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3.6: Absolute Value Functions - Mathematics

3.6: Absolute Value Functions - Mathematics


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Learning Objectives

  • Graph an absolute value function.
  • Solve an absolute value equation.

Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. In this section, we will investigate absolute value functions.

Understanding Absolute Value

Recall that in its basic form (f(x)=|x|), the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.

Absolute Value Function

The absolute value function can be defined as a piecewise function

[f(x)=|x|= egin{cases} x & ext{ if }x{geq}0 -x & ext{ if } x<0 end{cases}]

Example (PageIndex{1}): Determine a Number within a Prescribed Distance

Describe all values (x) within or including a distance of 4 from the number 5.

Solution

We want the distance between (x) and 5 to be less than or equal to 4. We can draw a number line, such as the one in , to represent the condition to be satisfied.

The distance from (x) to 5 can be represented using the absolute value as (|x−5|). We want the values of (x) that satisfy the condition (| x−5 |leq4).

Analysis

Note that

[egin{align*} -4&{leq}x-5 & x-5&leq4 [4pt] 1&{leq}x & x&{leq}9 end{align*}]

So (|x−5|leq4) is equivalent to (1{leq}xleq9).

However, mathematicians generally prefer absolute value notation.

Exercise (PageIndex{1})

Describe all values (x) within a distance of 3 from the number 2.

Answer

(|x−2|leq3)

Example (PageIndex{2}): Resistance of a Resistor

Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%.

Suppose we have a resistor rated at 680 ohms, ±5%. Use the absolute value function to express the range of possible values of the actual resistance.

Solution

5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance (R) in ohms,

[|R−680|leq34 onumber]

Exercise (PageIndex{2})

Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.

Answer

Using the variable (p) for passing, (| p−80 |leq20)

Graphing an Absolute Value Function

The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure (PageIndex{3}).

Figure (PageIndex{3}) shows the graph of (y=2|x–3|+4). The graph of (y=|x|) has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at ((3,4)) for this transformed function.

Example (PageIndex{3}): Writing an Equation for an Absolute Value Function

Write an equation for the function graphed in Figure (PageIndex{5}).

Solution

The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure (PageIndex{6}).

We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure (PageIndex{7}).

From this information we can write the equation

[egin{align*} f(x)&=2|x-3|-2, ;;;;;; ext{treating the stretch as a vertial stretch, or} f(x)&=|2(x-3)|-2, ;;; ext{treating the stretch as a horizontal compression.} end{align*}]

Analysis

Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.

Q & A

If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?

Answer

Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for (x) and (f(x)).

[f(x)=a|x−3|−2 onumber]

Now substituting in the point ((1, 2))

[egin{align*} 2&=a|1-3|-2 4&=2a a&=2 end{align*}]

Exercise (PageIndex{3})

Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units.

Answer

(f(x)=−| x+2 |+3)

Q & A

Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?

Answer

Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.

No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (Figure (PageIndex{8})).

Solving an Absolute Value Equation

Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as (8=|2x−6|), we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently.

[2x-6=8 quad ext{ or } quad 2x-6=-8 onumber]

[egin{align*} 2x &= 14 & 2x &= -2 x&=7 & x&=-1 end{align*}]

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.

An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example,

[egin{align*}|x|&=4, onumber [4pt] |2x−1| &=3, [4pt] |5x+2|−4 &=9. end{align*}]

Solutions to Absolute Value Equations

For real numbers (A) and (B), an equation of the form (|A|=B), with (Bgeq0), will have solutions when (A=B) or (A=−B). If (B<0), the equation (|A|=B) has no solution.

How To ...

Given the formula for an absolute value function, find the horizontal intercepts of its graph.

  1. Isolate the absolute value term.
  2. Use (|A|=B) to write (A=B) or (−A=B), assuming (B>0).
  3. Solve for (x).

Example (PageIndex{4}): Finding the Zeros of an Absolute Value Function

For the function (f(x)=|4x+1|−7), find the values of (x) such that (f(x)=0).

Solution

[egin{align*} 0&=|4x+1|-7 & & & ext{Substitute 0 for f(x).} 7&=|4x+1| & & & ext{Isolate the absolute value on one side of the equation.} 7&=4x+1 & ext{or} -7&=4x+1 & ext{Break into two separate equations and solve.} 6&=4x & -8&=4x & x&=frac{6}{4}=1.5 & x&=frac{-8}{4}=-2 end{align*}]

The function outputs 0 when (x=1.5) or (x=−2) (Figure (PageIndex{9})).

Exercise (PageIndex{4})

For the function (f(x)=|2x−1|−3), find the values of (x) such that (f(x)=0).

Solution

(x=−1) or (x=2)

Q & A

Should we always expect two answers when solving (|A|=B)?

Answer

No. We may find one, two, or even no answers. For example, there is no solution to (2+|3x−5|=1).

How To ...

Given an absolute value equation, solve it.

  1. Isolate the absolute value term.
  2. Use (|A|=B) to write (A=B) or (A=−B).
  3. Solve for (x).

Example (PageIndex{5}): Solving an Absolute Value Equation

Solve (1=4|x−2|+2).

Solution

Isolating the absolute value on one side of the equation gives the following.

[egin{align*} 1&=4|x-2|+2 -1&=4|x-2| -frac{1}{4}&=|x-2| end{align*}]

The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions.

Q & A

In Example (PageIndex{5}), if (f(x)=1) and (g(x)=4|x−2|+2) were graphed on the same set of axes, would the graphs intersect?

Answer

No. The graphs of (f) and (g) would not intersect, as shown in Figure (PageIndex{10}). This confirms, graphically, that the equation (1=4|x−2|+2) has no solution.

Find where the graph of the function (f(x)=−| x+2 |+3) intersects the horizontal and vertical axes.

(f(0)=1), so the graph intersects the vertical axis at ((0,1)). (f(x)=0) when (x=−5) and (x=1) so the graph intersects the horizontal axis at ((−5,0)) and ((1,0)).

Solving an Absolute Value Inequality

Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation of the form

[|A|B, onumber]

or

[ |A|{geq}B, onumber]

where an expression (A) (and possibly but not usually (B)) depends on a variable (x). Solving the inequality means finding the set of all (x) that satisfy the inequality. Usually this set will be an interval or the union of two intervals.

There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph.

For example, we know that all numbers within 200 units of 0 may be expressed as

[|x|<200 onumber]

or

[ −200

Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of values (x) such that the distance between (x) and 600 is less than 200. We represent the distance between (x) and 600 as (|x−600|).

[|x−600|<200]

or

[−200

[egin{align*} −200+600< &x−600+600<200+600 [4pt] 400< &x<800 end{align*}]

This means our returns would be between $400 and $800.

Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive.

How To ...

Given an absolute value inequality of the form (|x−A|{leq}B) for real numbers (a) and (b) where (b) is positive, solve the absolute value inequality algebraically.

  1. Find boundary points by solving (|x−A|=B).
  2. Test intervals created by the boundary points to determine where (|x−A|{leq}B).
  3. Write the interval or union of intervals satisfying the inequality in interval, inequality, or set-builder notation.

Example (PageIndex{6}): Solving an Absolute Value Inequality

Solve (|x −5|{leq}4).

Solution

With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where (|x−5|=4). We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve (|x−5|=4).

[egin{align*} x−5&=4 & ext{ or };;;;;;;; x&=9 x−5&=−4 & x&=1end{align*}]

After determining that the absolute value is equal to 4 at (x=1) and (x=9), we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:

[x<1,; 19. onumber]

To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table (PageIndex{1}).

Table (PageIndex{1})
Interval test (x)(f(x))(<4) or (>4)
(x<1)0(|0-5|=5)Greater than
(16(|6-5|=1)Less than
(x>9)11(|11-5|=6)Greater than

Because (1{leq}x{leq}9) is the only interval in which the output at the test value is less than 4, we can conclude that the solution to (|x−5|{leq}4) is (1{leq}x{leq}9), or ([1,9]).

To use a graph, we can sketch the function (f(x)=|x−5|). To help us see where the outputs are 4, the line (g(x)=4) could also be sketched as in Figure (PageIndex{11}).

We can see the following:

  • The output values of the absolute value are equal to 4 at (x=1) and (x=9).
  • The graph of (f) is below the graph of (g) on (1
  • The absolute value is less than or equal to 4 between these two points, when (1{leq}xleq9). In interval notation, this would be the interval ([1,9]).

Analysis

For absolute value inequalities,

[|x−A|C, −CC. onumber]

The (<) or (>) symbol may be replaced by (leq) or (geq).

So, for this example, we could use this alternative approach.

[egin{align*} |x−5|&{leq}4 −4&{leq}x−5{leq}4 & ext{Rewrite by removing the absolute value bars.} −4+5&{leq}x−5+5{leq}4+5 & ext{Isolate the x.} 1&{leq}xleq9 end{align*}]

Exercise (PageIndex{5})

Solve (|x+2| leq 6).

Answer

(-8 leq x leq 4)

How To ...

Given an absolute value function, solve for the set of inputs where the output is positive (or negative).

  1. Set the function equal to zero, and solve for the boundary points of the solution set.
  2. Use test points or a graph to determine where the function’s output is positive or negative.

Example (PageIndex{7}): Using a Graphical Approach to Solve Absolute Value Inequalities

Given the function (f(x)=−frac{1}{2}|4x−5|+3), determine the (x)-values for which the function values are negative.

Solution

We are trying to determine where (f(x)<0), which is when (−frac{1}{2}|4x−5|+3<0). We begin by isolating the absolute value.

[ egin{align*} -frac{1}{2}|4x−5|&<−3 ;;; ext{Multiply both sides by –2, and reverse the inequality.} |4x−5|&>6end{align*}]

Next we solve for the equality (|4x−5|=6).

[egin{align*} 4x-5&=6 & 4x-5&=-6 4x-6&=6 end{align*}]

or

[egin{align*} 4x&=-1 x&=frac{11}{4} & x&=-frac{1}{4} end{align*}]

Now, we can examine the graph of (f) to observe where the output is negative. We will observe where the branches are below the (x)-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at (x=−frac{1}{4}) and (x=frac{11}{4}) and that the graph has been reflected vertically. See Figure (PageIndex{12}).

We observe that the graph of the function is below the (x)-axis left of (x=−frac{1}{4}) and right of (x=frac{11}{4}). This means the function values are negative to the left of the first horizontal intercept at (x=−frac{1}{4}), and negative to the right of the second intercept at (x=frac{11}{4} ). This gives us the solution to the inequality.

[x<−frac{1}{4} ext{ or } x>1frac{1}{4} onumber]

In interval notation, this would be (( −infty,−0.25 )cup( 2.75,infty)).

Exercise (PageIndex{6})

Solve (−2|k−4|leq−6).

Answer

(kleq1) or (kgeq7); in interval notation, this would be (left(−infty,1 ight]cupleft[7,infty ight))

Key Concepts

  • The absolute value function is commonly used to measure distances between points.
  • Applied problems, such as ranges of possible values, can also be solved using the absolute value function.
  • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction.
  • In an absolute value equation, an unknown variable is the input of an absolute value function.
  • If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable.
  • An absolute value equation may have one solution, two solutions, or no solutions.
  • An absolute value inequality is similar to an absolute value equation but takes the form | A |B, or | A |≥B.It can be solved by determining the boundaries of the solution set and then testing which segments are in the set.
  • Absolute value inequalities can also be solved graphically.

Glossary

absolute value equation
an equation of the form (|A|=B), with (Bgeq0); it will have solutions when (A=B) or (A=−B)

absolute value inequality
a relationship in the form (|A|B), or (|A|{geq}B)


How to solve absolute value equations

The General Steps to solve an absolute value equation are:

  • Rewrite the absolute value equation as two separate equations, one positive and the other negative
  • Solve each equation separately
  • After solving, substitute your answers back into original equation to verify that you solutions are valid
  • Write out the final solution or graph it as needed

It's always easiest to understand a math concept by looking at some examples so, check outthe many examples and practice problems below.

You can always check your work with our Absolute value equations solver too

Practice Problems

Example Equation

Problem 1

Solve the equation: |X + 5| = 3

Click here to practice more problems like this one, questions that involve variables on 1 side of the equation.

Problem 2

Some absolute value equations have variables both sides of the equation. However, that will not change the steps we're going to follow to solve the problem as the example below shows:

Solve the equation: |3X| = X &minus 21

Problem 3

Solve the following absolute value equation: | 5X +20| = 80

Problem 4

Solve the following absolute value equation: | X | + 3 = 2X

This first set of problems involves absolute values with x on just 1 side of the equation (like problem 2).


Equations That have Absolute Value Sign on One Side

The absolute value equation |ax + b| = c (c &ge 0) can be solved by rewriting as two linear equations

and then solving each equation separately.

ABSOLUTE VALUES ALWAYS GIVE 2 EQUATIONS!

This equation has no solution, since an absolute value cannot be negative.

Since positive and negative 0 mean the same thing, we only need one equation

Exercise 1: Solve absolute value equations

Equations That have Absolute Value Signs on Both Sides

If we have absolute value signs on both sides of the equation, we can play the same game with two choices as follows.

3x + 4 = 2x - 3 or 3x + 4 = -(2x - 3)

3x + 4 = 2x - 3 or 3x + 4 = -2x + 3

Exercise 2: Solve absolute value equations


Algebra 2

I'm working on Lesson 6: Absolute Value Equations and Inequalities Algebra 2 A Unit 2: Expressions, Equations, and Inequalities and feel confident in my first few answers, but beyond that am lost. May I have some help?
**=the answer I believe to be true

1.)
The sides of a triangle are in the ratio 3:4:5. If the triangles perimeter 90cm, what is the length of each side?
a.)22.5 cm, 30 cm, 37.5**
b.)19.3, 25.7, 32.1
c.)7.5, 11.5, 32.1
d.)10.5, 11.5, 12.5

2.)
Is the following sometimes, always, or never true?

r is less than or equal to one fifth

4.)
Is the following sometimes always or never true?

I think the answer is always true

5.)
Solve the compound inequality

I think the answer is x<-5 or x>0 b.)

I think the answer is 1<= x < 4 c.)

7.) Solve the absolute value equation

I believe the answer is x=-1 or x=4 1/3 b.)

From here on out I am unsure or the answers. I imagine anybody looking at the questions is from Connections academy. If you would be so willing as to help me out I would greatly appreciate it. Have a lovely day.

#3, you list no choices, but the question itself is your answer:
r ≤ 1/5


Solutions for Chapter 3.6: ABSOLUTE VALUE FUNCTIONS

Since 42 problems in chapter 3.6: ABSOLUTE VALUE FUNCTIONS have been answered, more than 49540 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: College Algebra, edition: 1. College Algebra was written by and is associated to the ISBN: 9781938168383. Chapter 3.6: ABSOLUTE VALUE FUNCTIONS includes 42 full step-by-step solutions.

Tv = Av + Vo = linear transformation plus shift.

If diagonalizable, they share n eigenvectors.

dim(V) = number of vectors in any basis for V.

The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.

Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.

Constant along each antidiagonal hij depends on i + j.

Triangular matrix with one extra nonzero adjacent diagonal.

The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n - 1, volume of box = I det( A) I.

Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.

Vectors x with aT x = O. Plane is perpendicular to a =1= O.

A square matrix that has no inverse: det(A) = o.

(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.


Example 1: f is a function given by

  1. Find the x and y intercepts of the graph of f.
  2. Find the domain and range of f.
  3. Sketch the graph of f.

Solution to Example 1

  • a - The y intercept is given by
    (0 , f(0)) = (0 ,|-2|) = (0 , 2)
  • The x coordinate of the x intercepts is equal to the solution of the equation
    |x - 2| = 0
    which is
    x = 2
  • The x intercepts is at the point (2 , 0)
  • b - The domain of f is the set of all real numbers
    Since |x - 2| is either positive or zero for x = 2 the range of f is given by the interval [0 , +infinity).
  • c - To sketch the graph of f(x) = |x - 2|, we first sketch the graph of y = x - 2 and then take the absolute value of y.
    The graph of y = x - 2 is a line with x intercept (2 , 0) and y intercept (0 , -2). (see graph below)

  • We next use the definition of the absolute value to graph f(x) = |x - 2| = | y |.
    If y >= 0 then | y | = y , if y <0 then | y | = -y.
  • For values of x for which y is positive, the graph of | y | is the same as that of y = x - 2. For values of x for which y is negative, the graph of | y | is a reflection on the x axis of the graph of y. The graph of y = x - 2 above has y negative on the interval (-infinity , 2) and it is this part of the graph that has to be reflected on the x axis. (see graph below).

Example 2: f is a function given by

  1. Find the x and y intercepts of the graph of f.
  2. Find the domain and range of f.
  3. Sketch the graph of f.

Solution to Example 2

  • a - The y intercept is given by
    (0 , f(0)) = (0 ,(-2) 2 - 4) = (0 , 0)
  • The x coordinates of the x intercepts are equal to the solutions of the equation
    |(x - 2) 2 - 4| = 0
    which is solved
    (x - 2) 2 = 4
    Which gives the solutions
    x = 0 and x = 4
  • The x intercepts is at the point (0 , 0) and (4 , 0)
  • b - The domain of f is the set of all real numbers
    Since |(x - 2) 2 - 4| is either positive or zero for x = 4 and x = 0 the range of f is given by the interval [0 , +infinity).
  • c - To sketch the graph of f(x) = |(x - 2) 2 - 4|, we first sketch the graph of y = (x - 2) 2 - 4 and then take the absolute value of y.
    The graph of y = (x - 2) 2 - 4 is a parabola with vertex at (2,-4), x intercepts (0 , 0) and (4 , 0) and a y intercept (0 , 0). (see graph below)

  • The graph of f is given by reflecting on the x axis part of the graph of y = (x - 2) 2 - 4 for which y is negative. (see graph below).


How Can We Find Absolute Value Using abs() Function?

The abs() function in Python is used for obtaining the Python absolute value or the positive value of a number.

We can get the absolute value of an integer, complex number or a floating number using the abs() function. If the argument x (integral value) is a float or integer, then the resultant absolute value will be an integer or float respectively.

If the argument x (integral value) is a complex number, the return value will only be the magnitude part that can be a floating-point.


3.6: Absolute Value Functions - Mathematics

The next set of functions that we want to take a look at are exponential and logarithm functions. The most common exponential and logarithm functions in a calculus course are the natural exponential function, (<<f>^x>), and the natural logarithm function, (ln left( x ight)). We will take a more general approach however and look at the general exponential and logarithm function.

Exponential Functions

We’ll start off by looking at the exponential function,

We want to differentiate this. The power rule that we looked at a couple of sections ago won’t work as that required the exponent to be a fixed number and the base to be a variable. That is exactly the opposite from what we’ve got with this function. So, we’re going to have to start with the definition of the derivative.

Now, the ()is not affected by the limit since it doesn’t have any (h)’s in it and so is a constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives,

Now let’s notice that the limit we’ve got above is exactly the definition of the derivative of (fleft( x ight) = ) at (x = 0), i.e. (f'left( 0 ight)). Therefore, the derivative becomes,

[f'left( x ight) = f'left( 0 ight)]

So, we are kind of stuck. We need to know the derivative in order to get the derivative!

There is one value of (a) that we can deal with at this point. Back in the Exponential Functions section of the Review chapter we stated that (<f> = mbox <2.71828182845905>ldots ) What we didn’t do however is actually define where (f) comes from. There are in fact a variety of ways to define (f). Here are three of them.

Some Definitions of (f)

  1. (displaystyle <f> = mathop limits_ > ight)^n>)
  2. (displaystyle f) is the unique positive number for which (mathop limits_ frac<<<<f>^h> - 1>>= 1)
  3. (displaystyle <f> = sumlimits_^infty <>> )

The second one is the important one for us because that limit is exactly the limit that we’re working with above. So, this definition leads to the following fact,

Fact 1

For the natural exponential function, (fleft( x ight) = <<f>^x>) we have (f'left( 0 ight) = mathop limits_ frac<<<<f>^h> - 1>> = 1).

So, provided we are using the natural exponential function we get the following.

[fleft( x ight) = <<f>^x>hspace <0.5in>Rightarrow hspace<0.5in>f'left( x ight) = <<f>^x>]

At this point we’re missing some knowledge that will allow us to easily get the derivative for a general function. Eventually we will be able to show that for a general exponential function we have,

[fleft( x ight) = hspace <0.5in>Rightarrow hspace<0.5in>f'left( x ight) = ln left( a ight)]

Logarithm Functions

Let’s now briefly get the derivatives for logarithms. In this case we will need to start with the following fact about functions that are inverses of each other.

Fact 2

If (f(x)) and (g(x)) are inverses of each other then,

So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is!

So, if we have (fleft( x ight) = <<f>^x>) and (gleft( x ight) = ln x) then,

The last step just uses the fact that the two functions are inverses of each other.

Putting this all together gives,

Note that we need to require that (x > 0) since this is required for the logarithm and so must also be required for its derivative. It can also be shown that,

Using this all we need to avoid is (x = 0).

In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the derivative of the natural logarithm, which we just found, and the change of base formula. Using the change of base formula we can write a general logarithm as,

Differentiation is then fairly simple.

We took advantage of the fact that (a) was a constant and so (ln a) is also a constant and can be factored out of the derivative. Putting all this together gives,

Here is a summary of the derivatives in this section.

Okay, now that we have the derivations of the formulas out of the way let’s compute a couple of derivatives.

This will be the only example that doesn’t involve the natural exponential and natural logarithm functions.

Not much to this one. Just remember to use the product rule on the second term.

We’ll need to use the quotient rule on this one.

There’s really not a lot to differentiating natural logarithms and natural exponential functions at this point as long as you remember the formulas. In later sections as we get more formulas under our belt they will become more complicated.

Next, we need to do our obligatory application/interpretation problem so we don’t forget about them.

Does the object ever stop moving?

First, we will need the derivative. We need this to determine if the object ever stops moving since at that point (provided there is one) the velocity will be zero and recall that the derivative of the position function is the velocity of the object.

So, we need to determine if the derivative is ever zero. To do this we will need to solve,

Now, we know that exponential functions are never zero and so this will only be zero at (t = - 1). So, if we are going to allow negative values of (t) then the object will stop moving once at (t = - 1). If we aren’t going to allow negative values of (t) then the object will never stop moving.

Before moving on to the next section we need to go back over a couple of derivatives to make sure that we don’t confuse the two. The two derivatives are,

It is important to note that with the Power rule the exponent MUST be a constant and the base MUST be a variable while we need exactly the opposite for the derivative of an exponential function. For an exponential function the exponent MUST be a variable and the base MUST be a constant.

It is easy to get locked into one of these formulas and just use it for both of these. We also haven’t even talked about what to do if both the exponent and the base involve variables. We’ll see this situation in a later section.


3.6: Absolute Value Functions - Mathematics

⭐ Lesson 1: เทคนิคการทำข้อสอบ ARITHMETIC (เลขคณิต)

1.1 Basic Arithmetic Addition Subtraction, Multiplication and Division (พื้นฐานการบวก การลบ การคูณ และการหาร)

1.2 Decimals , Fractions , Ratios and Percentages (ทศนิยม เศษส่วน อัตราส่วน และ ร้อยละ)

1.4 Fraction, Ratio, Percentage Mixes (เศษส่วน อัตราส่วน และร้อยละ แบบยาก)

1.5 Powers and Square Root (เลขยกกำลังและรากที่สอง)

1.6 Negative Numbers (จำนวนลบ)

1.7 Divisibility (การหารลงตัว)

1.8 Even & Odd Numbers (จำนวนคู่และจำนวนคี่)

1.9 Absolute Value (ค่าสัมบูรณ์)

⭐ Lesson 2: เทคนิคการทำข้อสอบ GEOMETRY (เรขาคณิต)

2.1 Point , Lines and Angles (จุด เส้นตรง และ มุม)

2.2 Polygon (รูปหลายเหลี่ยม)

2.3 Triangles : 3 - Sided Polygons (รูปสามเหลี่ยม)

2.4 Quadrangles : 4 - Sided Polygons (รูปสี่เหลี่ยม)

2.6 Trigonometry (ตรีโกณมิติ)

2.7 Coordinate Geometry (เรขาคณิตในระบบพิกัดฉาก)

2.9 3 &ndash Dimensional Objects (รูปสามมิติ)

⭐ Lesson 3: เทคนิคการทำข้อสอบ ALGEBRA (พีชคณิต)

3.2 One Variable Simple Equation (สมการตัวแปรเดียว)

3.3 One Variable Inequalities (อสมการตัวแปรตัวเดียว)

3.4 Equation with Multiple Unknowns (โจทย์ปัญหาสมการ)

3.5 Equation with Powers (สมการเลขยกกำลัง)

3.6 Absolute Value Equation (สมการค่าสัมบูรณ์)

3.7 Inequalities With Absolute Value (อสมการค่าสัมบูรณ์)

3.8 Lines Equation (สมการเชิงเส้น)

3.9 Systems of Equation (ระบบสมการ)

3.10 Two Variables Inequalities (อสมการสองตัวแปร)

3.11 Proportionality (สัดส่วน)

3.13 Addition and Subtraction of Functions (การบวกและการลบของฟังก์ชัน)

3.14 Multiplication and Division of Functions (การคูณและการหารของฟังก์ชัน)

3.15 Linear Function (ฟังก์ชันเชิงเส้น)

3.16 Quadratic Functions (ฟังก์ชันกำลังสอง)

3.17 Remainder Theorem (ทฤษฎีเศษเหลือ)

3.18 Complex Number (จำนวนเชิงซ้อน)

⭐ Lesson 4: เนื้อหาข้อสอบ SAT MATH : COUNTING (การนับ)

4.1 Basic Counting (การนับเบื้องต้น)

4.2 Permutations (การเรียงสับเปลี่ยน)

4.3 Combinations (การจัดหมู่)

4.4 Independent Events (เหตุกาณ์อิสระต่อกัน)

4.5 Probability (ความน่าจะเป็น)

⭐ Lesson 5: เนื้อหาข้อสอบ SAT MATH : OTHER (อื่น ๆ)

5.2 Data Representation : Table , Pie Charts & Graphs (การนำเสนอข้อมูล : ตาราง แผนภูมิวงกลม และ กราฟ)


Common Syllabus for MATH 162

Review. Prerequisite Material from MATH 161 [1 Week]
Rapid review of differentiation rules.
More detailed review of integration (Ch. 5): areas and distances the definite integral the fundamental theorem of calculus.

Chapter 6. Applications of Integration [1.5 weeks 2 weeks if Section 6.4 is covered]
6.1 Area Between Curves
6.2 Volumes
6.3 Volumes by Cylindrical Shells
6.4 Optional:Work ( This Sections could be covered with Chapter 8)
6.5 Average Value of a Function

Chapter 7: Techniques of Integration [3 weeks]
7.1 Integration by Parts
7.2 Trigonometric Integrals
7.3 Trigonometric Substitution
7.4 Integration of Rational Functions by Partial Fractions
7.5 Strategy for Integration
7.6 Integration Using Tables and Computer Algebra Systems
7.7 Approximate Integration
7.8 Improper Integrals

Chapter 11: Infinite Sequences and Series [4 weeks]
11.1 Sequences
11.2 Series
11.3 The Integral Test and Estimates of Sums
11.4 The Comparison Tests
11.5 Alternating Series
11.6 Absolute Convergence and the Ratio and Root Tests
11.7 Strategy for Testing Series
11.8 Power Series
11.9 Representations of Functions as Power Series
11.10 Taylor and Maclauren Series
11.11 Optional: Applications of Taylor Polynomials

Chapter 10: Parametric Equations and Polar Coordinates [1-1.5 weeks]
10.1 Curves Defined by Parametric Equations
10.2 Optional: Calculus with Parametric Curves
10.3 Polar Coordinates
10.4 Optional: Areas and Lengths in Polar Coordinates
10.5 Optional: Conic Sections
10.6 Optional: COnic Sections in Polar Coordinates


Watch the video: OpenStax College Algebra Ch Absolute Value Function (May 2022).