1.5 algebraic and elementary functions

Elementary Formulas and Basic Graphs

Let’s start with evaluating log [ (2x + 1)/(sin1/3(x) + 1) ]
at x = 0.

at x = 0. The answer is 0.

at x = 0. The answer is 0. Now let’s analyze the steps
we would take in the calculation by evaluating at
x = 10o using a scientific calculator.
(i.e. 10 degrees)

x = 10o using a scientific calculator. (Ans ≈ 1.13).

Note that we use the following keys to execute it.
* number–keys * operation–keys +, – , *, /.

* the yx–key
* sin( )
* log( )
The formula–keys

The TI SE Calculator Keyboard
* the yx–key
* sin( )
* log( )
Here is a typical keyboard
layout for a scientific
calculator.
The formula–keys

The TI SE Calculator Keyboard
* the yx–key
* sin( )
* log( )
Here is a typical keyboard
layout for a scientific
calculator. The regions of the
mentioned above are noted.
The formula–keys

Among the many methods of recording measurements
such as printed tables, digital databases, graphs,
charts etc, using mathematical formulas is definitely
the most efficient method.

the most efficient method. Most of the mathematics
formulas used in the real world are “composed” with
members from the following three groups of formulas.

* the algebraic family
* the trigonometric family
* the exponential–log family

The three families of formulas
on every scientific calculator

An algebraic formula (in x) is a formula formed with
the variable x, real numbers,

the variable x, real numbers, repeated applications of
the algebraic operations +, – , *, /

the algebraic operations +, – , *, / and raising
fractional powers ( )p/q

the algebraic operations +, – , *, / and raising
fractional powers ( )p/q (which includes the root–
operations  , or  etc.. entered with the yx – key).3 4

Examples of algebraic expressions are
3x2 – 2x + 4,

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,
(x1/2 + π)1/3
(4x2 – (x + 4)1/2)1/4

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,
(x1/2 + π)1/3
(4x2 – (x + 4)1/2)1/4
We remind the readers that aNxN + aN-1xN-1...+ a1x + a0
where ai are numbers and N is a non–negative
integer are called polynomials (in x) as 3x2 – 2x + 4.

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,
(x1/2 + π)1/3
(4x2 – (x + 4)1/2)1/4
The algebraic expressions , where P and Q are
polynomials, are called rational expressions.
P
Q

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,
(x1/2 + π)1/3
(4x2 – (x + 4)1/2)1/4
P
Q
x2 + 3
3 x3 – 2x – 4
is a rational expression.

3x2 – 2x + 4,
x2 + 3
3 x3 – 2x – 4
,
(x1/2 + π)1/3
(4x2 – (x + 4)1/2)1/4
P
Q
x2 + 3
3 x3 – 2x – 4
x2 + 3
3x3 – 2x – 4
is not because of x3.
is a rational expression.

The trigonometric formulas such as sin(x), arctan(x)
and the exponential–log formulas such as ex, ln(x) are
called transcendental (non–algebraic) formulas.

Let’s call the polynomial formulas, the ( )p/q,

the six trig–formulas and their inverses,

the six trig–formulas and their inverses, logb(x) and bx,

and all their constant multiples the basic formulas.

Recall the “composition” operation i.e. input a formula
f(x) into another formula g(x) and get g(f(x)).

E.g. if f(x) = sin(x), g(x) = x2, then g(f(x)) = sin2(x).

A formula that may be built from the “basic formulas”
by applying the +, – , *, / and the composition
operations in finitely many steps is called an
elementary formula.

A formula that may be built from the “basic formulas”
by applying the +, – , *, / and the composition
operations in finitely many steps is called an
elementary formula. Basically, scientific calculators
handle calculations of elementary formulas.

The algebraic construction of a given elementary
formula is reflected in the corresponding successive
operations performed when we evaluate the formula.

Example A. Assemble the following formulas from
real numbers and the basic formulas using algebraic
operations: +, – , *, /, or the composition operation.
(All the basic formulas are shown in italic.)
a. 2x2 + cos(x)
b. ln(2x2 + cos(x))

a. 2x2 + cos(x)
b. ln(2x2 + cos(x))
Take 2x2
cos(x)

a. 2x2 + cos(x)
b. ln(2x2 + cos(x))
Take 2x2
cos(x)
+
2x2 + cos(x)

a. 2x2 + cos(x)
b. ln(2x2 + cos(x))
Take 2x2
cos(x)
+
2x2 + cos(x)
2x2
cos(x)
+
2x2 + cos(x)

a. 2x2 + cos(x)
b. ln(2x2 + cos(x))
Take 2x2
cos(x)
+
2x2 + cos(x)
2x2
cos(x)
+
2x2 + cos(x) ln(u) = ln(2x2 + cos(x))
composition
operation
u

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
+ 2x2 + cos(x) ln(u) = ln(2x2 + cos(x))
composition
operation

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
sin(x) eu = esin(x)
composition
operation

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
composition
operation

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
/ (divide)

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
/ (divide)
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
u1/2

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
/ (divide)
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
u1/2=
ln(2x2 + cos(x))
esin(x) – x2 ][
1/2

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
/ (divide)
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
u1/2=
ln(2x2 + cos(x))
esin(x) – x2 ][
1/2
Almost all the formulas we have in this course are
elementary formulas i.e. they are made in this manner
from basic formulas.

d.
ln(2x2 + cos(x))
esin(x) – x2[ ]
1/2
2x2
cos(x)
composition
operation
x2
sin(x) eu = esin(x)
– esin(x) – x2
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
/ (divide)
ln(2x2 + cos(x))
esin(x) – x2
composition
operation
u1/2=
ln(2x2 + cos(x))
esin(x) – x2 ][
1/2
Almost all the formulas we have in this course are
elementary formulas i.e. they are made in this manner
from basic formulas. Symbolic techniques in calculus
work precisely because elementary formulas are built
from the basic formulas with the above operations.

It’s useful to use the function notation to keep track of
the “deconstruction” of elementary formulas.

the “deconstruction” of elementary formulas. That is,
we would name the formulas as f(x), g(x), and h(x)
and perform algebraic operations +, – , * , / using
f(x), g(x) or h(x).

f(x), g(x) or h(x). Hence if f(x) = 2x2, g(x) = ln(x + 1),
then f(x) + 3g(x) = 2x2 + 3ln(x + 1).

(g f)(x) = g(f(x)); plug f(x) into g(x),
then f(x) + 3g(x) = 2x2 + 3ln(x + 1). Finally, we define

(g f)(x) = g(f(x)); plug f(x) into g(x), and
(f g)(x) = f(g(x)); plug g(x) into f(x)
for the composition operation.

Example B. Given that f(x) = 2x2 + 1, g(x) = ln(x),
h(x) = sin(x). Simplify the following.
a. (g f)(x)
b. (f g)(x)
c. [3h(x) + 2(g g)(x)]1/2

a. (g f)(x) = g( f(x) ) = g( 2x2 + 1)
b. (f g)(x)
c. [3h(x) + 2(g g)(x)]1/2

a. (g f)(x) = g( f(x) ) = g( 2x2 + 1) = ln(2x2 + 1)
b. (f g)(x)
c. [3h(x) + 2(g g)(x)]1/2

a. (g f)(x) = g( f(x) ) = g( 2x2 + 1) = ln(2x2 + 1)
b. (f g)(x) = f( g(x)) = f( ln(x) )
c. [3h(x) + 2(g g)(x)]1/2

a. (g f)(x) = g( f(x) ) = g( 2x2 + 1) = ln(2x2 + 1)
b. (f g)(x) = f( g(x)) = f( ln(x) ) = 2ln2(x) + 1
c. [3h(x) + 2(g g)(x)]1/2

a. (g f)(x) = g( f(x) ) = g( 2x2 + 1) = ln(2x2 + 1)
b. (f g)(x) = f( g(x)) = f( ln(x) ) = 2ln2(x) + 1
c. [3h(x) + 2(g g)(x)]1/2 = 3sin(x) + 2ln( ln(x) )

Example C.
a. Express sin(9 + x2) as the composition of
a polynomial and a trig–function (basic formulas).

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
b. Express 4sin3(9 + x2) + 5 using +, – , * , /, and the
composition operation with basic formulas.

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
Let f(x) = 9 + x2, g(x) = sin(x),

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
Let f(x) = 9 + x2, g(x) = sin(x),
then (g f)(x) = g(f(x)) = sin(9 + x2).

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
Let f(x) = 9 + x2, g(x) = sin(x),
then (g f)(x) = g(f(x)) = sin(9 + x2).
Let h(x) = 4x3 + 5,

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
Let f(x) = 9 + x2, g(x) = sin(x),
then (g f)(x) = g(f(x)) = sin(9 + x2).
Let h(x) = 4x3 + 5,
then h(g f) = h(sin(9 + x2)) = 4sin3(9 + x2) + 5.

Example C.
Let f(x) = 9 + x2, g(x) = sin(x)
Then (g f)(x) = g(f(x)) = sin(9 + x2)
Let f(x) = 9 + x2, g(x) = sin(x),
then (g f)(x) = g(f(x)) = sin(9 + x2).
Let h(x) = 4x3 + 5,
then h(g f) = h(sin(9 + x2)) = 4sin3(9 + x2) + 5.
The calculations in calculus are based on the fact
that the elementary functions are built in steps in this
manner from the basic formulas.

Set the function output f(x) as y,
the plot of all the points (x, y = f(x))
is the graph of the function f(x).

y= f(x)

The coordinate of a generic point on
the graph is (x, f(x)) or (x, y) as shown. x
(x, f(x))
y= f(x)
f(x) or y

(x, f(x))
y= f(x)
f(x) or y
Graphs of 1st and 2nd degree functions

(x, f(x))
y= f(x)
f(x) or y
The graphs of linear functions
y = f(x) = mx + b
are non–vertical lines where
m is the slope, and (0, b) is the y
intercept.

(x, f(x))
y= f(x)
f(x) or y
y = f(x) = mx + b
intercept. The graph of
y = f(x) = 2x + 6 is drawn here by
plotting the x and y intercepts.

(x, f(x))
y= f(x)
f(x) or y
y = f(x) = mx + b
(0, 6)
(–3, 0)
y = 2x + 6

(x, f(x))
y= f(x)
f(x) or y
y = f(x) = mx + b
(0, 6)
(–3, 0)
(x, 2x+6)
y = 2x + 6

The graphs of quadratic (2nd degree) functions
y = f(x) = ax2 + bx + c are parabolas with
vertex at x = – b/(2a) and y–intercept is at (0, c).

Parabolas
, #)
2a
–b(vertex =

Parabolas
The parabola opens up if a >0, opens down if a < 0.
, #)
2a
–b(vertex =

To plot a 2nd degree function, plot the vertex point at
x = – b/(2a).
Parabolas
, #)
2a
–b(vertex =

x = – b/(2a). Then plot the y–intercept (0, c) and it’s
reflection across the line of symmetry.
Parabolas
, #)
2a
–b(vertex =

x = – b/(2a). Then plot the y–intercept (0, c) and it’s
reflection across the line of symmetry. Trace the
parabola with these points.
Parabolas
, #)
2a
–b(vertex =

Example D.
Sketch the graph of y = f(x) = – x2 – 2x + 8.

It’s a parabola that opens down with vertex at
x = – b/(2a) = –1 or (–1, f(–1) = 9)
Example D.

x = – b/(2a) = –1 or (–1, f(–1) = 9), y–intercept is (0, 8).
Example D.

Plot these points.
Example D.

Plot these points.
Example D.
(0, 8)
(–1, 9)
x

Plot these points.
Example D.
The reflection of (0, 8) across
the line of symmetry is (–2, 8). (0, 8)
(–1, 9)
(–2 , 8)
x

Plot these points.
Example D.
the line of symmetry is (–2, 8).
Trace the parabola with these
points.
(0, 8)
(–1, 9)
(–2 , 8)
x

Plot these points.
Example D.
points.
(0, 8)
(–1, 9)
(–2 , 8)
Finally, we solve f(x) = 0 to find
the x–intercept. x

Plot these points.
Example D.
The reflection of (0, 8)across
points.
(0, 8)
(–1, 9)
(–2 , 8)
Finally, we solve f(x) = 0 to find
the x–intercept. Hence
–x2 – 2x + 8 = 0 or x = 4, – 2.
Label these points on the graph.
(2, 0)(–4, 0)
x

Graphs of Simple Algebraic Functions y = f(x)
y = 1
y = x/3 + 1
y = –x + 1
y = –x2 + 2x + 8 y = x2/10 + 8
(–1, 8.01) (1, 8.01)
(0, 8)
y = x3 y = x3 – x
y = x1/2
y = 1/x
Vertical
Asymptote
at x = 0
y = x1/3
Horizontal
Asymptote
at y = 0

There are two other common operations for functions.

I. The “Absolute Value Operation”

The “Absolute Value Operation” applies to individual
function f(x) and changes it into |f(x)|.

function f(x) and changes it into |f(x)|. The outputs of
|f(x)| are always nonnegative.

|f(x)| are always nonnegative. The graph of |f(x)| is
easily obtained from the graph of f(x) by reflecting all
the negative f(x)’s above the x–axis as shown here.

y = x2– 4
(2, 0)(–2, 0)
(0, –4)

y = x2– 4
(2, 0)(–2, 0)
→
(0, –4)
y = |x2– 4|
(0, 4)
(2, 0)(–2, 0)
Absolute Value

II. The “Cut and Paste Operation”

The “Cut and Paste Operation” pieces many parts
from different functions to make new functions. The
results are known as the piecewise functions.

f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
For example,

f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
For example,
f(x) is pieced together with three
parts and it’s graph is drawn here.

f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
For example,
parts and it’s graph is drawn here. x
y
y = f(x)

f(x) =
1 if 0 < x
0 if 0 = x
–1 if x < 0
For example,
parts and it’s graph is drawn here.
We often construct these as
conceptual examples–such an f(x)
as this is not elementary and its
graph is broken in a manner that
we will address later.
x
y
y = f(x)

1.5 algebraic and elementary functions

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1.5 algebraic and elementary functions