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Question: Simplify the expression |x^7+5x|/x where x is negative. Then let h(x) represent the simplified expression and determine its domain, and then decide whether this simplified expression is equal to the original expression.

My attempt: I am very confused. When the question says "where x is negative" does it mean that the negative sign is outside of the absolute value brackets as in the stuff on the numerator is less than 0. OR does it mean to literally plug in -x into the expression and then simplify it.

From the latter mentioned guess, I would say the simplified expression is not equal to the original because it only includes the positive half of the function when graphed out and the domain is x ≠ 0. I'm not certain if this is correct, could you guys solve this question and compare with mine?

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In very much detail, the question is asking you something like "if you knew for a fact that $x$ were negative, how can you simplify the expression?"

As an example of the idea, consider the question asked for the expression $\sqrt{x^2}$ instead. Normally, there isn't a way to simplify this (you can't say it's simply $x$ because $\sqrt{(-5)^2} = 5 \neq -5$), but for all negative $x$ you in fact can say it just equals $-x$ (why not $x$?).

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First, "where $x$ is negative" means "where we pretend we are in a universe in which $x$ can only take negative values". So for instance, we allow $x$ to be $-5$, but we do not allow $x$ to be five.

Then, this matters for absolute values because their definition explicitly depends on whether their argument is negative or non-negative: $$ |z| = \begin{cases} z, & z \geq 0 \\\ -z, & z < 0 \end{cases} \text{.} $$

Finally, we use this to make progress on your problem. First we use a property of absolute values: $$ |a b| = |a|\,|b| \text{.} $$ This let's us write $$ \frac{|x^7 + 5x|}{x} = \frac{|(x^6 + 5)x|}{x} = \frac{|x^6 + 5| \ |x|}{x} \text{.} $$ Since we are in a universe where $x < 0$, the "$|x|$" in the numerator uses the second line from the definition of absolute value above, $|x| = -x$. (If we imagine $x$ is $-5$ this is just saying $|-5| = -(-5) = 5$, as we expect.) This gives us $$ \frac{|x^6 + 5| \ |x|}{x} = \frac{|x^6 + 5| (-x)}{x} = \frac{|x^6 + 5| (-1)(x)}{x} \text{.} $$

Now that we have $x$ as a factor of the entire numerator and as a factor of the entire denominator, we can cancel them to obtain $$ \frac{|x^6 + 5| (-1)}{1} = -|x^6 + 5| \text{,} $$ a simpler expression than the one we started with and a reasonable candidate for $h$.

(Hint for the domain part of the question: We usually only have to remove something from the domain when it results in division by zero or results in taking an even root of a negative number. Since we don't have any roots in this problem, we should be thinking about choices of $x$ that cause division by zero in the original expression but no in $h$ or in $h$ but not in the original expression.)

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  • $\begingroup$ @Sameer123 : If I know that $x$ is negative, then I know that $x$ is not zero. $\endgroup$ – Eric Towers Sep 7 at 23:19
  • $\begingroup$ The original expression has a "hole" at $x=0$, but the simplified expression has no such hole. If we are being pedantic, the domain of the simplified expression is $(-\infty, 0)$, so it (the domain) doesn't have a hole at zero, it just stops at zero. $\endgroup$ – Eric Towers Sep 7 at 23:20
  • $\begingroup$ I just don't see what you are getting at. The domain of the original expression is $(-\infty,0) \cup (0, \infty)$ and the domain of the simplified expression is $(-\infty,0)$. You seem to be saying that at some point I wrote $\frac{|x^6 + 5|(-1)(x)}{x} = \frac{|x^6 + 5|(-1)}{1}$, which is an identity, but not an equality of functions. If you look back, I never write such a thing. Also, we are asked to find an expression for $h$, not the function $h$, so our work is not supposed to produce a domain. We do that afterwards (using that our simplifications required $x < 0$). $\endgroup$ – Eric Towers Sep 7 at 23:25
  • $\begingroup$ The simplified expression cannot be equal to the original expression since they disagree on some choices of $x$. (Notice that we are comparing expressions, not functions so we can ignore their domains.) Try $x = 1$. It is entirely reasonable for two expressions to agree in some places and disagree in others. We only expect these two expressions to agree on the intersection of the domain of the starting expression and the domain of $h$. That intersection does not contain $1$, so while the expressions are not equal, we have an identity of functions. $\endgroup$ – Eric Towers Sep 7 at 23:39
  • $\begingroup$ @Sameer123 : In this case, that is correct. It would not be correct if we had $-|x^5 + 5|$, since, even though we know $x < 0$, we do not know the sign of $x^5 + 5$. Here, we know that $x^6 \geq 0$ no matter what $x$ is and so we know $x^6 + 5 > 0$. Therefore, we know which piece of the definition of the absolute value applies to this expression. $\endgroup$ – Eric Towers Sep 7 at 23:49

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