Absolute Value Inequalities - 1 Quadratic Term

Absolute value inequalities are an invaluable tool in the practice of STEM disciplines. Notably, the $\varepsilon - \delta$ definition of a limit makes use of inequalities of this form.

NOTE: Produce Python images to display domains.

Introduction - General

Recall the definition of absolute value:

For any real valued function $f(x)$ , its absolute value is defined as

$|f(x)| = \begin{cases} f(x) & \textrm{if } f(x) > 0 \\ 0 & \textrm{if } f(x) = 0 \\ -[f(x)] & \textrm{if } f(x) < 0. \\ \end{cases}$

We seek to solve inequalities of the form

$| \alpha_1 x^2 + \beta_1 x + \gamma_1 | > f(x)$
$| \alpha_1 x^2 + \beta_1 x + \gamma_1 | <f(x)$
$| \alpha_1 x^2 + \beta_1 x + \gamma_1 | \leq |f(x)|$
$| \alpha_1 x^2 + \beta_1 x + \gamma_1 | \geq |f(x)|$

such that coefficients $( \alpha_1, \beta_1, \gamma_2)$ exist in the reals and $( \alpha_1) \neq (0)$ . In order to process such a collection of coefficients, I suggest one develops in their minds eye a coordinate systems containing the coefficients $( \alpha_1, \beta_1, \gamma_1) .$ Notice that when one considers points in either space to be in the form $( 0, \beta_1, \gamma_1)$ , you are effectively considering inequalities in the form addressed in absolute value inequalities.

NOTE: Insert python generated GIF of random continuous walk through $( \alpha_1, \beta_1, \gamma_1)$ and $( \alpha_2, \beta_2, \gamma_2)$ and resulting $\big(x, f_1(x, \alpha_1, \beta_1, \gamma_1)\big)$ and $\big(x, f_2(x, \alpha_2, \beta_2, \gamma_2)\big).$

Problem Solving Method:

Solving equations of this type require the problem solver to conduct minimally four steps:

Enact the action of the absolute value function. That being a transformation of the function(s) being acted on into a piecewise function(s) likely containing $\big(-[f(x)], 0, f(x)\big)$ . Extract the domains of each term and each element of the piecewise function in the equation.
Consider the intersection of domains of included functions and construct the standard inequality for each case.
Evaluate the inequality over each intersection and extract the domain for which the inequality holds true on that intersection.
Reconstitute the domains on which the inequality holds true.

Problem Solving 1: $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | > C$

In solving inequalities containing absolute value functions, we must first enact the action of the absolute value function on the composed function. Only after this process can we address the question of the inequality. Provided $|f(x)|$ , we must analyze $f(x)$ under the three conditions

$\begin{aligned} (f(x) < 0) &\implies -[f(x)]\\ (f(x) = 0) &\implies 0\\ (f(x) > 0) &\implies f(x),\end{aligned}$

and extract the domain for each case.

Given the inequality $|x^2 - 1| < 1$ , extract the absolute value function, separate it into the three cases described in the definition of the absolute value function, and state the domain for each case.

Case 1. $(f(x) < 0) \implies -[f(x)]$
We have $\begin{aligned} x^2 - 1 &< 0\\ x^2 &< 1\\ x &< \pm 1. \end{aligned}$ Therefore, $|f(x)| = -[f(x)]$ for $-1 < x < 1$ .

Case 2. $(f(x) = 0) \implies 0$
We have $\begin{aligned} x^2 - 1 &= 0\\ x^2 &= 1\\ x &= \pm 1. \end{aligned}$ Therefore, $|f(x)| = 0$ for $x = 1 \cup x = -1$ .

Case 3. $(f(x) > 0) \implies f(x)$
We have $\begin{aligned} x^2 - 1 &> 0\\ x^2 &> 1\\ x &> \pm 1. \end{aligned}$ Therefore, $|f(x)| = f(x)$ for $x > 1 \cup x < -1$ . $_\square$

Now that we've discovered the domain of each component of our piecewise function, we are now in a position to address the question posed by the equation.

Let us collect the sub-domains of $|f_{1}(x)|$ and $f_{2}(x)$ into sets $D_{1,n}$ and $D_{2,n},$ respectively:

$\begin{aligned} D_{1,n} &= { \big((-1 < x < 1)\big), \big((x = 1) \cup (x = -1)\big), \big((x > 1) \cup (x < -1)\big)} \\ &= { D_{1,1}, D_{1,2}, D_{1,3} } \\\\ D_{2,n} &= { (-\infty < x < \infty) } \\&= { D_{2,1} }. \end{aligned}$

Remember that each domain corresponds to an element of each piecewise function. We now must check for intersections of each domain and construct and solve the corresponding inequality:

Case 1. $D_{1,1} \cap D_{2,1} = (-1 < x < 1) \implies -[f_{1}(x)] < f_{2}(x)$
Case 2. $D_{1,2} \cap D_{2,1} = \big((x = 1) \cup (x = -1)\big) \implies 0 < f_{2}(x)$
Case 3. $D_{1,3} \cap D_{2,1} = \big((x > 1) \cup (x < -1)\big) \implies f_{1}(x) < f_{2}(x)$ .

Let us now address the question of the inequality $|x^2 - 1| < 1:$

Case 1. $(-1 < x < 1)$ : $1 - x^2 < 1 \implies \big(( 1 > x > 0) \cup (-1 < x < 0)\big)$
Case 2. $\big((x = 1) \cup (x = -1)\big)$ : $0 < 1 \implies \big((x = 1) \cup (x = -1)\big)$
Case 3. $\big((x > 1) \cup (x < -1)\big)$ : $x^2 - 1 < 1 \implies \Big(\big(1 < x < \sqrt{2}\big) \cup \big(-1 > x > -\sqrt{2}\big)\Big)$ .

Finally, we can union the domains on which the inequality holds:

$\begin{aligned} \Big(\big(1 < x < \sqrt{2}\big) \cup \big(-1 > x > -\sqrt{2}\big)\Big) &\cup \big(( 1 > x > 0) \cup (-1 < x < 0)\big) \cup \big((x = 1) \cup (x = -1)\big) \\\\ \Rightarrow \big( -\sqrt{2} < x < 0\big) &\cup \big( 0 < x < \sqrt{2}\big). \end{aligned}$

Problem Solving 2: $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | < f(x)$

In the following example, we will confront an absolute value inequality of the form $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | < f(x)$ . As we have done previously, we shall first enact the action of the absolute value function.

Given the inequality $|x^2 - 7x + 10| < 3x - 6$ , extract the absolute value function, separate it into the three cases described in the definition of the absolute value function, and state the domain for each case.

Case 1. $(f(x) < 0) \implies -[f(x)]$
We have $\begin{aligned} x^2 - 7x + 10 &< 0\\ (x-5)(x-2) &< 0\\ 2 &< x < 5. \end{aligned}$ Therefore, $|f(x)| = -[f(x)]$ for $2 < x < 5$ .

Case 2. $(f(x) = 0) \implies 0$
We have $\begin{aligned} x^2 - 7x + 10 &= 0\\ (x-5)(x-2) &= 0\\ x &= 2 \cup x = 5. \end{aligned}$ Therefore, $|f(x)| = 0$ for $x = 2 \cup x = 5$ .

Case 3. $(f(x) > 0) \implies f(x)$
We have $\begin{aligned} x^2 - 7x + 10 &> 0\\ (x-5)(x-2) &> 0\\ x &< 2 \cup x > 5. \end{aligned}$ Therefore, $|f(x)| = f(x)$ for $x < 2 \cup x > 5$ . $_\square$

Now that we've discovered the domain of each component of our piecewise function, we are now in a position to address the question posed by the equation.

Let us collect the sub-domains of $|f_{1}(x)|$ and $|f_{2}(x)|$ into sets $D_{1,n}$ and $D_{2,n},$ respectively:

$\begin{aligned} D_{1,n} &= { (2 < x < 5) , (x = 2 \cup x = 5), (x < 2 \cup x > 5)} \\&= { D_{1,1}, D_{1,2}, D_{1,3} } \\\\ D_{2,n} &= { (-\infty < x < \infty) } \\&= { D_{2,1} }. \end{aligned}$

Remember that each domain corresponds to an element of each piecewise function. We now must check for intersections of each domain and construct and solve the corresponding inequality:

Case 1. $D_{1,1} \cap D_{2,1} = (2 < x < 5) \implies -[f_{1}(x)] < f_{2}(x)$
Case 2. $D_{1,2} \cap D_{2,1} = \big((x = 2) \cup (x = 5)\big) \implies 0 < f_{2}(x)$
Case 3. $D_{1,3} \cap D_{2,1} = \big((x < 2) \cup (x > 5)\big) \implies f_{1}(x) < f_{2}(x)$ .

Let us now address the question of the inequality $|x^2 - 7x + 10| < 3x - 6:$

Case 1. $(2 < x < 5)$ : $-[x^2 - 7x + 10] < 3x - 6 \implies (2 < x < 5)$
Case 2. $\big((x = 2) \cup (x = 5)\big)$ : $0 < 3x + 6 \implies (x = 5)$
Case 3. $\big((x < 2) \cup (x > 5)\big)$ : $x^2 - 7x + 10 < 3x - 6 \implies (5 < x < 8)$ .

Finally, we can union the domains on which the inequality holds:

$\begin{aligned} (2 < x < 5) &\cup (x = 5) \cup (5 < x < 8) \\\\ \Rightarrow (2 &< x < 8). \end{aligned}$

Problem Solving 3: $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | \geq |f(x)|$

In the following example, we will confront an absolute value inequality of the form $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | \geq |f(x)|,$ or most generally, they may be called $|f_{1}(x)| \geq |f_{2}(x)|$ . There exists two absolute value functions in this inequality, and both must be addressed.

Given the inequality $|9 - x^2| \geq |x|$ , extract the absolute value functions, separate them into the three cases described in the definition of the absolute value function, and state the domain for each case.

Before we continue, let us call $9 - x^2 = f_{1}(x)$ and $x = f_{2}(x)$ .

Consider $f_{1}(x)$ :

Case 1. $f_{1}(x) < 0) \implies -[f(x)]$
We have $\begin{aligned} 9 - x^2 &< 0\\ x^2 &> 9\\ x &> \pm 3. \end{aligned}$ Therefore, $|f_{1}(x)| = -[f_{1}(x)]$ for $x < -3 \cup x > 3$ .

Case 2. $(f_{1}(x) = 0) \implies 0$
We have $\begin{aligned} 9 - x^2 &= 0\\ x^2 &= 9\\ x &= \pm 3. \end{aligned}$ Therefore, $|f(x)| = 0$ for $x = -3 \cup x = 3$ .

Case 3. $(f_{1}(x) > 0) \implies f(x)$
We have $\begin{aligned} 9 - x^2 &> 0\\ x^2 &< 9\\ x &< \pm 3. \end{aligned}$ Therefore, $|f(x)| = f(x)$ for $-3 < x < 3$ .

Consider $f_{2}(x)$ :

Case 1. $(f_{2}(x) < 0) \implies -[f(x)]$
We have $x < 0.$ Therefore, $|f_{2}(x)| = -[f_{2}(x)]$ for $x < 0$ .

Case 2. $(f_{2}(x) = 0) \implies 0$
We have $x = 0.$ Therefore, $|f(x)| = 0$ for $x = 0$ .

Case 3. $(f_{2}(x) > 0) \implies f(x)$
We have $x > 0.$ Therefore, $|f(x)| = f(x)$ for $x > 0$ . $_\square$

As opposed to the absolute value functions we've dealt with previously, we are now left with six sub domains: three from $|f_{1}(x)|$ and three from $|f_{2}(x)|$ . We must now work to find the intersections of those sub-domains so that we may address the question of the inequality.

Let us collect the sub-domains of $|f_{1}(x)$ and $|f_{2}(x)$ into sets $D_{1,n}$ and $D_{2,n}$ respectively:

$\begin{aligned} D_{1,n} &= { (x < -3 \cup x > 3) , (x = -3 \cup x = 3), (-3 < x < 3)} \\ &= { D_{1,1}, D_{1,2}, D_{1,3} } \end{aligned}$

Test Image

and

$\begin{aligned} D_{2,n} &= { (x < 0) , (x = 0), (x < 0)} \\ &= { D_{2,1}, D_{2,2}, D_{2,3} }. \end{aligned}$

Remember that each domain corresponds to an element of each piecewise function. We now must check for intersections of each domain and construct and solve the corresponding inequality:

Case 1. $D_{1,1} \cap D_{2,1} = (x < -3) \implies -[f_{1}(x)] \geq -[f_{2}(x)]$
Case 2. $D_{1,1} \cap D_{2,2} = \varnothing \implies$ no intersection
Case 3. $D_{1,1} \cap D_{2,3} = (x > 3) \implies -[f_{1}(x)] \geq f_{2}(x)$
Case 4. $D_{1,2} \cap D_{2,1} = (x = -3) \implies 0 \geq -[f_{2}(x)]$
Case 5. $D_{1,2} \cap D_{2,2} = \varnothing \implies$ no intersection
Case 6. $D_{1,2} \cap D_{2,3} = (x = 3) \implies 0 \geq f_{2}(x)$
Case 7. $D_{1,3} \cap D_{2,1} = (-3 < x < 0) \implies f_{1}(x) \geq -[f_{2}(x)]$
Case 8. $D_{1,3} \cap D_{2,2} = (x = 0) \implies f_{1}(x) \geq 0$
Case 9. $D_{1,3} \cap D_{2,3} = (0 < x < 3) \implies f_{1}(x) \geq f_{2}(x)$ .

Let us now address the question of the inequality: $|9 - x^2| \geq |x|:$

Case 1. $(x < -3)$ : $x^2 - 9 \geq -x \implies \left(x \geq \frac{-1 \pm \sqrt{37}}{2}\right)$
Case 2. no intersection
Case 3. $(x > 3)$ : $x^2 - 9 \geq x \implies \left(x \geq \frac{1 \pm \sqrt{37}}{2}\right)$
Case 4. $(x = -3)$ : $0 \geq -x \implies$ equality unsatisfied
Case 5. no intersection
Case 6. $(x = 3)$ : $0 \geq x \implies$ equality unsatisfied
Case 7. $(-3 < x < 0)$ : $9 - x^2 \geq -x \implies \left(x \geq \frac{1 \pm \sqrt{37}}{2}\right)$
Case 8. $(x = 0)$ : $9 - x^2 \geq 0 \implies (9 \geq 0)$
Case 9. $(0 < x < 3)$ : $9 - x^2 \geq x \implies \left(x \leq \frac{-1 \pm \sqrt{37}}{2}\right)$ .

Finally, we can union the domains on which the inequality holds:

$\begin{aligned} \left(x \leq \frac{-1 - \sqrt{37}}{2}\right) &\cup \left(x \geq \frac{1 + \sqrt{37}}{2}\right) \cup \left(0 > x \geq \frac{1 - \sqrt{37}}{2}\right) \cup (x = 0) \cup \left(0 < x \leq \frac{-1 + \sqrt{37}}{2}\right)\\\\ \Rightarrow \left(x \leq \frac{-1 - \sqrt{37}}{2}\right) &\cup \left(x \geq \frac{1 + \sqrt{37}}{2}\right) \cup \left( \frac{1 - \sqrt{37}}{2} \leq x \leq \frac{-1 + \sqrt{37}}{2}\right). \end{aligned}$

Problem Solving 4: $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | < f(x)$

In the following example, we will confront an absolute value inequality of the form $| \alpha_1 x^2 + \beta_1 x + \gamma_1 | < f(x)$ . As we have done previously, we shall first enact the action of the absolute value function.

Given the inequality $\frac{1}{|-x^2 + 7|} \leq \frac{2} {x+10}$ , extract the absolute value function, separate it into the three cases described in the definition of the absolute value function, and state the domain for each case.

Note that $\frac{1}{|-x^2 + 7|} \leq \frac{2} {x+10} \implies |-x^2 + 7| \geq \frac{x+10} {2} \implies |f_{1}(x)| \geq f_{2}(x)$ . Then we have the following 3 cases:

Case 1. $(f_{1}(x) < 0) \implies -[f(x)]$
We have $\begin{aligned} 7 - x^2 &< 0\\ -x^2 &< -7\\ x &> \pm \sqrt{7}. \end{aligned}$ Therefore, $|f(x)| = -[f(x)]$ for $\Big(\big(x > \sqrt{7}\big) \cup \big(x < -\sqrt{7}\big)\Big)$ .

Case 2. $(f(x) = 0) \implies 0$
We have $\begin{aligned} 7 - x^2 &= 0\\ -x^2 &= -7\\ x &= \pm \sqrt{7}. \end{aligned}$ Therefore, $|f(x)| = 0$ for $\Big(\big(x = \sqrt{7}\big) \cup \big(x = -\sqrt{7}\big)\Big)$ .

Case 3. $(f(x) > 0) \implies f(x)$
We have $\begin{aligned} 7 - x^2 &> 0\\ -x^2 &> -7\\ x &< \pm \sqrt{7}. \end{aligned}$ Therefore, $|f(x)| = f(x)$ for $\big(-\sqrt{7} < x < \sqrt{7}\big)$ . $_\square$

Now that we've discovered the domain of each component of our piecewise function, we are now in a position to address the question posed by the equation.

Let us collect the sub-domains of $|f_{1}(x)|$ and $f_{2}(x)$ into sets $D_{1,n}$ and $D_{2,n},$ respectively:

$\begin{aligned} D_{1,n} &= { ((x > \sqrt{7}) \cup (x < -\sqrt{7})) , ((x = \sqrt{7}) \cup (x = -\sqrt{7})), (-\sqrt{7} < x < \sqrt{7})} \\ &= { D_{1,1}, D_{1,2}, D_{1,3} } \\\\ D_{2,n} &= { (-\infty < x < \infty) } \\ &= { D_{2,1} }. \end{aligned}$

Remember that each domain corresponds to an element of each piecewise function. We now must check for intersections of each domain and construct and solve the corresponding inequality:

Case 1. $D_{1,1} \cap D_{2,1} = \Big(\big(x > \sqrt{7}\big) \cup \big(x < -\sqrt{7}\big)\Big) \implies -[f_{1}(x)] \geq f_{2}(x)$
Case 2. $D_{1,2} \cap D_{2,1} = \Big(\big(x = \sqrt{7}\big) \cup \big(x = -\sqrt{7}\big)\Big) \implies 0 \geq f_{2}(x)$
Case 3. $D_{1,3} \cap D_{2,1} = \big(-\sqrt{7} < x < \sqrt{7}\big) \implies f_{1}(x) < f_{2}(x)$ .

Let us now address the question of the inequality: $|x^2 - 1| < 1:$

Case 1. $\Big(\big(x > \sqrt{7}\big) \cup \big(x < -\sqrt{7}\big)\Big)$ : $x^2 - 7 \geq \frac{x+10} {2} \implies \left(x \leq \frac{1 - \sqrt{193}}{4}\right) \cup \left(x \geq \frac{1 + \sqrt{193}}{4}\right)$
Case 2. $\Big(\big(x = \sqrt{7}\big) \cup \big(x = -\sqrt{7}\big)\Big)$ : $0 \geq \frac{x+10} {2} \implies$ unsatisfied on given domain
Case 3. $\big(-\sqrt{7} < x < \sqrt{7}\big)$ : $7 - x^2 \geq \frac{x+10} {2} \implies \left(\frac{-1 - \sqrt{33}}{4} \leq x \leq \frac{-1 + \sqrt{33}}{4}\right)$ .

Finally, we can union the domains on which the inequality holds:

$\left(x \leq \frac{1 - \sqrt{193}}{4}\right) \cup \left(x \geq \frac{1 + \sqrt{193}}{4}\right) \cup \left(\frac{-1 - \sqrt{33}}{4} \leq x \leq \frac{-1 + \sqrt{33}}{4}\right).$