Questions: Consider the following functions. f(x)=(x+8)/(x-7), g(x)=(7x+8)/(x-1) (a) Verify that f and g are inverse functions algebraically. f(g(x))=f((7x+8)/(x-1)) =((square)+8)/((square)-7) =x g(f(x))=g(square) =(7(square)+8)/(square-1) =x

Consider the following functions.
f(x)=(x+8)/(x-7), g(x)=(7x+8)/(x-1)
(a) Verify that f and g are inverse functions algebraically.
f(g(x))=f((7x+8)/(x-1))
=((square)+8)/((square)-7)
=x
g(f(x))=g(square)
=(7(square)+8)/(square-1)
=x

Transcript text: Consider the following functions. \[ f(x)=\frac{x+8}{x-7}, \quad g(x)=\frac{7 x+8}{x-1} \] (a) Verify that $f$ and $g$ are inverse functions algebraically. \[ \begin{array}{l} f(g(x))=f\left(\frac{7 x+8}{x-1}\right) \\ =\frac{\square+8}{\square-7} \\ =x \\ g(f(x))=g(\square) \\ =\frac{7(\square)+8}{\square-1} \\ =x \end{array} \]

Solution

Solution Steps

To verify that $ f $ and $ g $ are inverse functions, we need to show that $ f(g(x)) = x $ and $ g(f(x)) = x $. This involves substituting $ g(x) $ into $ f(x) $ and simplifying, and then substituting $ f(x) $ into $ g(x) $ and simplifying.

Step 1: Define the Functions

We are given the functions: \[ f(x) = \frac{x + 8}{x - 7} \] \[ g(x) = \frac{7x + 8}{x - 1} \]

Step 2: Verify $ f(g(x)) = x $

First, we substitute $ g(x) $ into $ f(x) $: \[ f(g(x)) = f\left(\frac{7x + 8}{x - 1}\right) = \frac{\left(\frac{7x + 8}{x - 1}\right) + 8}{\left(\frac{7x + 8}{x - 1}\right) - 7} \]

Simplify the expression: \[ f(g(x)) = \frac{\frac{7x + 8 + 8(x - 1)}{x - 1}}{\frac{7x + 8 - 7(x - 1)}{x - 1}} = \frac{\frac{7x + 8 + 8x - 8}{x - 1}}{\frac{7x + 8 - 7x + 7}{x - 1}} = \frac{\frac{15x}{x - 1}}{\frac{15}{x - 1}} = x \]

Thus, we have: \[ f(g(x)) = x \]

Step 3: Verify $ g(f(x)) = x $

Next, we substitute $ f(x) $ into $ g(x) $: \[ g(f(x)) = g\left(\frac{x + 8}{x - 7}\right) = \frac{7\left(\frac{x + 8}{x - 7}\right) + 8}{\left(\frac{x + 8}{x - 7}\right) - 1} \]

Simplify the expression: \[ g(f(x)) = \frac{\frac{7(x + 8) + 8(x - 7)}{x - 7}}{\frac{x + 8 - (x - 7)}{x - 7}} = \frac{\frac{7x + 56 + 8x - 56}{x - 7}}{\frac{15}{x - 7}} = \frac{\frac{15x}{x - 7}}{\frac{15}{x - 7}} = x \]

Thus, we have: \[ g(f(x)) = x \]