domain of p(x) 2-9x/ 6x^2+49x+8

Question

 domain of p(x) 2-9x/ 6x^2+49x+8

Answer 1

The domain is the set of all possible input values of x.

The restriction on this is that we cannot divide by zero.

Thus we must find all values of x that make the denominator zero.

 

Assuming we are missing the grouping parenthesis which MUST BE INCLUDED:

 

(2-9x)/(6x²+49x+8)

 

= (2-9x)/(6x²+48x+x+8)

 

= (2-9x)/[6x(x+8) + (x+8)]

 

= (2-9x)/[(6x+1)(x+8)]

 

Now that we have factored the denominator, look for the values that

would make that equal to zero:

 

6x+1=0         x+8=0

6x = -1          x = -8

x = -1/6

 

The domain of x is all real numbers EXCEPT -1/6 and -8

 

D = (-∞, -8)U(-8, -1/6)U(-1/6, ∞)