(adsbygoogle = window.adsbygoogle || []).push({}); 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 power^{6}

-> ##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.

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# Homework Help: Sign of a function

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