Help required in this inequality:
If $$\frac {x^2-|x|-12}{x-3} \ge 0 $$ prove that $x \in [-4,3) \cup [4,\infty) \quad $.
Problem I'm facing:
I know that $x^2$ can be made $|x|^2$, but what about the $x $ in denominator? If that can be changed to $|x|$, then I can easily solve by wavy curve method.
N.B.: I'm a beginner in pure maths and will perhaps remain so for the rest of my life. So, please explain simply and don't complicate things.
On
Hint: write it as $\;\cfrac{(\,|x|-4\,)(\,|x|+3\,)}{x-3} \ge 0\,$ and check the signs of each factor. For example, $\,|x|+3 \gt 0\,$ for $\forall x \in \mathbb{R}$ which leaves only two factors to consider.
$|x|-4 \ge 0$ and $x-3 \gt 0\,$, or
$|x|-4 \le 0$ and $x-3 \lt 0\,$
The first case, for example, implies $|x| \ge 4$ and $x \gt 3\,$, giving:
$$x \in \big( (-\infty,-4] \cup [4,\infty) \big) \;\cap\; (3,\infty) \,=\, [4,\infty)\,$$
The second case can be worked out similarly for $|x| \le 4$ and $x \lt 3\,$, giving:
$$x \in [-4,4] \;\cap\; (-\infty, 3) \,=\, [-4,3)\,$$
Putting the two together, the final answer is $\, [-4,3) \,\cup\,[4,\infty)\,$.
On
Solve the equation on three intervals:
On
Taking hint from @Dynamo, I did it in this way:
For finding critical points:
Factorising, numerator = $(|x|-4)(|x|+3) $.
Now, $|x|+3=0$ means $|x| =-3, which is meaningless and hence discarded.
$|x|-4 $ has two critical points for $x $, viz. 4 & -4, both closed. Denominator has critical point 3, open.
Now I've put it on the curve and solved. Thanks @dynamo.
HINT:
For Wavy Curve method always try to find zeroes of the equation(numerator and denominator separately) and then plot them on number lines.
from numerator, the zeroes are $4$ and $-4$. and from denominator $3$.
You cant take $3$ as the function will not be defined on $3$.
Now construct a number line and check for what values of $x$ function is positive.That would be the desired result.