Domain of a squared polynomial?

Question

I understand that a function is undefined when the denominator equates to zero. So in theory, I guess, just always set it to zero to find the given equation's domain. When I do this problem [1/(x^2-7x)^(1/4)] I just set the denominator to zero, square root both sides twice to cancel out the (1/4) exponent. after that i'm left with just x^2-7x=0 so its the quadratic formula all the way down getting x = 0 & 7. So I assume the domain is undefined at these two x values only. But when the graph the function, the domain is all real numbers except from [0,7] when I had it as

(-infinity,0]U[0,7]U[7,infinity) what am i doing wrong?

Answer 1

you have reasoned well.all real numbers sort of includes infinity.

Answer 2

1/(x^2 - 7x)^(1/4)

the 1/4 power is the fourth root, which is even

because this is an even root, that means that x^2 - 7x >= 0 (just like you can't take the square root of a negative number, you also can't take the fourth root, or any other even root, of a negative number)

because it is in the denominator, that means that x^2 - 7x <> 0

combining these two parts, the domain is x^2 - 7x > 0

this will be "outside" the two zeros of the parabola

x^2 - 7x = 0

x( x - 7) = 0

therefore, the zeros of the parabola are 0 and 7

y > 0 when x is in the intervals (- inf , 0) U (7 , inf)

this is the domain: (- inf , 0) U (7 , inf)

Answer 3

x^2-7x ^ (1/4) cannot equal 0

x^2 - 7x cannot equal 0

x(x-7) cannot equal 0

x cannot equal 0 or 7

(-infinity,0) U (0,7) U (7, infinity)

I think your solution is correct.