Find the domain of the function $g(x) =\sqrt{x^2 − 4x − 32}$.

Find the domain of the function: $g(x) = \sqrt{ x^2 − 4x − 32}.$

I got the square root of $(x-8) (x+4).$ Intervals $(-4,8).$

Since I am asked to find the domain, I know that my domain must be >/ (greater or equal) to zero, thus no negative integers.

The answer in the book is $(-\infty, -4] \cup [8, \infty).$

How did it come to that solution?

I also just (arbitrarily) plugged in $-5$ into the original solution and got $13.$ What does that mean?

• Just to make clear, is it: $g(x) = \sqrt{x^2-4x-32}$?
– Eff
Jun 15 '14 at 17:47
• What do you mean "Intervals $(-4,8)$"? Jun 15 '14 at 18:13

As you remarked, we can write $g(x)$ as: $$g(x)=\sqrt{x^2-4x-32}=\sqrt{(x-8)(x+4)}.$$ We know that the square root of a number is defined only and only if this number is positive. So for our particular example, we need to figure out when the expression inside the square root is positive, and thus we need to consider solving the inequality: $(x-8)(x+4)\geqslant0.$ How can we do it? One of the possibilities, is to use a table, where we study where each 1 degree polynomial is positive and negative, and then we conclude when their product is positive:

$\;\;\;$ From it we can see that the product $(x+4)(x-8)$ is positive if and only if: $$x\in\big(-\infty,-4\big]\cup\big[8,+\infty\big).$$ Therefore, the domain of our function is just the set: $${\rm D}_g=\big(-\infty,-4\big]\cup\big[8,+\infty\big).$$

And since $-5$ is in the domain of our function, and since the range of the square root function is $[0,+\infty)$, then it is natural that $g(5)\in[0,+\infty).$

• Ok. got it. I think I fully understand now. Jun 15 '14 at 18:34
• @Utvecklaochförenkla Excellent! With some practice those problems will be very easy to handle in the future. Jun 15 '14 at 18:36

You have $$g(x)=\sqrt{(x-8)(x+4)}.$$ For it to be defined you need $(x-8)(x+4) \geq 0$. Now split the real line into three parts by using the points $x=8$ and $x=-4$ (the zeros of your function).

Now consider the first region: When $\mathbf{x \geq 8}$:

Then both $x-8 \geq 0$ and $x+4 \geq 0$. Thus $(x-8)(x+4) \geq 0$. Thus anything in $[8,\infty)$ is good region for you. Now test the remaining two regions.

You know that if the thing inside the square root (a.k.a the radicand) is $\geq 0$, then it is on the graph. If the radicand is $< 0$, then it is not a real number and therefore not on the graph. We basically need to solve the inequality $x^2-4x-32\geq 0$

$$x^2-4x-32\geq 0$$ $$(x+4)(x-8)\geq 0$$

The critical points (values that will make the left hand side equal $0$) are $-4$ and $8$.

There are now $3$ cases to consider.

Case $1$: $x \geq 8$

Chose any number that is greater than or equal to $8$. I will chose $10$.

Now, plug in this number (in this case, $10$) to the inequality.

$$(10+4)(10-8)$$ $$=14(2)$$ $$=28$$

Since it is positive, $x \geq 8$ is a solution.

Case $2$: $-4 < x < 8$

I will chose $-1$ for this one.

$$(-1+4)(-1-8)$$ $$=3(-9)$$ $$=-27$$ Since it is negative, $x$ is not in the region $-4<x<8$

Case $3$: $x\leq -4$

I will choose $-5$ for this one. $$(-5+4)(-5-8)$$ $$=(-1)(-13)$$ $$=13$$ Since it is positive, $x\leq -4$ is a solution.

The solutions to the inequality are: $$x\geq 8 \ \text{or} \ x \leq -4$$ In interval notation: $$[-\infty, \ -4]\cup [8, \ \infty]$$

These values make the radicand in the original function positive. In other words, these values are on the graph. The domain is therefore: $$x\geq 8 \ \text{or} \ x\leq -4$$