find the domain of sqrt(9-x^2)/(x^3-4x)

Question

 Need help finding domain of f(x)=sqrt(9-x^2)/(x^3-4x)

Answer 1

This is a two part kind of problem. We need to know where both the numerator and denominator are defined. Additionally we must remember that this is a square root function we are dealing with (i.e. no negatives will be defined in this function)

 

For the numerator, we know that 3 is an issue. At x=±3 we have the function equal to zero. 

 

For the denominator, we have to know what points will make this function undefined. This occurs at 0,±2.

 

So now we just need to know what endpoints are inclusive, and what happens after our trouble points. 

 

As x becomes larger than 3, our function becomes negative so this will not do. We know the upper bound of the domain is x≤3. As x becomes negative (less than -3) our function is still defined thanks to the denominator. So, our lower bound for the domain is -∞≤x≤-3.

 

Now we just need to know what occurs between -3 and 3. 

 

At x=2 and x=-2 the function is undefined....so we cannot have inclusive endpoints. (i.e. -2<x<2). Remembering that we also have to worry about 0, we see the function is also undefined there. (i.e. x≠0). 

 

 

Putting all of this together, we get ∞≤x≤-3, -2<x<0, and 2<x≤3.

 

Formally, we can say {x∈R: x≤-3, -2<x<0, 2<x≤3}.

Answer 2

Domains represent a list of all x-values that are allowed. In order to make that list, we first have to find what x-values are not allowed. This particular expression is a fraction and also a square root function. Let's start with the fraction part of it.

 

For a fraction, denominators cannot be equal to zero, so find what value of x would make the denominator zero, and this will give one of the restrictions.

 

x³ - 4x = 0

 

Factor out an x.

 

x³ - 4x³ = 0

x(x² - 4x) = 0

 

That parenthetical term sure looks like a difference of squares.

 

x(x² - 4x) = 0

x(x + 2)(x - 2) = 0

 

For this particular equation, x = -2, 0, and 2. These three values of x will make the denominator zero, and because a zero denominator makes a function undefined, these are values of x that are on the restricted list.

 

Now recall that we said that this expression is a combination of a fraction and a square root function. We can't take the square root of a negative number, so the expression under the square root can't be negative. Set the expression under the radical to be greater than or equal to zero, in that case.

 

9 - x² > 0

 

This sure looks like yet another difference of squares.

 

9 - x² > 0

(3 + x)(3 - x) > 0

 

This is a quadratic inequality, which can get kind of weird. The zeroes of the quadratic expression are at x = -3 and x = 3. When dealing with quadratic inequalities, plot these on a number line to do something called line analysis.

 

<----(-3)----(3)---->

 

The quadratic inequality wants the expression to be greater than or equal to zero. Another way to say "greater than zero" would just be to say "positive." So we're going to need to find what values of x make the expression positive. The line analysis will help us out. The left part of the line represents values less than -3. Choose a number less than that, say -4, and plug it in the inequality.

 

(3 + x)(3 - x) > 0

(3 + (-4))(3 - (-4)) > 0

(-1)(7) > 0

-7 > 0

 

A number less than -3 will make the expression negative.

 

   (-)

<----(-3)----(3)---->

 

Now pick a value between -3 and 3 to test. 0 is an easy one to use, so use that.

 

(3 + x)(3 - x) > 0

(3 + 0)(3 - 0) > 0

(3)(3) > 0

9 > 0

 

A value between -3 and 3 will make this expression positive.

 

   (-)      (+)
<----(-3)----(3)---->

 

Finally choose a value greater than 3 to test. 4 will do the trick.

 

(3 + x)(3 - x) > 0

(3 + 4)(3 - 4) > 0

(7)(-1) > 0

-7 > 0

 

Values greater than 3 will make the expression negative.

 

   (-)      (+)      (-)
<----(-3)----(3)---->

 

Remember that we set the radicand greater than or equal to zero. -3 and 3 will make the radicand exactly zero, so those are allowed in the domain. The radicand can also be greater than zero, or in other words, positive. Based on this line analysis, the radicand is positive on the interval between -3 and 3.

 

Therefore, put that all together. The radicand is "greater than or equal to zero" on the interval -3 < x < 3. Also, remember way back in the problem, when we found the restrictions based on the denominator. x cannot be -2, 0, and 2.

 

Combine these, and we find that the domain is all values in the interval -3 < x < 3, except -2, 0, and 2. Here are some ways to write that.

 

-3 < x < -2, -2 < x < 0, 0 < x < 2, 2 < x < 3

 

Here's another way.

 

[-3, -2), (-2, 0), (0, 2), (2, 3]

 

And one last way.

 

[-3, 3] - {-2, 0, 2}

 

Hopefully one of these interval notations represents the way your teacher wants you to write it. Hope this helps, and good luck with the rest of your math work!