what would the domain be if i had a problem with the denominator as x^3?
YOu cant have x=0 in the domain.
it is just basic factorising but im not sure how u do it.
x2-5x =
by the way that is meant to be x squared, minus 5 x.
also 3a2 - 6a =
First, if you have a question, it is much better to put it in as a separate post in <Post a New Question> rather than attaching it to a previous question, where it is more likely to be overlooked.
Use a carat (^) to indicate a power.
x^2-5x = x(x-5)
3a^2-6a = 3a(a-2)
I hope this helps. Thanks for asking.
To determine the domain of a function, you need to identify any values of the independent variable (x) that would result in an undefined expression. In this case, the denominator is x^3.
For a fraction to be defined, the denominator cannot be zero. So, the values for x that would make the denominator zero need to be excluded from the domain.
Setting the denominator equal to zero, we have:
x^3 = 0
To find the values of x that satisfy this equation, we can solve it by taking the cube root of both sides:
∛(x^3) = ∛(0)
x = 0
Hence, x = 0 is the only value that makes the denominator zero. Therefore, the domain for the function would be all real numbers except x = 0. In interval notation, it can be expressed as (-∞, 0) U (0, +∞).