The functions $f$ and $g$ are defined as follows: \[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}.\]Explain why the functions $f$ and $g$ are not the same function. Please help, I can’t seem to find the domain of f :(

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codiepienagoya codiepienagoya · Answer 1 · 2020-07-24T10:19:35+02:00

Answer:

The answer is "domain and range are different".

Step-by-step explanation:

Given:

[tex]f(x) = \sqrt{\frac{x+1}{x-1}}\\\\g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\[/tex]

Solve f(x) to find domain and range:

for element x: R:

⇒ [tex]x \leq -1 \ \ \ and \ \ \ x > 1[/tex] range:

[tex]for \ \ f(x) \times element : 0 \leq f(x)< 1 \ \ \ or \ \ f(x)>1[/tex]

[tex]g(x) =\frac{\sqrt{x+1}}{\sqrt{x-1}}\\[/tex]

Solve for domain:

⇒ [tex]\ when \ x \ element \ R: \ \ \ x>1[/tex]

Solve for range:

⇒ [tex]g(x) \times element[/tex] R : [tex]g(x)>1[/tex]

So, the value of the method f(x) and g(x) (range and domain) were different.

MrRoyal MrRoyal · Answer 2 · 2022-03-21T16:38:52+01:00

Functions f(x) and g(x) are not the same because they do not have the same domain

The functions are given as:

[tex]\[f(x) = \sqrt{\dfrac{x+1}{x-1}}\quad\text{and}\quad g(x) = \dfrac{\sqrt{x+1}}{\sqrt{x-1}}.\][/tex]

Next, we plot the graphs of functions f(x) and g(x).

From the graph (see attachment), we have:

The domain of f(x) is [tex]\[x \le -1}\quad\text{and}\quad x > 1[/tex]

The domain of g(x) is [tex]x > 1[/tex]

Hence, functions f(x) and g(x) are not the same because they do not have the same domain