Asked by Sammy

Find the domain of:
6/(9-4x)
7/(x^2-8x)
(x^2+9x)/(x^3-13x^2+40x)

Answers

Answered by Kuai

6/(9-4x)

9-4x =0
9-4x +4x = 0+4x
9= 4x
x =9/4
Domain of x is 9/4

7/(x^2-8x)

x^2 -8x = 0
x(x-8) =0

x =0
x -8 =0
x -8+8 = 0
x =8

domain of x are 0, 8


(x^2+9x)/(x^3-13x^2+40x))

x^3-13x^2+40x =0

x (x^2-13x +40) =0

x(x-5)(x-8) =0

x =0
x = 5
x = 8

Domain of x are 0, 5, and 8


Answered by Steve
the special values are correct, but the domain is all real numbers <b>except</b> those values.
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