# Calculate x in 3/(x^2-6x+8)-6/(x^2-16)<=0 ?

Apr 17, 2017

See below.

#### Explanation:

If $\frac{3}{{x}^{2} - 6 x + 8} \le \frac{6}{{x}^{2} - 16}$ then

$\frac{6}{{x}^{2} - 16} - \frac{3}{{x}^{2} - 6 x + 8} \ge 0$ or

$\frac{6}{\left(x - 4\right) \left(x + 4\right)} - \frac{3}{\left(x - 2\right) \left(x - 4\right)} \ge 0$ or

$\frac{1}{x - 4} \left(\frac{6}{x + 4} - \frac{3}{x - 2}\right) \ge 0$ or

(x-4)(3(x-8))/((x+4)(x-2))) ge 0

Now we have the sign dependence

$\left(- \infty\right) \left(\text{+++")(-4)("----")(2)("+++")(4)("----")(8)("++++}\right) \left(\infty\right)$

Then the inequality is true for

$\left(- \infty , - 4\right] \cup \left(2 , 4\right) \cup \left[8 , \infty\right)$

Apr 17, 2017

$\left(- \infty , - 4\right) \cup \left(2 , 4\right) \cup \left[8 , + \infty\right)$

#### Explanation:

$\text{Express the rational functions as a single rational function}$

$\frac{3}{{x}^{2} - 6 x + 8} - \frac{6}{{x}^{2} - 16} \le 0$

$\text{factorise the denominators}$

$\frac{3}{\left(x - 2\right) \left(x - 4\right)} - \frac{6}{\left(x - 4\right) \left(x + 4\right)} \le 0$

$\Rightarrow \frac{3 \left(x + 4\right)}{\left(x - 2\right) \left(x - 4\right) \left(x + 4\right)} - \frac{6 \left(x - 2\right)}{\left(x - 2\right) \left(x - 4\right) \left(x + 4\right)} \le 0$

$\Rightarrow \frac{- 3 \left(x - 8\right)}{\left(x - 2\right) \left(x - 4\right) \left(x + 4\right)} \le 0$

$\text{the zero's of the numerator/denominator are}$

$\text{ numerator " x=8, "denominator } x = 2 , x = 4 , x = - 4$

These indicate where the rational function may change sign and which values x cannot be on the denominator.

$\text{the intervals for consideration are}$

$x < - 4 , \textcolor{w h i t e}{x} - 4 < x < 2 , \textcolor{w h i t e}{x} 2 < x < 4 , 4 < x \le 8 , x > 8$

$\text{ Consider " color(blue)" a test point " " in each interval}$

We want to find where the function is negative, that is < 0

Substitute each test point into the function and consider it's sign.

$\textcolor{red}{x = - 6} \to \frac{+}{-} \to \textcolor{b l u e}{\text{ negative}}$

$\textcolor{red}{x = - 2} \to \frac{+}{+} \to \textcolor{red}{\text{ positive}}$

$\textcolor{red}{x = 3} \to \frac{+}{-} \to \textcolor{b l u e}{\text{ negative}}$

$\textcolor{red}{x = 6} \to \frac{+}{+} \to \textcolor{red}{\text{ positive}}$

$\textcolor{red}{x = 8} \to \frac{0}{+} \to \textcolor{m a \ge n t a}{\text{ x-intercept}}$

$\textcolor{red}{x = 10} \to \frac{-}{+} \to \textcolor{b l u e}{\text{ negative}}$

$\Rightarrow \left(- \infty , - 4\right) \cup \left(2 , 4\right) \cup \left[8 , + \infty\right)$
graph{(-3(x-8))/((x-2)(x-4)(x+4)) [-10, 10, -5, 5]}