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What is the domain of the function $f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$?

Guest Mar 15, 2018
 #1
avatar+14579 
+1

(x+6) /(sqrt(x^2-3x-4) )

 

 

Has a domain for ALL x such that the denominator does NOT equal Zero..AND we do not have a NEGATIVE sqrt

let's find what those values are

 

sqrt(x^2-3x-4)  = sqrt(x-4)(x+1)       This WILL equal zero when x = 4 or -1 <------ NOT allowed !

 

ALSO from    - infinity to +4 is invalid as a negative sqrt will result

 

to NET it all out , the domain is         4< x  thru  + infinity

ElectricPavlov  Mar 15, 2018
 #2
avatar
0

What is that in interval notation?

Guest Mar 15, 2018
 #3
avatar+14579 
0

(4, infinity)

ElectricPavlov  Mar 15, 2018
edited by ElectricPavlov  Mar 15, 2018
 #4
avatar+92843 
+1

\(f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}\)

 

Note that     x^2 - 3x - 4    must be >  0

 

Set  this to 0

 

x^2  -3x - 4  = 0

 

(x - 4) ( x + 1)  = 0

 

Setting each factor to  0   and solving for x gives us

 

x = -1    and x  = 4

 

We have three  possible intervals

 

(-inf , -1)  U (-1, 4)   U (4, inf)

 

Testing x  = 0   in the middle interval makes the polynomial negative....so....this interval does not work

 

The domain is the other two intervals ⇒    (-inf, -1)  U  ( 4, inf )

 

 

Here's a graph  :   https://www.desmos.com/calculator/axfthxpp5t

 

 

cool cool cool

CPhill  Mar 16, 2018

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