( g(x) = \frac{1}{x+1} \). Substitute \( x = \frac{1}{2} \):\[ g(1/2) = \frac{1}{(1/2) + 1} = \frac{1}{3/2} = \frac{2}{3} \]

Step 2: Compute f(g(1/2))

Now, we have \( g(1/2) = \frac{2}{3} \). Let's compute \( f(g(1/2)) = f(\frac{2}{3}) \). The function \( f(x) = x - 1 \). Substitute \( x = \frac{2}{3} \):\[ f(\frac{2}{3}) = \frac{2}{3} - 1 = \frac{2}{3} - \frac{3}{3} = -\frac{1}{3} \]

Step 3: Calculate f(1/2)

First, find the value of \( f(1/2) \). The function is \( f(x) = x - 1 \). Substitute \( x = \frac{1}{2} \):\[ f(1/2) = \frac{1}{2} - 1 = \frac{1}{2} - \frac{2}{2} = -\frac{1}{2} \]

Step 4: Compute g(f(1/2))

Now, we have \( f(1/2) = -\frac{1}{2} \). Let's compute \( g(f(1/2)) = g(-\frac{1}{2}) \). The function \( g(x) = \frac{1}{x+1} \). Substitute \( x = -\frac{1}{2} \):\[ g(-\frac{1}{2}) = \frac{1}{-\frac{1}{2} + 1} = \frac{1}{\frac{1}{2}} = 2 \]

Step 5: Compute f(g(x))

Now let's find the composite function \( f(g(x)) \). First, express \( g(x) = \frac{1}{x+1} \), then substitute this into \( f(x) = x - 1 \):\[ f(g(x)) = f\left(\frac{1}{x+1}\right) = \frac{1}{x+1} - 1 = \frac{1}{x+1} - \frac{x+1}{x+1} = \frac{1 - (x+1)}{x+1} = \frac{-x}{x+1} \]

Step 6: Compute g(f(x))

Next, we need to compute \( g(f(x)) \). First, express \( f(x) = x - 1 \), then substitute this into \( g(x) = \frac{1}{x+1} \):\[ g(f(x)) = g(x-1) = \frac{1}{(x-1) + 1} = \frac{1}{x} \]

Step 7: Calculate f(f(2))

First, compute the inner function \( f(2) \). The function is \( f(x) = x - 1 \). Substitute \( x = 2 \):\[ f(2) = 2 - 1 = 1 \]Now, compute \( f(f(2)) = f(1) \):\[ f(1) = 1 - 1 = 0 \]

Step 8: Calculate g(g(2))

First, compute the inner function \( g(2) \). The function is \( g(x) = \frac{1}{x+1} \). Substitute \( x = 2 \):\[ g(2) = \frac{1}{2+1} = \frac{1}{3} \]Now, compute \( g(g(2)) = g\left(\frac{1}{3}\right) \):\[ g\left(\frac{1}{3}\right) = \frac{1}{\frac{1}{3} + 1} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

Step 9: Express f(f(x))

To find \( f(f(x)) \), substitute \( f(x) = x - 1 \) into itself:\[ f(f(x)) = f(x-1) = (x-1) - 1 = x - 2 \]

Step 10: Express g(g(x))

To find \( g(g(x)) \), substitute \( g(x) = \frac{1}{x+1} \) into itself:\[ g(g(x)) = g\left(\frac{1}{x+1}\right) = \frac{1}{\frac{1}{x+1} + 1} = \frac{1}{\frac{x+2}{x+1}} = \frac{x+1}{x+2} \]

Key Concepts

Function Composition

Function Composition is like making a sandwich with functions. You take the result of one function and "plug it" into another function. This is shown as \( f(g(x)) \) or \( g(f(x)) \), where

• \( f \) and \( g \) are the functions involved
• You first compute \( g(x) \) and then use that result in \( f \), hence \( f(g(x)) \)

In step-by-step problems, you'll see something like \( f(g\left(\frac{1}{2}\right)) \). Here, you compute \( g\left(\frac{1}{2}\right) \) first, then use that result as the input for \( f \). Simple, right?

Inverse Functions

Inverse Functions are like function "undoers." If \( f(x) \) does something to \( x \), then \( f^{-1}(x) \) will do the opposite to bring you back to your original \( x \).

Consider \( f(x) = x - 1 \); its inverse \( f^{-1}(x) \) should "add 1" to get back to \( x \). That means:

• If \( f(x) = y \), then \( f^{-1}(y) = x \)

Functions \( f \) and \( f^{-1} \) "cancel each other out," and that's what makes them inverse.

Algebraic Functions

Algebraic Functions are the heart of algebra, using operations like addition, subtraction, multiplication, or division with variables and exponents. In exercises like ours:

• \( f(x) = x - 1 \) and \( g(x) = \frac{1}{x+1} \) are classic examples of algebraic functions

These functions let us calculate values for any input \( x \) by substituting numbers and following simple algebra steps, as seen in the solutions given.

Function Notation

Function Notation is like having a shorthand for math works. It uses letters like \( f \), \( g \), or \( h \) to represent functions. If \( f(x) = x - 1 \), everybody knows exactly what \( f(x) \) does to \( x \): it subtracts 1 from \( x \).

• Function notation makes it simple to apply functions, see relationships, and track what's being computed
• Just use the letter and the input in parentheses \( f(input) \)

It helps keep math efficient and lets us easily see how different functions interact within problems, just like in the exercise above.