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FUNCTIONS (LIVE)
04 MAY 2015
Section A: Summary Notes
Functions can be one-to-one relations or many-to-one relations. A many-to-one relation associates
two or more values of the independent (input) variable with a single value of the dependent (output)
variable. The domain is the set of values to which the rule is applied (A) and the range is the set of
values (also called the images or function values) determined by the rule.
Example of a one-to-one function: 𝑦 = 𝑥 + 1
Example of a many-to-one function: 𝑦
= 𝑥2
Function Notation
For the function y = f (x), y is the dependent variable, because the value of y (output) depends on the
value of x (input). We say x is the independent variable, since we can choose x to be any number.
Similarly, if g (t) = 2t + 1, then t is the independent variable and g is the function name.
Vertical Line Test
Given the graph of a relation, there is a simple test for whether or not the relation is a function. This
test is called the vertical line test. If it is possible to draw any vertical line (a line of constant x) which
crosses the graph of the relation more than once, then the relation is not a function. If more than one
intersection point exists, then the intersections correspond to multiple values of y for a single value of
x (one-to-many). If any vertical line cuts the graph only once, then the relation is a function (one-toone or many-to-one).
Inverse Functions
An inverse function is a function which does the “reverse” of a given function, it is the reflection of the
-1
graph in the line 𝑦 = 𝑥. More formally, if f is a function with domain x, then f is its inverse function if
-1
and only if f (f (x)) = x for every x 2 X.
y = f(x): indicates a function
-1
f (y) = x : indicates the inverse function
A function must be a one-to-one relation if its inverse is to be a function. If a function f has an inverse
-1
function f , then f is said to be invertible
-1
Given the function f(x), we determine the inverse f (x) by:



interchanging x and y in the equation;
making y the subject of the equation;
expressing the new equation in function notation.
Note: if the inverse is not a function then it cannot be written in function notation. For example, the
inverse of 𝑓 𝑥 = 3𝑥 2 cannot be written as 𝑓 −1 𝑥 = ±
inverse as 𝑦 = ±
1
3
1
3
𝑥 as it is not a function. We write the
𝑥 and conclude that f is not invertible.
Logarithms
A logarithm is a way of writing an exponential equation with the exponent as the subject of the
formula. A logarithmic function is the inverse function of the exponential function.
If 𝑥 = 𝑏 𝑦 𝑡𝑕𝑒𝑛 𝑦 = 𝑙𝑜𝑔𝑏 𝑥 𝑤𝑕𝑒𝑟𝑒 𝑏 ≠ 1 𝑎𝑛𝑑 𝑥 > 0
Section B: Exam practice questions
Question 1
In the diagram, the graphs of the following functions have been sketched:
f ( x)  a ( x  p ) 2  q
and
g ( x) 
a
q
x p
The two graphs intersect at A(2 ; 4) and the turning point of the parabola lies at the point of
intersection of the asymptotes of the hyperbola. The line
parabola.
x 1
is the axis of symmetry of the
A(2 ; 4)
T
y2
x 1
f ( x)
1.1
Determine the equation of
in the form
1.2
Determine the equation of g ( x) in the form
1.3
Write down the range for the graph of f.
1.4
Write down the values of x for which
y  a( x  p)2  q
(3)
a
q
x p
(3)
y
g ( x)  0
(1)
(2)
Question 2
Given:
f ( x)  2( x  1)2  8
2.1
Sketch the graphs of h and f
and
h( x )  4 x
Indicate ALL intercepts with the axes and any turning points.
2.2
Show, algebraically, that
(9)
1

h  x    2h( x ) .
2

(3)
Question 3
The diagram below shows the graphs of 𝑓 𝑥 = 𝑎𝑥 2 and 𝑔 𝑥 =
−2
𝑥
.
The point 𝑀 1; −2 is the point of intersection of 𝑓 and 𝑔.
a) Determine the value of 𝑎.
b) Sketch the graph of 𝑕 𝑥 =
(1)
−2
𝑥−2
+1
(3)
Question 4
𝑎
The diagram below shows the graphs of 𝑓 𝑥 = 4𝑥 and 𝑔 𝑥 = . The point 𝑃 1; 4 is the point of
𝑥
intersection of 𝑓 and 𝑔.
4.1
Write down the equation of 𝑓 −1 in the form 𝑦 = ⋯
(2)
4.2
Is 𝑓 −1 a function? Substantiate your answer.
(2)
4.3
Determine the equation of 𝑕(𝑥), the resultant function when 𝑓(𝑥) is reflected about
the y-axis.
(2)
4.4
Determine the value of 𝑎 in 𝑔(𝑥).
(2)
4.5
Determine the equation of 𝑚 𝑥 , the resultant function when 𝑔 𝑥 is shifted horizontally
2 units to the right and vertically 1 unit down.
(2)
4.6
Calculate the intercepts of 𝑚 𝑥 with the axes.
(3)
Question 5
Given:
1
g ( x)   
2
x
-1
5.1
Write the inverse of g in the form g (x) =.......
(2)
5.2
-1
Sketch the graph of g .
(2)
5.3
Determine graphically the values of x for which
log 1 x  0
2
(1)
Question 6
The graph of
f : x  loga x
passes through the point (16 ; 2).
6.1
Calculate the value of a.
6.2
Write down the equation of the inverse in the form
6.3
(3)
-1
f 1 ( x)  .....
Sketch the graphs of f and f on the same set of axes.
(2)
(4)
Section C: Solutions
Question 1
1.1
f ( x)  a ( x  p ) 2  q
Substitute the turning point T(1; 2) :
For the graph of
y  a( x  1)2  2
y  a( x  1)2  2
 a2
2
 y  2( x  1)  2

(3)
Substitute the point A(2 ; 4) :
2
4  a(2  1)  2
4  a  2
a  2
The equation of the parabola is therefore:
2
y  2( x  1)  2
1.2
a
q
x

p
For the graph of
The vertical asymptote is x  1 and the horizontal asymptote is
y 2.
g ( x) 
y 
a
2
x 1
a
2
x 1

 a2
2
y
2
x 1

y
(3)
Substitute the point A(2 ; 4) :
a
2
2 1
4  a  2
a  2
4 
The equation of the hyperbola is therefore:
y
1.3
1.4
2
2
x 1
y 2 ; 
y 2 ; 
Range:

(1)
g ( x)  0
0  x  1
 answers
(2)
Question 2
2.1
f ( x)  2( x  1)2  8
Turning point: (1;  8)
For f ( x)  2( x  1)
 turning point
 shape
 axis of symmetry
 y-intercept
 x-intercepts
x
For h( x)  4
 y-intercept
 shape
 coordinates
x-intercepts of parabola:
0  2( x  1) 2  8
8  2( x  1) 2
4  ( x  1) 2
2  x  1 or  2  x  1
x3
or
x  1
2
8
(9)
y-intercept of parabola:
y  2(0  1)  8  6
2
(1;4)
1
1
6
2.2
1
(1;  8)
1

h x  
2

4
x
1
2

1
4 x.4 2
3

 
 4 x .2  2h( x)

4
x
1
2
1
.4 2
x
4
x
 4 .2  2h( x)
 
(3)
Question 3
Question 4
Question 5
5.1
1
y 
2
x
Remember that the inverse of a graph
is determined by interchanging x and y
in the equation of the original graph.
y
1
x 
2
 log 1 x  y

1
x 
2
 g
1
y
( x)  log 1 x
2
(2)
2
 g 1 ( x)  log 1 x
2
 shape
5.2
Don’t forget to indicate the
coordinates of the intercept
with the axes. The y-axis is
the asymptote of the graph.
(1;0)
 (1;0)
(2)
5.3
log 1 x  0
2
for
 x 1
x 1
(1)
Question 6
6.1
 2  log a 16

2  log a 16
 a 2  16


a 2  16
a4
a  4
6.2
(3)
y  log 4 x

 x  log 4 y

(2)
 4x  y
x  log 4 y
f 1( x)  4 x
 f 1 ( x)  4 x
6.3
f 1 ( x)  4 x
y
yx
f ( x)  log 4 x
1
1
x
 Graph of f ( x)
 y-intercept of f
f 1 ( x)
 Graph of
f 1
 x-intercept of
(4)