# Math Help - common point of the graph of function

1. ## common point of the graph of function

The function f(x) = 2 + sqrt(x+4) is one to one. Find the common point of the graph of y = f(x) and y = f^-1(x)

2. ## Re: common point of the graph of function

for future reference ... the posts you've made in this forum are not advanced algebra.

3. ## Re: common point of the graph of function

Originally Posted by rcs
The function f(x) = 2 + sqrt(x+4) is one to one. Find the common point of the graph of y = f(x) and y = f^-1(x)
1. $D_f = [-4,+\infty)$ and the range is $R_f=[2, +\infty)$

Therefore $D_{f^{-1}}= [2, +\infty)$ and the range of $f^{-1}$ is $R_{f^{-1}} = [-4,+\infty)$

2. The common point of both graphs must be located onn the line y = x (why?)

Therefore solve for x:

$x = 2 + \sqrt{x+4}~,~x\ge 2$

Spoiler:
You should come out with (5, 5)

4. ## Re: common point of the graph of function

Originally Posted by earboth
1. $D_f = [-4,+\infty)$ and the range is $R_f=[2, +\infty)$

Therefore $D_{f^{-1}}= [2, +\infty)$ and the range of $f^{-1}$ is $R_{f^{-1}} = [-4,+\infty)$

2. The common point of both graphs must be located onn the line y = x (why?)

Therefore solve for x:

$x = 2 + \sqrt{x+4}~,~x\ge 2$

Spoiler:
You should come out with (5, 5)
i got the answer that does not have solution...

it says (5,5)
it's wacky and confusing to follow your step sir... is there any other way?

5. ## Re: common point of the graph of function

it's wacky and confusing to follow your step sir... is there any other way?
why? ... it would help if you were to show how you solved the equation given to you by earboth.