Given the function #g(x)=(x^2-3x-4)/(x-5)#, how do you find the domain?

Question

Avery Alexander · Accepted Answer

#x inRR,x!=5#

Since equating the denominator to zero and solving for x yields the value that x cannot be, the denominator of g(x) cannot be zero as this would render g(x) undefined.

#"solve "x-5=0rArrx=5larrcolor(red)"excluded value"#

#"domain is "x inRR,x!=5#

#(-oo,5)uu(5,oo)larrcolor(blue)"in interval notation"# graph{(x^2-3x-4)/(x-5 [-40, 40, -20, 20]}

Hazel Anglin · Answer

#x inRR, x!=5#

The only #x# value that will make this function undefined is when the denominator is set to zero. We see that this value is #x=5#.

Therefore, we can say that the domain is #x inRR, x!=5#. This is just a fancy way of saying #x# can be any real number except #5#.

We also see this graphically, as we have a vertical asymptote at #x=5#.

graph{[-74, 86, -36.8, 43.2]};x^2-3x-4)/(x-5)

I hope this is useful.

Eli Assouline · Answer

undefined To find the domain of the function g(x) = (x^2 - 3x - 4) / (x - 5), you need to identify any values of x that would make the denominator zero, since division by zero is undefined. Set the denominator equal to zero and solve for x to find any restrictions on the domain. In this case, x - 5 = 0, so x = 5. Therefore, the domain of the function is all real numbers except x = 5.