# Math Help - I'm trying to understand why this is the domain of this function

1. ## I'm trying to understand why this is the domain of this function

I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....

but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...

2. Originally Posted by jonjon1324
I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....

but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...
Draw the graph of y = 9 - x^2. For what values of x is $y \geq 0$ ....?

3. Originally Posted by mr fantastic
Draw the graph of y = 9 - x^2. For what values of x is $y \geq 0$ ....?
Yeah, in the graph I can see that, but if I'm actually writing out the problem, and I find the square root of 9, how does the inequality flip from x <= + and -3 to x >= -3?

4. Originally Posted by jonjon1324
Yeah, in the graph I can see that, but if I'm actually writing out the problem, and I find the square root of 9, how does the inequality flip from x <= + and -3 to x >= -3?
Unless it's a linear inequality my advice for solving inequalities is to use a graph.

However, if you must use algebra, then I suggest you consider $(3 - x)(3 + x) \geq 0$ and then solve the following two cases for x:

Case 1: $3 - x \geq 0$ AND $3 + x \geq 0$.

Case 2: $3 - x \leq 0$ AND $3 + x \leq 0$.

5. Originally Posted by jonjon1324
I'm trying to find the domain for the function: square root of (9-x^2)...So what I do is solve for 9-x^2 >= 0 (b/c you can't have a negative number under a radical)...and get x is less than or equal to +3 and -3....
This is your error. You cannot "solve" an inequality like this just by taking the square root of both sides. Specifically, $9- x^2 \ge 0$ does give $9\ge x^2$ since you can add $x^2$ to both sides with changing the inequality sign. But you cannot then say " $\pm 3\ge x$".

When you are squaring, or taking the square root of both sides of an inequality, that is equivalent to multipying or dividing by an unknown- specifically you do not know whether that unknown is positive or negative and remember that multiplying or dividing both sides of an inequality by a negative number, the inequality sign reverses.

Once you have $9\ge x^2$, I would recommend first solving the associated equation. That is, $9= x^2$ which does, in fact, have x= 3 and x= -3 as solutions. That divides the number line into 3 intervals, $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$ on each interval of which the inequality is either true for every point in the interval or false for every point in the interval. (That is true because for any continuous function, f(x), f(x)< 0 can only change to f(x)> 0, and vice-versa, where f(x)= 0.)

Knowing that the three intervals are $(-\infty, -3)$, $(-3, 3)$, and $(3, \infty)$. We need only look at a single point in each interval. x= -4 is in $(-\infty, -3)$ and $(-4)^2= 16> 9$ so $x^2> 9$ for all x in $(-\infty, -3)$. x= 0 is in $(-3, 3)$ and $(0)^2= 0< 9$ so $x^2< 9$ for all x in $(-3, 3)$. Finally, $x= 4$ is in (3, \infty) and $4^2= 16> 9$ so $x^2> 9$ for all x in $(3, \infty)$. Add to this the fact that $3^2= (-3)^2= 9$ and we have that $x^2\le 9$ if and only if x is in [-3, 3] or $-3\le x\le 3$.

Going back to $9- x^2= (3- x)(3+ x)= -(x- 3)(x+ 3)\ge 0$, another and perhaps simpler way to see that is this: $a- b> 0$ if and only if $a> b$ and $a- b< 0$ if and only if a< b. If x< -3 then it is also less than 3 so both of x- 3 and x+ 3= x-(3) are negative. Their product (the product of two negative numbers) is positive and so -(x- 3)(x+ 3) is [b]negative[/tex]. If -3< x< 3, then x- 3 is still negative but x+ 3= x-(-3) is now positive. Their product (the product of a positive and negative number) is negative so -(x- 3)(x+ 3) is positive. Finally, if x> 3, it is also greater than -3 so both x- 3 and x+ 3 are positive. The product of two positive numbers is positive so -(x- 3)(x+ 3) is negative. Again, we have that $x^2- 9\ge 0$ if and only if $-3\le x\le 3$.

but the domain is supposedly [-3,3]...which would mean x > or = -3....but when I solved for it I got that x < or = -3...IT DOESN'T MAKE SENSE...