common point of the graph of function

**rcs** · December 22nd 2012, 06:29 AM

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)

**skeeter** · December 22nd 2012, 06:49 AM

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

**earboth** · December 22nd 2012, 06:53 AM

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:

**rcs** · December 22nd 2012, 07:17 AM

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:

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?

**skeeter** · December 22nd 2012, 08:02 AM

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.

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