Domain question

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October 4th 2012, 04:27 PM
kingsolomonsgrave

Domain question

$f(x)=(5x+sqrt(x^2+2x))/4+x$

I figure that the domain can't include -4 and $x^2+2$ is greater than or equal to zero

working from there I get

$x^2+2x$ , is greater than or equal to zero so

$x^2$ is greater than or equal to $-2x$

and therefore x is greater than or equal to -2

however I have been told the answer is actually x equal to or LESS than -2

I don't understand, where did x become less than -2?
October 4th 2012, 05:21 PM
skeeter

Re: Domain question

that is not the correct method to solve a quadratic inequality.

$x^2+2x \ge 0$

$x(x+2) \ge 0$

now think about the graph of the parabola y = x(x+2) ... its zeros are x = 0 and x = -2 ... where (x-values) are the y-values > 0 ?
October 4th 2012, 05:26 PM
Prove It

Re: Domain question

Quote:

Originally Posted by kingsolomonsgrave

$f(x)=(5x+sqrt(x^2+2x))/4+x$

I figure that the domain can't include -4 and $x^2+2$ is greater than or equal to zero

working from there I get

$x^2+2x$ , is greater than or equal to zero so

$x^2$ is greater than or equal to $-2x$

and therefore x is greater than or equal to -2

however I have been told the answer is actually x equal to or LESS than -2

I don't understand, where did x become less than -2?

You need to use brackets where they're needed. Anyway, I assume what you meant was $\displaystyle \begin{align*} f(x) = \frac{5x + \sqrt{x^2 + 2x}}{4 + x} \end{align*}$ .

Yes, you are correct that $\displaystyle \begin{align*} x \neq -4 \end{align*}$ . Yes, you are correct that the function is only defined where $\displaystyle \begin{align*} x^2 + 2x \geq 0 \end{align*}$ . Solving this inequality we have

$\displaystyle \begin{align*} x^2 + 2x &\geq 0 \\ x^2 + 2x + 1 &\geq 1 \\ (x + 1)^2 &\geq 1 \\ |x + 1| &\geq 1 \\ x + 1 \leq -1 \textrm{ or } x + 1 &\geq 1 \\ x \leq -2 \textrm{ or } x &\geq 0 \end{align*}$

So the domain of this function is $\displaystyle \begin{align*} x \in (-\infty, -4) \cup (-4, -2] \cup [0, \infty) \end{align*}$ .