Absolute Value Functions - Revision history

Khanh at 04:39, 14 November 2021

2021-11-14T04:39:08Z

Lila at 17:33, 18 October 2021

2021-10-18T17:33:21Z

Lila: /* Absolute Value Equations */

2021-10-18T17:29:02Z

Absolute Value Equations

Lila: /* Inequalities with Absolute Values */

2021-10-18T17:24:15Z

Inequalities with Absolute Values

Lila: /* Domain and Range */

2021-10-18T17:20:08Z

Domain and Range

Lila: /* Absolute Value Equations */

2021-10-18T17:13:30Z

Absolute Value Equations

Show changes

Lila: /* Even or odd? */

2021-10-18T17:06:27Z

Even or odd?

Lila: /* Graphing Absolute Value Functions */

2021-10-18T17:04:20Z

Graphing Absolute Value Functions

Lila: /* Resources */

2021-10-18T16:59:55Z

Resources

Lila: /* Properties of the Absolute Value Function */

2021-10-18T16:57:23Z

Properties of the Absolute Value Function

@@ Line 587: / Line 587: @@
 ==Resources==
 * [https://math.libretexts.org/Bookshelves/Precalculus/Precalculus_(OpenStax)/01%3A_Functions/1.06%3A_Absolute_Value_Functions Absolute Value Functions], LibreTexts
 * [https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:absolute-value-piecewise-functions/x2f8bb11595b61c86:graphs-of-absolute-value-functions/a/absolute-value-equations-and-graphs-review Absolute Value Graphs Review], Khan Academy
 * [https://www.youtube.com/watch?v=sCcsnuwihEw How To Solve Absolute Value Equations], The Organic Chemistry Tutor
 * [https://www.youtube.com/watch?v=Tj_tQxtI3gs Solving an Equation with Two Absolute Value Expressions], patrickJMT

@@ Line 358: / Line 358: @@
 :<math>|3x-3|=2x-7</math>
 Because the absolute value is isolated, we can begin with our generalized procedure. Assuming <math>2x-7>0</math>, we may begin by denoting these two equations:
-{{Equation|<math>3x-3=2x-7</math>|1|center}}
+: <math>3x-3=2x-7</math>
-{{Equation|<math>3x-3=-(2x-7)</math>|2|center}}
+: <math>3x-3=-(2x-7)</math>
 These are only true if <math>2x-7>0</math>. For now, <u>assume</u> this condition is true. Let us solve for <math>x</math> with each respective equation:
@@ Line 371: / Line 371: @@
 :<math>x=2</math>
 We have two <u>potential</u> solutions to the equation. Try to answer why we said potential here based on what you know so far about this problem.
-{{Question-answer|question=Why did we state we had two <u>potential</u> solutions?|answer=Because we had to <u>assume</u> that <math>2x-7>0</math> and <math>|3x-3|=2x-7</math> is true for the provided <math>x</math>.}}
 Because of this, we have to verify the solutions to this equation exist. Therefore, let us substitute those values into the equation:
 :<math>|3(-4)-3|=2(-4)-7</math>. Notice that the right-hand side is negative. Also, the left-hand side and the right-hand side are not equivalent. Therefore, this is not a solution.
@@ Line 384: / Line 386: @@
 :<math>\left\vert5a+\frac{1}{2}\right\vert=|9a+9|</math>
 Looking at the second equation might be the first declaration of absurdity. However, an application of the fundamental properties of absolute values is enough to do this problem.
-{{Equation|<math>5a+\frac{1}{2}=|9a+9|</math>|3|center}}
+: <math>5a+\frac{1}{2}=|9a+9|</math>
-{{Equation|<math>5a+\frac{1}{2}=-|9a+9|</math>|4|center}}
+: <math>5a+\frac{1}{2}=-|9a+9|</math>
 Peel the problem one layer at a time. For this one, we will categorize equations based on where they come from; this should hopefully explain the dashes: 3-1 is first equation formulated from <math>5a+\frac{1}{2}=|9a+9|</math>, for example.
-{{Equation|<math>9a+9=5a+\frac{1}{2}</math>|3-1|center}}
+: <math>9a+9=5a+\frac{1}{2}</math>|3-1|center
-{{Equation|<math>9a+9=-\left(5a+\frac{1}{2}\right)</math>|3-2|center}}
+: <math>9a+9=-\left(5a+\frac{1}{2}\right)</math>
-{{Equation|<math>-(9a+9)=5a+\frac{1}{2}</math>|4-1|center}}
+: <math>-(9a+9)=5a+\frac{1}{2}</math>
-{{Equation|<math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math>|4-2|center}}
+: <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math>
 We can demonstrate that some equations are equivalents of the other. For example, <math>9a+9=5a+\frac{1}{2}</math> and <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> are equivalent, since dividing both sides of <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> by <math>-1</math> gives <math>9a+9=5a+\frac{1}{2}</math>. Further, <math>9a+9=-\left(5a+\frac{1}{2}\right)</math> and <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> are equivalent (multiply both sides of equation <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> by <math>-1</math>). After determining all the equations that are equivalent, distribute <math>-1</math> to the corresponding parentheses.
-{{Equation|<math>9a+9=5a+\frac{1}{2}</math>|5|center}}
+: <math>9a+9=5a+\frac{1}{2}</math>
-{{Equation|<math>9a+9=-5a-\frac{1}{2}</math>|6|center}}
+: <math>9a+9=-5a-\frac{1}{2}</math>
 Now all that is left to do is solve the equations. We will leave this step as an exercise for the reader. There are two potential solutions: <math>a=-\frac{19}{28},-\frac{17}{8}</math>. All that is left to do is verify that the equation in the question is true when looking at these specific values of <math>a</math>:
 :<math>a=-\frac{19}{28}</math>
@@ Line 469: / Line 472: @@
 The next problem typically requires some trigonometry to solve easily. However, with one extra piece of information, one can use the properties of the absolute value property to solve the following problem.
-Example 2.0(h): Tiling a Roof (adapted from [[Trigonometry/Worked Example: Area of a Roof|Trigonometry Book 1]])
+Example 2.0(h): Tiling a Roof
 [[File:Roof Plan for Trigonometry.svg|thumb|'''Figure 2''': The plan for a roof is given in the image above. We want to find the area of the figure using only what is given with absolutely no trigonometry.|290x290px]]
 An engineer is planning to make an roof with a <math>\displaystyle 30</math> m. frame base and <math>\displaystyle 100</math> m. perimeter. The angle of the slope of the roof to the base is <math>\displaystyle\theta</math>. The sloped sides are congruent. A reference image ('''Figure 2''') of the sloped-roof (with no cartesian plane) is provided. Given the area of a triangle is <math>\displaystyle\frac{1}{2}bh</math>, and the distance formula is <math>\displaystyle d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}</math>, find the area of the triangular cross section of the roof.

@@ Line 362: / Line 362: @@
 These are only true if <math>2x-7>0</math>. For now, <u>assume</u> this condition is true. Let us solve for <math>x</math> with each respective equation:
-Equation {{Eqnref|1}}
+Equation <math>3x-3=2x-7</math>
 :<math>3x-3=2x-7\qquad\text{Add }3\text{ and subtract }2x\text{ on both sides.}</math>
 :<math>x=-4</math>
-Equation {{Eqnref|2}}
+Equation <math>3x-3=-(2x-7)</math>
 :<math>3x-3=-(2x-7)\qquad\text{Distribute }-1\text{.}</math>
 :<math>3x-3=-2x+7\qquad\quad\text{Add }3\text{ and add }2x\text{ on both sides.}</math>
@@ Line 386: / Line 386: @@
 {{Equation|<math>5a+\frac{1}{2}=|9a+9|</math>|3|center}}
 {{Equation|<math>5a+\frac{1}{2}=-|9a+9|</math>|4|center}}
-Peel the problem one layer at a time. For this one, we will categorize equations based on where they come from; this should hopefully explain the dashes: 3-1 is first equation formulated from {{Eqnref|3}}, for example.
+Peel the problem one layer at a time. For this one, we will categorize equations based on where they come from; this should hopefully explain the dashes: 3-1 is first equation formulated from <math>5a+\frac{1}{2}=|9a+9|</math>, for example.
 {{Equation|<math>9a+9=5a+\frac{1}{2}</math>|3-1|center}}
 {{Equation|<math>9a+9=-\left(5a+\frac{1}{2}\right)</math>|3-2|center}}
 {{Equation|<math>-(9a+9)=5a+\frac{1}{2}</math>|4-1|center}}
 {{Equation|<math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math>|4-2|center}}
-We can demonstrate that some equations are equivalents of the other. For example, {{Eqnref|3-1}} and {{Eqnref|4-2}} are equivalent, since dividing both sides of {{Eqnref|4-2}} by <math>-1</math> gives {{Eqnref|3-1}}. Further, {{Eqnref|3-2}} and {{Eqnref|4-2}} are equivalent (multiply both sides of equation {{Eqnref|4-2}} by <math>-1</math>). After determining all the equations that are equivalent, distribute <math>-1</math> to the corresponding parentheses.
+We can demonstrate that some equations are equivalents of the other. For example, <math>9a+9=5a+\frac{1}{2}</math> and <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> are equivalent, since dividing both sides of <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> by <math>-1</math> gives <math>9a+9=5a+\frac{1}{2}</math>. Further, <math>9a+9=-\left(5a+\frac{1}{2}\right)</math> and <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> are equivalent (multiply both sides of equation <math>-(9a+9)=-\left(5a+\frac{1}{2}\right)</math> by <math>-1</math>). After determining all the equations that are equivalent, distribute <math>-1</math> to the corresponding parentheses.
 {{Equation|<math>9a+9=5a+\frac{1}{2}</math>|5|center}}
 {{Equation|<math>9a+9=-5a-\frac{1}{2}</math>|6|center}}

@@ Line 525: / Line 525: @@
 == Inequalities with Absolute Values ==
 It is important to keep in mind that any function can be less than any other function. For example, <math>2x-5<54-13x</math> has any solutions for <math>x<\frac{59}{13}=3+\frac{14}{15}</math>. So long as the value for <math>x</math> is within that range, the function <math>2x-5</math> is less than the output of <math>54-13x</math>. The algebra for inequalities of <math>f(x)=|x|</math> requires a bit more of demonstration to understand. While the methods we use will not be proven, per se, our examples and explanations should give a good intuition behind the idea of find the inequalities of absolute values.
-{{ExampleRobox|theme=3|title='''Example 3.0(a)''': <math>|10-20x|<50</math>}}
 First, let us simplify the following expression through the method we demonstrated in the previous section (factoring the inside of the absolute value and bringing the constant out). Keep in mind, since we are switching the sides for which we view the equation, 50 is the left instead of right, we must also "flip" the inequality to be consistent with the original equation.
 :<math>\begin{align}
@@ Line 553: / Line 555: @@
 ::: <math>\Leftrightarrow x<3</math>
 We have found a possible distribution of values that allows the following equation to be true, where <math>|2x-5|<5</math>, and it is for values of <math>x</math> in between <math>-2</math> and <math>3</math>, non-inclusive.
-{{Robox/Close}}
 The above example is an intuition behind how solving for inequalities work. Technically speaking, we could make a proof for why we have to "operate" (take the steps seen above) absolute value inequalities this way. However, this will be a little too technical and involve a lot of generalization that could potentially confuse students rather than enlighten. If the student feels the challenge is worth, then one may try the proof of the steps we derived below. This is considered standard procedure (according to many High School textbooks).
 # Simplify until only the "absolute value bar term" is left.
@@ Line 559: / Line 561: @@
 # Rewrite <math>x</math> into necessary notatioon.
 Although the procedure may seem to be confusing, we are really only trying to make the algorithm as specific as possible. In reality, we will show just how easy it is to apply this algorithm for the problem above.
-{{ExampleRobox|theme=1|title='''Example 3.0(a) (REPEAT)''': <math>|10-20x|<50</math>}}
 Let us skip to the most simplified form.
 :<math>|1-2x|<5</math>
@@ Line 573: / Line 576: @@
 ::: <math>1-2x>-5</math>
 ::: <math>\Leftrightarrow x<3</math>
-{{Robox/Close}}
 There are two possible reasons why this procedure exists. For one, it allows us to quickly solve for <math>x</math> in the "right-hand" equation without the need for double the amount of multiplications necessary to solve for <math>x</math> (it lessens the amount of times we have to flip the inequality). Next, it allows us to focus more on the idea behind absolute value equations (the value inside will be positive, and hence, we want to find all values that allow us to find all possible solutions).
 Nevertheless, keep in mind how we found this procedure, and it was through applying the function definition of absolute values. In reality, we did the exact same thing for absolute value ''equations''. The only difference in application of algorithm applies to the inequality, which further "complicates" matters by introducing a new concept to the non-injective absolute value function. Through finding two solutions, we gave two possible ranges for values of <math>x</math>.
-Hopefully, this example should further shine a light into what many high schoolers think to be "black magic" among finding solutions to absolute value inequalities and equalities. The next examples should only hopefully further the concepts learned. Keep in mind, if one does not like the algorithm presented in the repeat example above, one is perfectly fine to use the other algorithm. The benefit of multiple choice is the ability to use any method, and only the correctness of your answer will be considered.
+Hopefully, this example should further shine a light into what many high schoolers think to be "black magic" among finding solutions to absolute value inequalities and equalities.
 ==Resources==

@@ Line 40: / Line 40: @@
 \end{cases}</math>.
 Because it can only be the case that <math>y=-x\text{ if }x<0</math> and <math>y=x\text{ if }x\ge0</math>, it is not possible for <math>|x|<0</math>. However, since <math>x</math> has no restriction, the domain, <math>A</math>, has no restriction. Thus, if <math>B</math> represents the range of the function, then  <math>A=\{x\in\mathbb{R}\}</math> and <math>B=\{y\ge0 | y\in\mathbb{R}\}</math>.
 '''Domain and Range:'''
 : Let <math>f(x)=|x|</math> whose mapping is <math>f:\mathbb{R}\to\mathbb{R}</math> represent the absolute value function. If <math>A</math> is the domain and <math>B</math> is the range, then <math>A=\{x\in\mathbb{R}\}</math> and <math>B=\{y\ge0 | y\in\mathbb{R}\}</math>.

@@ Line 60: / Line 60: @@
 :<math>\Rightarrow f(-x)=f(x)\blacksquare</math>
-Because <math>f(x)</math> is even, it is also the case that it is ''symmetrical''. A review of this can be found [[CLEP College Algebra/Graphing Functions|here (Graphs and Their Properties)]].
+Because <math>f(x)</math> is even, it is also the case that it is ''symmetrical''.
 ==== One-to-one and onto? ====

@@ Line 118: / Line 118: @@
 This subsection is absolutely not optional. You will be asked these questions very explicitly, so it is a good idea to understand this section. If you didn't read the previous subsection, you are not going to understand how any of this makes sense.
-Fortunately, the idea behind graphing any arbitrary function is mostly dependent on what you know about the function. Therefore, we can easily be able to graph functions. These examples should hopefully be further confirmation of what you learned in [[Algebra of Functions]].
+Fortunately, the idea behind graphing any arbitrary function is mostly dependent on what you know about the function. Therefore, we can easily be able to graph functions.
-{{ExampleRobox|theme=3|title='''Example 1.2(a)''': Graph the following absolute value function:
+'''Example 1.2(a)''': Graph the following absolute value function:
-<center><math>f(x)=\frac{1}{2}|2x+6|-5</math></center>}}
+<center><math>f(x)=\frac{1}{2}|2x+6|-5</math></center>
 :'''Method 1''': Follow procedure from Algebra of Functions
@@ Line 158: / Line 158: @@
 ''To be continued.''
 === Practice Problems ===

@@ Line 40: / Line 40: @@
 \end{cases}</math>.
 Because it can only be the case that <math>y=-x\text{ if }x<0</math> and <math>y=x\text{ if }x\ge0</math>, it is not possible for <math>|x|<0</math>. However, since <math>x</math> has no restriction, the domain, <math>A</math>, has no restriction. Thus, if <math>B</math> represents the range of the function, then  <math>A=\{x\in\mathbb{R}\}</math> and <math>B=\{y\ge0 | y\in\mathbb{R}\}</math>.
-{{Calculus/Def|title=Definition: Domain and Range|
+'''Domain and Range:'''
-text=Let <math>f(x)=|x|</math> whose mapping is <math>f:\mathbb{R}\to\mathbb{R}</math> represent the absolute value function. If <math>A</math> is the domain and <math>B</math> is the range, then <math>A=\{x\in\mathbb{R}\}</math> and <math>B=\{y\ge0 | y\in\mathbb{R}\}</math>.}}
+: Let <math>f(x)=|x|</math> whose mapping is <math>f:\mathbb{R}\to\mathbb{R}</math> represent the absolute value function. If <math>A</math> is the domain and <math>B</math> is the range, then <math>A=\{x\in\mathbb{R}\}</math> and <math>B=\{y\ge0 | y\in\mathbb{R}\}</math>.
 By the above definition, there exists an ''absolute minimum'' to the parent function, and it exists at the origin, <math>O(0,0)</math>
@@ Line 48: / Line 48: @@
 :If <math>x\in A</math> and <math>f(-x)=f(x)</math>, then <math>f</math> is even.
 :If <math>x\in A</math> and <math>f(-x)=-f(x)</math>, then <math>f</math> is odd.
-{{ExampleRobox|theme=2|title=Proof: <math>f(x)=|x|</math> is even}}
 Let <math>f:\mathbb{R}\to\mathbb{R}:x\mapsto |x|</math>. By definition,
 :<math>f(x)=|x|=\begin{cases}
@@ Line 58: / Line 59: @@
 :<math>f(-x)=-(-x)=x</math>
 :<math>\Rightarrow f(-x)=f(x)\blacksquare</math>
-{{Robox/Close}}
 Because <math>f(x)</math> is even, it is also the case that it is ''symmetrical''. A review of this can be found [[CLEP College Algebra/Graphing Functions|here (Graphs and Their Properties)]].
@@ Line 65: / Line 66: @@
 :If <math>u,v\in A</math>, and <math>f(u)=f(v)\Rightarrow u=v</math>, then <math>f(x)</math> is injective.
 :If for all <math>b\in B</math> there is an <math>a\in A</math> such that <math>f(a)=b</math>, then <math>f(x)</math> is surjective.
-{{ExampleRobox|theme=2|title=Proof: <math>f(x)=|x|</math> is non-injective}}
 Suppose <math>u,v\in\mathbb{R}</math> and <math>f(u)=f(v)</math>. By the previous proof, we showed <math>f(x)</math> is even. As such, we can use the value <math>v=-u</math> to make the following statement:
 :<math>f(u)=f(v)\Rightarrow u\ne v</math>
 Therefore, <math>f(x)</math> is non-injective.
-{{Robox/Close}}
 Because we have not established how to prove these statements through algebraic manipulation, we will be deriving properties as we go to gain a further understanding of these new functions. Establishing whether a function is surjective is simply through checking the definition (negating if otherwise to establish it as non-surjective).
-{{ExampleRobox|theme=2|title=Proof: <math>f(x)=|x|</math> is non-surjective}}
 Suppose <math>b\in\mathbb{R}</math>. There exists an element <math>b=-1\in\mathbb{R}</math>, for which <math>f(x)=|x|\ne -1</math> for all <math>x\in\mathbb{R}</math>.<math>\blacksquare</math>
 ==== Intercepts and Inflections of the Parent Function ====
@@ Line 93: / Line 94: @@
 Many times, one will not be working with the parent function. Many real life applications of this function involve at least some manipulation to either the input or the output: vertical stretching/contraction, horizontal stretching/contraction, reflection about the <math>x</math>-axis, reflection about the <math>y</math>-axis, and vertical/horizontal shifting. Luckily, not much changes when it comes to the manipulation of these functions. The exceptions will be talked about in more detail:
-{{Calculus/Def|title=Vertical Expansion/Contraction/Flipping|
+'''Vertical Expansion/Contraction/Flipping'''
-text=Let <math>f(x)=|x|</math> and <math>g(x)=A\cdot f(x)</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0},Ay_{0}\right)\in g(x)</math>. Thus,
+Let <math>f(x)=|x|</math> and <math>g(x)=A\cdot f(x)</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0},Ay_{0}\right)\in g(x)</math>. Thus,
 * If <math>A>1</math>, then <math>g(x)</math> is an expansion of <math>f(x)</math> by a factor of <math>A</math>.
 * If <math>0<A<1</math>, then <math>g(x)</math> is a contraction of <math>f(x)</math> by a factor of <math>A</math>.
-* If <math>A<0</math>, then <math>g(x)</math> is a reflection of <math>f(x)</math> about the <math>x</math>-axis.}}
+* If <math>A<0</math>, then <math>g(x)</math> is a reflection of <math>f(x)</math> about the <math>x</math>-axis.
-{{Calculus/Def|title=Vertical Shift|
+'''Vertical Shift'''
-text=Let <math>f(x)=|x|</math> and <math>g(x)=f(x)+b</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0},y_{0}-b\right)\in g(x)</math>. Thus,
+Let <math>f(x)=|x|</math> and <math>g(x)=f(x)+b</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0},y_{0}-b\right)\in g(x)</math>. Thus,
 * If <math>b>0</math>, then <math>g(x)</math> is an upward shift of <math>f(x)</math> by <math>b</math>.
-* If <math>b<0</math>, then <math>g(x)</math> is a downward shift of <math>f(x)</math> by <math>b</math>.}}
+* If <math>b<0</math>, then <math>g(x)</math> is a downward shift of <math>f(x)</math> by <math>b</math>.
-{{Calculus/Def|title=Horizontal Shift|
+'''Horizontal Shift'''
-text=Let <math>f(x)=|x|</math> and <math>g(x)=f(x+a)</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0}-a, y_{0}\right)\in g(x)</math>. Thus,
+Let <math>f(x)=|x|</math> and <math>g(x)=f(x+a)</math>. There must be an <math>\left(x_{0},y_{0}\right)\in f(x)\Leftrightarrow \left(x_{0}-a, y_{0}\right)\in g(x)</math>. Thus,
 * If <math>a>0</math>, then <math>g(x)</math> is a leftward shift of <math>f(x)</math> by <math>a</math>.
-* If <math>a<0</math>, then <math>g(x)</math> is a rightward shift of <math>f(x)</math> by <math>a</math>.}}
+* If <math>a<0</math>, then <math>g(x)</math> is a rightward shift of <math>f(x)</math> by <math>a</math>.
-The properties not listed above are exceptions to the general rule about functions found in the chapter [[Clep College Algebra/Algebra of Functions|Algebra of Functions]]. The exceptions are not anything substantial. The only difference with what we found generally versus what we have provided above are simply a result of what we found in the previous section.
+The properties not listed above are exceptions to the general rules about functions. The exceptions are not anything substantial. The only difference with what we found generally versus what we have provided above are simply a result of what we found in the previous section.
 * There is no reflection about the <math>y</math>-axis because the function is even and symmetrical.
 * There is no horizontal expansion and contraction because it gives the same result as vertical expansion and contraction (this will be proven later).