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Functions Review
What is a function
• A function is a set of numbers in which every x
has a y value
• The x values are called the functions DOMAIN
• The y functions are called the called the
RANGE
• Functions are usually written as f(x) (but they
can also be written as g(x), h(x) etc)
• If you have to evaluate a function, remember
it’s a substitution
Evaluating functions
f ( x)  2 x  1
f (3) 
f ( 0) 
f ( 5) 
15
g ( x) 
x 3
g (1) 
g (7 ) 
g (3) 
h( x )  4  x
h(0) 
h(1) 
h(5) 
Domain
• Hey, Remember that we said that Domain was
the x values of an equation so
• Look for which values you can graph,
• Look for x values which make the bottom of a
fraction zero – Not allowed
• Look for values which give you the square root
of a negative number – Not allowed either
Find the domain of the following functions:
A)
B)
15
g ( x) 
x3
Domain is all real numbers but x  3
C)
Square root is real only for nonnegative numbers.
Graph of a function
The graph of the function f(x) is the set of
points (x,y) in the xy-plane that satisfy the
relation y = f(x).
Domain and Range from the Graph of a function
y
5
4
3
2
1
0
-3
-2
-1
0
1
2
3
x
-1
Domain = {x /
Range = {y /
 3  or
x  1
}
1 x  3
or
}
3 y 5
1  y  2
Determine the domain, range, and intercepts of the
following graph.
y
4
0
-4
(2, 3)
(1, 0)
(0, -3)
(4, 0)
(10, 0)
x
Theorem Vertical Line Test
A set of points in the xy - plane is the
graph of a function if and only if a
vertical line intersects the graph in at
most one point.
y
x
Not a function.
y
x
Function.
Is this a graph of a function?
y
4
0
-4
(2, 3)
(1, 0)
(0, -3)
(4, 0)
(10, 0)
x
Even functions
A function f is an even function if
f ( x)  f ( x)
for all values of x in the domain of f.
Example: f ( x)  3x 2  1 is even because
f ( x)  3( x) 2  1  3x 2  1  f ( x)
Odd functions
A function f is an odd function if
f ( x)   f ( x)
for all values of x in the domain of f.
Example: f ( x)  5 x 3  xis odd because
f ( x)  5( x)3  x  5x3  x  (5x3  x)   f ( x)
Determine if the given functions are even or odd
1) f ( x )  x  x  1
4
2
2) g ( x )  x
3) h( x ) | x | 1
3
4) k ( x )  x  x
3
2
Graphs of Even and Odd functions
The graph of an even function is symmetric with
respect to the x-axis. (i.e. you can flip it with
an axis of symmetry SIDEWAYS)
The graph of an odd function is symmetric with
respect to the origin. (i.e. you can rotate it 180
and get the same graph!)
y
6
5
4
3
2
1
0
-3
-2
-1
0
-1
-2
-3
-4
-5
-6
1
2
3
x
Determine if the function is even or odd?
y
5
2.5
0
-5
-2.5
0
2.5
5
x
-2.5
-5
Determine if the function is even or odd?
y
5
3.75
2.5
1.25
0
-5
-2.5
0
2.5
5
x
Determine if the function is even or odd?
y
100
50
0
-5
-2.5
0
2.5
5
x
-50
-100
Continuous Functions
y
100
• Consider the graph of f(x) = x3
50
0
-5
-2.5
0
2.5
5
x
-50
-100
• We can see that there are no "gaps" in the curve. Any
value of x will give us a corresponding value of y. We
could continue the graph in the negative and positive
directions, and we would never need to take the pencil
off the paper.
• Such functions are called continuous functions
Discontinuous Functions
y
5
2.5
• Consider the graph of y= 2/x:
0
-5
-2.5
0
2.5
5
x
-2.5
-5
• We note that the curve is not continuous at x = 1.
Such functions are called discontinuous functions
• In simple English: The graph of a continuous
function can be drawn without lifting the pencil
from the paper.
Example of limits at infinity
• The function can converge
The function
converges to a single
value (1), called the limit
of f.
We write
limx + f(x) = 1
Example of limits at infinity
• The function can converge
The function
converges to a single
value (0), called the limit
of f.
We write
limx + f(x) = 0
Example of limits at infinity
• The function can diverge
The function doesn’t
converge to a single value
but keeps growing.
It diverges.
We can write
limx + f(x) = +
Example of limits at infinity
• The function can diverge
The function doesn’t
converge to a single value
but
its amplitude
keeps growing.
It diverges.
Example of limits at infinity
• The function may neither converge nor
diverge!
Example of limits at infinity
• The function can do all this either at + infinity
or - infinity
The function converges at
- and diverges at + .
We can write
limx + f(x) = +
limx - f(x) = 0
Example of limits at infinity
• The function can do all this either at + infinity
or - infinity
The function converges at +
and diverges at -.
We can write
limx + f(x) = 0
Examples of limits at x=0
(x becomes very small!)
• The function can have asymptotes (it
diverges). The limit at 0 doesn’t exist…
Examples of limits at x=0
• The function can have a gap! The limit at 0
doesn’t exist…
Examples of limits at x=0
• The function can behave in a complicated
(exciting) way.. (the limit at 0 doesn’t exist)
Examples of limits at x=0
• But most functions at most points behave in a
simple (boring) way.
The function has a limit
when x tends to 0 and that
limit is 0.
We write
limx  0 f(x) = 0
Evaluating Limits
• Have a look at this video here. It’s not bad.
• (the last bit on division might mean something
to 3unit students but not 2 unit!)
A trigonometric function is a ratio of certain parts of a triangle. The
names of these ratios are: The sine, cosine, tangent, cosecant, secant,
cotangent.
Let us look at this triangle…
B
a
C
Sinθ=
Cos θ=
Tan θ=
Given the assigned letters to the sides and
angles, we can determine the following
trigonometric functions.
c
The Cosecant is the inversion of the
sine, the secant is the inversion of
the cosine, the cotangent is the
ө
A
inversion of the tangent.
b
Side Opposite =
Hypothenuse
Side Adjacent =
Hypothenuse
Side Opposite =
Side Adjacent
a
c
b
c
a
b
With this, we can find the sine of
the value of angle A by dividing side
a by side c. In order to find the
angle itself, we must take the sine of
the angle and invert it (in other
words, find the cosecant of the sine
of the angle).
Try finding the angles of the following triangle from the side lengths using
the trigonometric ratios from the previous slide.
Click for the Answer…
B
6
α
The first step is to use the trigonometric
functions on angle A.
10
Sin θ =6/10
β
C
θ
Sin θ =0.6
A
8
Csc0.6~36.9
Angle A~36.9
Because all angles add up to 180,
B=90-11.537=53.1
B
The measurements have changed. Find side BA and side AC
α
β
C
0.559=2/BA
The Pythagorean theorem
when used in this triangle
states that…
0.559BA=2
BC2+AC2=AB2
Sin34=2/BA
2
34º
A
BA=2/0.559
BA~3.578
AC2=AB2-BC2
AC2=12.802-4=8.802
AC=8.8020.5~3
When solving oblique triangles, simply using trigonometric functions
is not enough. You need…
The Law of Sines
a
b
c


sin A sin B sin C
B
c
a
The Law of Cosines
a2=b2+c2-2bc cosA
b2=a2+c2-2ac cosB
c2=a2+b2-2ab cosC
A
C
b
It is useful to memorize these laws. They
can be used to solve any triangle if
enough measurements are given.
Solve this triangle
Click for answers…
B
c=6
a=4
28º
C
A
b
Because this triangle has an angle given, we can use the law of sines to solve it.
a/sin A = b/sin B = c/sin C and subsitute: 4/sin28º = b/sin B = 6/C. Because we know nothing about
b/sin B, lets start with 4/sin28º and use it to solve 6/sin C.
Cross-multiply those ratios: 4*sin C = 6*sin 28, divide 4: sin C = (6*sin28)/4.
6*sin28=2.817. Divide that by four: 0.704. This means that sin C=0.704. Find the Csc of 0.704 º.
Csc0.704º =44.749. Angle C is about 44.749º. Angle B is about 180-44.749-28=17.251.
The last side is b. a/sinA = b/sinB, 4/sin28º = b/sin17.251º, 4*sin17.251=sin28*b,
(4*sin17.251)/sin28=b. b~2.53.
Solve this triangle:
Hint: use the law of cosines
B
c=5.2
a=2.4
A
b=3.5
C
Start with the law of cosines because there are no angles given.
a2=b2+c2-2bc cosA. Substitute values. 2.42=3.52+5.22-2(3.5)(5.2) cosA,
5.76-12.25-27.04=-2(3.5)(5.2) cos A, 33.53=36.4cosA, 33.53/36.4=cos A, 0.921=cos A, A=67.07.
Now for B.
b2=a2+c2-2ac cosB, (3.5)2=(2.4)2+(5.2)2-2(2.4)(5.2) cosB, 12.25=5.76+27.04-24.96 cos B.
12.25=5.76+27.04-24.96 cos B, 12.25-5.76-27.04=-24.96 cos B. 20.54/24.96=cos B. 0.823=cos B.
B=34.61.
C=180-34.61-67.07=78.32.
When solving a triangle, you must remember to choose the correct law to
solve it with.
Whenever possible, the law of sines should be used. Remember that at
least one angle measurement must be given in order to use the law of
sines.
The law of cosines in much more difficult and time consuming method than
the law of sines and is harder to memorize. This law, however, is the only
way to solve a triangle in which all sides but no angles are given.
Only triangles with all sides, an angle and two sides, or a side and two
angles given can be solved.
What is the sine of 60º? 0.866. What is the cosine of 30º? 0.866. If you look at
the name of cosine, you can actually see that it is the cofunction of the sine
(co-sine). The cotangent is the cofunction of the tangent (co-tangent), and the
cosecant is the cofunction of the secant (co-secant).
Sine60º=Cosine30º
Secant60º=Cosecant30º
tangent30º=cotangent60º
The following trigonometric identities are useful to remember.
Sin θ=1/cosec θ
Cos θ=1/sec θ
Tan θ=1/cot θ
Cosec θ=1/sin θ
Sec θ=1/cos θ
Tan θ=1/cot θ
(sin θ)2 + (cos θ)2=1
1+(tan θ)2=(sec θ)2
1+(cot θ)2=(cosec θ)2