Solve the following $\frac{3x}{x+6} \ge 0 $

Question

Solve $$\frac{3x}{x+6} \geq 0 $$

My work

$$(x+6) / 3x <0 $$

$$1/3 + 6/x <0 $$

$$ 6/x <-1/3 $$

$$ x >-18 $$

is that correct

Answer 1

If $\dfrac{3x}{x+6}=0, x=0$

Else $\dfrac{3x}{x+6}>0\iff x(x+6)>0$

Now $(x-a)(x-b)>0$ where $a<b$

we can prove that we need either $x<a$ or $x>b$

Answer 2

You have three errors. The first and largest is going from $\frac {3x}{x+6} \ge 0$ to $(x+6)/3x \lt 0$. When you invert a number its sign stays the same, so it should be $(x+6)/(3x) \ge 0$. Second, you should also have parentheses around the $3x$ as I do-as you wrote it the $x$ could be in the numerator. This does not affect the rest because you are consistent. Third, when you went from $(x+6)/3x \lt 0$ to $1/3 + 6/x \lt 0$ you dropped a factor $3$ in the second term. It should be $1/3 + 6/(3x) \lt 0$ or $1/3+2/x \lt 0$

Answer 3

The equality holds only when $x=0$.

$$\frac{3x}{x+6}\cdot\frac{x+6}{x+6} >0 \iff \frac{3x\cdot (x+6)}{(x+6)^2}>0$$

Since $3>0 $ and $(x+6)^2$ are positive for any $x$ , $$x(x+6)>0$$

So we need to find $x$ such that $x(x+6)$ is positive.

What are the cases ?

$1)$ If $x>0$ , then $(x+6)$ must be positive.That is $x>0$ and $x>-6$

$2)$ If $x<0$ , then $(x+6)$ must be negative.That is $x<0$ and $x<-6$

By $1)$ We have that $x>0$.

By $2)$ We have that $x<-6$.

Answer 4

The rational function $f(x)=\frac{3x}{x+6}$ is continuous over its domain. The numerator vanishes only at $x=0$, the denominator only at $x=-6$, hence we have three intervals to study: $I_1=(-\infty,-6)$, $I_2=(-6,0)$ and $I_3=(0,+\infty)$. Pick a point in $I_k$ (any point!), evaluate $f$ at that point and you will have the sign of $f$ over $I_k$. In your case $f\geq 0$ over $\mathbb{R}\setminus[-6,0)$.

Answer 5

Let $\ast\in\{\leq,\geq,<,>\}$. Generally, $$ \frac{f(x)}{g(x)}\ast 0 \Longleftrightarrow f(x)g(x)\ast 0\ \text{and}\ g(x)\neq 0 $$ To prove this euivalent multiply by $f(x)^2$ which is non-negative and won't change ''$\ast$"

In your example $$ \frac{3x}{x+6}\geq 0 \Longleftrightarrow 3x(x+6)\geq 0\ \text{and}\ x+6\neq 0 \Longleftrightarrow (x\geq 0\ \text{or}\ x\leq -6)\ \text{and}\ x\neq -6 $$ that is $$ x\geq 0\ \text{or}\ x<-6 $$