# Homework Help: Sign of a function

1. Nov 1, 2017

### Alex126

1. The problem statement, all variables and given/known data
$y = \sqrt {x+5} + \sqrt[3] {\frac {x} {4}}$

Find when the function is positive/negative/zero.

We were actually supposed to only calculate the domain in this exercise, but we had done some simpler and more basic positive/negative calculations before, and I was curious about this one.
3. The attempt at a solution
The way I tried to solve it was as follows:

0. (Calculate the domain)
1. I put both roots under the same index. If my reasoning is correct, this leads to two separate cases. Correct me if I'm wrong, but I would say that the starting function is equivalent to the union of the following:

f(x) =
\begin{cases}
y = \sqrt[6] {(x+5)^3} + \sqrt[6] {\frac {x^2} {4^2}} & \text{if } x \geq 0 \\
y = \sqrt[6] {(x+5)^3} - \sqrt[6] {\frac {x^2} {4^2}} & \text{if } x < 0
\end{cases}

That's, I think, because $\sqrt {x+5} = \sqrt[6] {(x+5)^3}$ since they share the same domain, whereas $\sqrt[3] {\frac {x} {4}}$ is equivalent to $\sqrt[6] {\frac {x^2} {4^2}}$ when the radicand is positive (so $x \geq 0$), and equivalent to $-\sqrt[6] {\frac {x^2} {4^2}}$ when the radicand is negative ($x<0$).

Assuming all of this is correct, I then proceeded solving the inequalities. Again, not sure if what I did next is legitimate or not. I thought the solutions would be the union of two systems:

System A:
1. $\sqrt[6] {(x+5)^3} + \sqrt[6] {\frac {x^2} {4^2}} \geq 0$
2. $x \geq 0$
3. $x \geq -5$ (Domain)

System B:
1. $\sqrt[6] {(x+5)^3} + \sqrt[6] {\frac {x^2} {4^2}} \geq 0$
2. $x<0$
3. $x \geq -5$ (Domain)

To solve them, I did (System A):
1. $\sqrt[6] {(x+5)^3} \geq - \sqrt[6] {\frac {x^2} {4^2}} 0$
-> elevated both terms to the power6
-> $x^3+125+15x^2+75x \geq \frac {x^2} {16}$
-> multiply by 16
-> $16x^3+2000+240x^2+1200x \geq x^2$
-> solve the "equation" (Ruffini's method), which then becomes (x=-4 is a solution)
-> $(x+4)(16x^2+175x+500) \geq 0$

The second piece ($16x^2+175x+500$) has Δ < 0, so it's always positive, so the solution for the inequality is: $x \geq -4$

Since this is the first system, the final solution of System A is then $x \geq 0$

For System B, the first part of the solution is the same ($x \geq -4$), and the final solution of System B is then $-4 \leq x \lt 0$

Final overall solution is then the union of the two, and therefore $x \geq -4$.

I confronted this solution with a graph drawer, and it seems to be correct.

So, did I do the passages correctly or did I just fluke the right solution?
And, is there a better/easier way to come to this solution? The numbers in this one were pretty simple, and that x = -4 solution was God-sent, but I bet a lot of other functions will have much less appealing numbers to work with, if this is the procedure to follow.

Last edited: Nov 1, 2017
2. Nov 1, 2017

### Ray Vickson

Trying to put all terms under the "same index" is a waste of time, and might even produce incorrect solutions. Just determine the domain of each term separately, then take their intersection to find the overall domain.

Also, both terms are strictly increasing in $x$, so if there is a point $x=x_0$ where $y=0$, then we have $y < 0$ for $x < x_0$ and $y > 0$ for $x > x_0$.

You have arrived at a correct solution, but I think you are just lucky in this case: your process of putting everything under a common index is suspect, and can sometimes go disatrously wrong.