# How to find domain for f(x) = sqrt(x^2 - 5x)?

Apr 19, 2018

(-infinity,0] U [5,infinity)

#### Explanation:

I often find it easiest to see what the limiting factor is before I start with domain problems.

Can you square any number from negative to positive infinity? Yes.

Can you multiply any number from negative to positive infinity? Yes.

Can you square root any number from negative to positive infinity? No.

Recall that you cannot take the square root of a negative number, so this means that everything under the radical must always be positive. Now we have the beginnings of a domain: $0 \le {x}^{2} - 5 x$

Factor and find the zeros of the quadratic. Note: Do NOT divide by x! You will end up losing a solution! The factored form of the equation should look like $0 = x \left(x - 5\right)$ Set each term equal to zero and you should get x=0,5.

Now, it is necessary to test for whether the function is positive or negative at these values. You know that since the graph is a parabola, it should pass through the x axis at most twice (as it does here).
graph{x^2-5x [-10, 10, -10, 10]}

When a graph passes over the x axis, its outputs change sign, positive to negative or vice versa. Here, I would advise setting up a number line to test for values before x=0, between x=0 and x=5, and after x=5. Specific numbers aren't needed, you just need to know if it is positive or negative.

Before x=0: you can use some very big negative number, say -100. ${\left(- 100\right)}^{2}$ is a very large positive number and $- 5 \left(- 100\right)$ is also a positive number. A positive plus a positive is always a positive. You can confirm that from negative infinity to x=0 the graph within the square root is always positive, and thus the function always defined.

Next you have the area between x=0 and x=5. Plug in a number, say 3. ${3}^{2} = 9$ and $- 5 \left(3\right) = - 15$. 9-15 is a negative number! So you can safely assume that the x values between 0 and 5 are all negative and the function is not defined here.

Finally numbers greater than x=5. Put in 100. ${100}^{2}$ is a huge positive number while $- 5 \left(100\right)$ is a much smaller negative number. A huge positive number plus a smaller negative number is always a positive number. You can assume that the function is always above 0 and thus the function is defined on all values from x=5 to positive infinity.

Combining all the domains and you get (-infinity,0] U [5,infinity) brackets are used for the 0 and 5 because they also lead to non negative numbers within the radical. The U symbol means union, and it just combines two domains or sets.