#1**+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

#4**+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

CPhill Mar 16, 2018