Question

Consider \(f(x)=1-x\) and \(g(x)=\frac{x}{1-x}, x \neq 1\) a) Show that \(g(f(x))=\frac{1}{g(x)}\). b) Does \(f(g(x))=\frac{1}{f(x)} ?\)

Short answer

a) Yes, g(f(x)) = \frac{1}{g(x)}.b) No, f(g(x)) eq \frac{1}{f(x)}.

Step-by-step solution

- Find g(f(x))

First, we need to find the composition of functions where we input the function f(x) into g(x). Since f(x) = 1 - x, replace x in g(x) with 1 - x.So,\[ g(f(x)) = g(1 - x) = \frac{1 - x}{1 - (1 - x)} \]Simplify the denominator:\[ g(1 - x) = \frac{1 - x}{x} \]

- Simplify g(f(x))

Simplify the expression obtained in Step 1:\[ g(f(x)) = \frac{1 - x}{x} = \frac{1}{x} - 1 \]

- Find \frac{1}{g(x)}

Next, we need to determine \frac{1}{g(x)} where \[ g(x) = \frac{x}{1 - x} \].So, \[ \frac{1}{g(x)} = \frac{1}{\frac{x}{1 - x}} = \frac{1 - x}{x} \]

- Compare \frac{1}{g(x)} and g(f(x))

From Step 2, we have \[ g(f(x)) = \frac{1 - x}{x} \]From Step 3, we also have \[ \frac{1}{g(x)} = \frac{1 - x}{x} \]Therefore, \[ g(f(x)) = \frac{1}{g(x)} \] as required.

- Find f(g(x))

Next, we need to find the composition of functions where we input g(x) into f(x). Since g(x) = \frac{x}{1 - x}, replace x in f(x) with \frac{x}{1 - x}.So, \[ f(g(x)) = f\left(\frac{x}{1 - x}\right) = 1 - \frac{x}{1 - x} \]

- Simplify f(g(x))

Simplify the expression obtained in Step 5:\[ f\left(\frac{x}{1 - x}\right) = \frac{(1 - x) - x}{1 - x} = \frac{1 - 2x}{1 - x} \]

- Find \frac{1}{f(x)}

Next, we need to determine \frac{1}{f(x)} where \[ f(x) = 1 - x \].So, \[ \frac{1}{f(x)} = \frac{1}{1 - x} \]

- Compare \frac{1}{f(x)} and f(g(x))

From Step 6, we have \[ f(g(x)) = \frac{1 - 2x}{1 - x} \]From Step 7, we have \[ \frac{1}{f(x)} = \frac{1}{1 - x} \]Since \[ f(g(x)) eq \frac{1}{f(x)} \], the equation \[ f(g(x)) = \frac{1}{f(x)} \] does not hold.

Key concepts

Composite Functions

Composite functions combine two functions to create a new function. When we have two functions, say \(f(x)\) and \(g(x)\), the composite function \(g(f(x))\) is formed by substituting \(f(x)\) into \(g(x)\). It's like feeding \(x\) into \(f(x)\) first and then taking that result and putting it into \(g(x)\).
For example, in our problem:
If we have \(f(x) = 1 - x\) and \(g(x) = \frac{x}{1 - x}\), then \