Exponential function

Tomasz Lechowski Batory A & A HL December 7, 2021 1 / 20

We will analyse functions f (x) = ax , where a ∈ R+ , i.e. a is a positive real
number.

Tomasz Lechowski Batory A & A HL December 7, 2021 2 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .
Each of the above is of the form f (x) = ax , but we divide them into 3
categories:

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .
Each of the above is of the form f (x) = ax , but we divide them into 3
categories:
f (x) = ax , where a > 1, examples (i) and (iii),

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .
Each of the above is of the form f (x) = ax , but we divide them into 3
categories:
f (x) = ax , where a > 1, examples (i) and (iii),
f (x) = ax , where 0 < a < 1, example (ii),

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .
Each of the above is of the form f (x) = ax , but we divide them into 3
categories:
f (x) = ax , where a > 1, examples (i) and (iii),
f (x) = ax , where 0 < a < 1, example (ii),
f (x) = ax , where a = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

Introduction

These are some examples of an exponential function:

i f (x) = 3x ,
ii f (x) = (0.2)x ,
iii f (x) = (1.3)x ,
iv f (x) = 1x .
Each of the above is of the form f (x) = ax , but we divide them into 3
categories:
f (x) = ax , where a > 1, examples (i) and (iii),
f (x) = ax , where 0 < a < 1, example (ii),
f (x) = ax , where a = 1.
We will analyse them separately.

Tomasz Lechowski Batory A & A HL December 7, 2021 3 / 20

a>1

We will start with f (x) = ax , where a > 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 4 / 20

a>1

We will start with f (x) = ax , where a > 1. Examples include: f (x) = 2x ,

g (x) = 3x , h(x) = 5x .

Tomasz Lechowski Batory A & A HL December 7, 2021 4 / 20

a>1

We will start with f (x) = ax , where a > 1. Examples include: f (x) = 2x ,

g (x) = 3x , h(x) = 5x .

We will start with the primary school approach. Substitute some value for
x and organize the results into a table:

Tomasz Lechowski Batory A & A HL December 7, 2021 4 / 20

a>1

We will start with f (x) = ax , where a > 1. Examples include: f (x) = 2x ,

g (x) = 3x , h(x) = 5x .

We will start with the primary school approach. Substitute some value for
x and organize the results into a table:

x -2 -1 0 1 2 3 4
f(x) 0.25 0.5 1 2 4 8 16
g(x) 0.(1) 0.(3) 1 3 9 27 81
h(x) 0.004 0.02 1 5 25 125 625

Tomasz Lechowski Batory A & A HL December 7, 2021 4 / 20

We can use the table to draw the graphs:

Tomasz Lechowski Batory A & A HL December 7, 2021 5 / 20

We can use the table to draw the graphs:

Tomasz Lechowski Batory A & A HL December 7, 2021 5 / 20

a>1
What observations can we make?

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.).

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

lim f (x) = ∞.
x→∞

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

lim f (x) = ∞.
x→∞

As x approaches minus infinity, the values of the function approach 0.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

lim f (x) = ∞.
x→∞

As x approaches minus infinity, the values of the function approach 0.

lim f (x) = 0.
x→−∞

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

lim f (x) = ∞.
x→∞

As x approaches minus infinity, the values of the function approach 0.

lim f (x) = 0.
x→−∞

The function is always positive.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
What observations can we make?
For x = 0, the value is 1. No surprises here since f (0) = a0 = 1.

The greater the argument, the greater the value. So the function is
increasing.

We can substitute any value for x (fraction, 0, negatives, etc.). So

the domain is all real numbers.

As x approaches infinity, then so does the values of the function.

lim f (x) = ∞.
x→∞

As x approaches minus infinity, the values of the function approach 0.

lim f (x) = 0.
x→−∞

The function is always positive. The range is ]0, ∞[.

Tomasz Lechowski Batory A & A HL December 7, 2021 6 / 20

a>1
Based on these observations we can do some exercises.

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1),

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1), so it’s an increasing function,

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1), so it’s an increasing function, so the greater
the argument, the greater the value.

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1), so it’s an increasing function, so the greater
the argument, the greater the value. We will organize the arguments first:

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1), so it’s an increasing function, so the greater
the argument, the greater the value. We will organize the arguments first:
√ √ √ √
− 6< 2< 3<2<2 2

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

a>1
Based on these observations we can do some exercises. Remember
however that we are only considering f (x) = ax , where a > 1.

Arrange the following in ascending order:

√ √ √ √
7 3
, 7 2
, 72 , 7− 6
, 72 2

We can think of the function f (x) = 7x , it’s an exponential function

f (x) = ax with a > 1 (7 > 1), so it’s an increasing function, so the greater
the argument, the greater the value. We will organize the arguments first:
√ √ √ √
− 6< 2< 3<2<2 2

so we have: √ √ √ √
7− 6
<7 2
<7 3
< 72 < 72 2

Tomasz Lechowski Batory A & A HL December 7, 2021 7 / 20

Exercise 1

2
Find the range of f (x) = .
3x +1

Tomasz Lechowski Batory A & A HL December 7, 2021 8 / 20

Exercise 1

2
Find the range of f (x) = .
3x +1
In the denominator we have a function 3x , whose range is ]0, ∞[.

Tomasz Lechowski Batory A & A HL December 7, 2021 8 / 20

Exercise 1

2
Find the range of f (x) = .
3x +1
In the denominator we have a function 3x , whose range is ]0, ∞[.
So the range of values of the denominator is ]1, ∞[.

Tomasz Lechowski Batory A & A HL December 7, 2021 8 / 20

Exercise 1

2
Find the range of f (x) = .
3x +1
In the denominator we have a function 3x , whose range is ]0, ∞[.
So the range of values of the denominator is ]1, ∞[. The denominator is
then always positive, so the greater the denominator, the smaller the
whole fraction and vice versa.

Tomasz Lechowski Batory A & A HL December 7, 2021 8 / 20

Exercise 1

2
Find the range of f (x) = .
3x +1
In the denominator we have a function 3x , whose range is ]0, ∞[.
So the range of values of the denominator is ]1, ∞[. The denominator is
then always positive, so the greater the denominator, the smaller the
whole fraction and vice versa.
So the range of the function will be ]0, 2[ (0 when the denominator
approaches ∞, and 2 when the denominator approaches 1).

Tomasz Lechowski Batory A & A HL December 7, 2021 8 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1
We will rearrange the function:

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1
We will rearrange the function:
2x + 4 2x + 1 + 3 3
f (x) = x
= x
=1+ x
2 +1 2 +1 2 +1

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1
We will rearrange the function:
2x + 4 2x + 1 + 3 3
f (x) = x
= x
=1+ x
2 +1 2 +1 2 +1
Now the problem is similar to the previous one.

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1
We will rearrange the function:
2x + 4 2x + 1 + 3 3
f (x) = x
= x
=1+ x
2 +1 2 +1 2 +1
Now the problem is similar to the previous one. 2x + 1 has range of ]1, ∞[,
3
so x has range of ]0, 3[,
2 +1

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 2

2x + 4
Find the range of f (x) = .
2x + 1
We will rearrange the function:
2x + 4 2x + 1 + 3 3
f (x) = x
= x
=1+ x
2 +1 2 +1 2 +1
Now the problem is similar to the previous one. 2x + 1 has range of ]1, ∞[,
3
so x has range of ]0, 3[,
2 +1
We add 1 so in the end the range of the function is ]1, 4[.

Tomasz Lechowski Batory A & A HL December 7, 2021 9 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you.

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic.

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ).

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ). Now we analyse the
quadratic:

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ). Now we analyse the
quadratic: a = −1 < 0, so arms downwards.

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ). Now we analyse the
quadratic: a = −1 < 0, so arms downwards. No roots.

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ). Now we analyse the
quadratic: a = −1 < 0, so arms downwards. No roots. Y-intercept
(0, −5).

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

Find the range of f (x) = −36x − 4 · 6x − 5.

This should look familiar to you. It’s a disguised quadratic. We set t = 6x

and we get:
f (t) = −t 2 − 4t − 5
With t ∈]0, ∞[ (since this is the range of 6x ). Now we analyse the
quadratic: a = −1 < 0, so arms downwards. No roots. Y-intercept
(0, −5). The vertex is (−2, −1).

Tomasz Lechowski Batory A & A HL December 7, 2021 10 / 20

Exercise 3

The graph looks like this

Tomasz Lechowski Batory A & A HL December 7, 2021 11 / 20

Exercise 3

The graph looks like this

Tomasz Lechowski Batory A & A HL December 7, 2021 11 / 20

Exercise 3

The graph looks like this

But we’re interested in the blue part only (since t ∈]0, ∞[), so in the end
the range is ] − ∞, −5[.

Tomasz Lechowski Batory A & A HL December 7, 2021 11 / 20

Short but important note must be made here. The blue part of the graph
of the quadratic is not the graph of f (x) (in particular the domain of f (x)
is all real numbers), but the ranges of these functions are the same.

Tomasz Lechowski Batory A & A HL December 7, 2021 12 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse,

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse, we just need to
find its domain.

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse, we just need to
find its domain.
Since x ∈ [−1, 1], then t = −x 2 + 9 ∈ [8, 9] (this is a simple quadratic, if
you struggle to understand, where these values came from, sketch the
function with the domain [−1, 1]).

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse, we just need to
find its domain.
Since x ∈ [−1, 1], then t = −x 2 + 9 ∈ [8, 9] (this is a simple quadratic, if
you struggle to understand, where these values came from, sketch the
function with the domain [−1, 1]).
We go back to f (t) = 2t ,

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse, we just need to
find its domain.
Since x ∈ [−1, 1], then t = −x 2 + 9 ∈ [8, 9] (this is a simple quadratic, if
you struggle to understand, where these values came from, sketch the
function with the domain [−1, 1]).
We go back to f (t) = 2t , the domain is t ∈ [8, 9] and 2t is an increasing
function,

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

Exercise 4

2 +9
Find the range of f (x) = 2−x for x ∈ [−1, 1].

We will set t = −x 2 + 9 to simplify things.

We get the function f (t) = 2t , which is easy to analyse, we just need to
find its domain.
Since x ∈ [−1, 1], then t = −x 2 + 9 ∈ [8, 9] (this is a simple quadratic, if
you struggle to understand, where these values came from, sketch the
function with the domain [−1, 1]).
We go back to f (t) = 2t , the domain is t ∈ [8, 9] and 2t is an increasing
function, so the range is [28 , 29 ], so [256, 512].

Tomasz Lechowski Batory A & A HL December 7, 2021 13 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.Examples
include f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x .

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.Examples
include f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x .

We can do what we did in the case a > 1, namely create a table and based
on that draw the graph.

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.Examples
include f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x .

We can do what we did in the case a > 1, namely create a table and based
on that draw the graph.
We will however look at this differently. Let’s compare f1 (x) = (0.5)x and
f2 (x) = 2x ,

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.Examples
include f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x .

We can do what we did in the case a > 1, namely create a table and based
on that draw the graph.
We will however look at this differently. Let’s compare f1 (x) = (0.5)x and
f2 (x) = 2x , we have:
 x
1
f1 (x) = = (2−1 )x = 2−x = f2 (−x)
2

What does this mean?

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1

Now we will consider the case f (x) = ax , where 0 < a < 1.Examples
include f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x .

We can do what we did in the case a > 1, namely create a table and based
on that draw the graph.
We will however look at this differently. Let’s compare f1 (x) = (0.5)x and
f2 (x) = 2x , we have:
 x
1
f1 (x) = = (2−1 )x = 2−x = f2 (−x)
2

What does this mean? It means that the graph of f1 (x) is a reflection of
the graph of f2 (x) in the y -axis.

Tomasz Lechowski Batory A & A HL December 7, 2021 14 / 20

0<a<1
So the graphs of f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x look as
follows (dotted lines represent graphs of 2x , 3x and 5x ):

Tomasz Lechowski Batory A & A HL December 7, 2021 15 / 20

0<a<1
So the graphs of f (x) = (0.5)x , g (x) = ( 13 )x , h(x) = (0.2)x look as
follows (dotted lines represent graphs of 2x , 3x and 5x ):

Tomasz Lechowski Batory A & A HL December 7, 2021 15 / 20

0<a<1

What do we see?

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

lim f (x) = 0.
x→∞

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

lim f (x) = 0.
x→∞

As x approaches minus infinity, the values of the function approach

infinity.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

lim f (x) = 0.
x→∞

As x approaches minus infinity, the values of the function approach

infinity. lim f (x) = ∞.
x→−∞

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

lim f (x) = 0.
x→∞

As x approaches minus infinity, the values of the function approach

infinity. lim f (x) = ∞.
x→−∞

The function is always positive

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

0<a<1

What do we see? The observation are similar:

For x = 0, the value of the function is 1.

The larger the argument, the smaller the value. So the function is
decreasing.

The domain is all real numbers.

As x approaches infinity, the values of the function approach 0.

lim f (x) = 0.
x→∞

As x approaches minus infinity, the values of the function approach

infinity. lim f (x) = ∞.
x→−∞

The function is always positive The range of values is ]0, ∞[.

Tomasz Lechowski Batory A & A HL December 7, 2021 16 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4

Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4
 x
1
We consider the function f (x) = , since 0 < 14 < 1, the function is
4
decreasing.

Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4
 x
1
We consider the function f (x) = , since 0 < 14 < 1, the function is
4
decreasing. We arrange the arguments first:

Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4
 x
1
We consider the function f (x) = , since 0 < 14 < 1, the function is
4
decreasing. We arrange the arguments first:
1 √ √
−1 < − < 3 < 2 < 5 < 3
2

Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4
 x
1
We consider the function f (x) = , since 0 < 14 < 1, the function is
4
decreasing. We arrange the arguments first:
1 √ √
−1 < − < 3 < 2 < 5 < 3
2
Since the function is decreasing (the larger the argument, the smaller the
value) we have:

Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20

Exercise 5
Arrange in ascending order:
 √5  √3  −1  − 1  3  2
1 1 1 1 2 1 1
, , , , ,
4 4 4 4 4 4
 x
1
We consider the function f (x) = , since 0 < 14 < 1, the function is
4
decreasing. We arrange the arguments first:
1 √ √
−1 < − < 3 < 2 < 5 < 3
2
Since the function is decreasing (the larger the argument, the smaller the
value) we have:
 3  √5  2  √3  − 1  −1
1 1 1 1 1 2 1
< < < < <
4 4 4 4 4 4
Tomasz Lechowski Batory A & A HL December 7, 2021 17 / 20
Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .
We need to find its domain.

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .
We need to find its domain. Since x ∈ [0, 3], then t = x 2 − 2x + 1 ∈ [0, 4]
(t = 0 for x = 1 and t = 4 for x = 3).

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .
We need to find its domain. Since x ∈ [0, 3], then t = x 2 − 2x + 1 ∈ [0, 4]
(t = 0 for x = 1 and t = 4 for x = 3).
f (t) is decreasing so we will get the least value for t = 4,

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .
We need to find its domain. Since x ∈ [0, 3], then t = x 2 − 2x + 1 ∈ [0, 4]
(t = 0 for x = 1 and t = 4 for x = 3).
f (t) is decreasing
√
so we will get the least value for t = 4,
3 4 1
f (4) = ( 3 ) = 9 and the greatest value for t = 0, f (0) = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

Exercise 6

 √ x 2 −2x+1
3
Find the set of values of f (x) = for x ∈ [0, 3].
3
√
3 t
We let t = x 2 − 2x + 1 and we get a much simpler function f (t) = ( 3 ) .
We need to find its domain. Since x ∈ [0, 3], then t = x 2 − 2x + 1 ∈ [0, 4]
(t = 0 for x = 1 and t = 4 for x = 3).
f (t) is decreasing
√
so we will get the least value for t = 4,
3 4 1
f (4) = ( 3 ) = 9 and the greatest value for t = 0, f (0) = 1.
So in the end the range is [ 19 , 1].

Tomasz Lechowski Batory A & A HL December 7, 2021 18 / 20

a=1

Finally the case f (x) = ax , where a = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 19 / 20

a=1

Finally the case f (x) = ax , where a = 1. This is a trivial case

f (x) = ax = 1x = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 19 / 20

a=1

Finally the case f (x) = ax , where a = 1. This is a trivial case

f (x) = ax = 1x = 1. So we have a constant function, whose graph is a
horizontal line y = 1.

Tomasz Lechowski Batory A & A HL December 7, 2021 19 / 20

a=1

Finally the case f (x) = ax , where a = 1. This is a trivial case

f (x) = ax = 1x = 1. So we have a constant function, whose graph is a
horizontal line y = 1. No more needs to be said about this case.

Tomasz Lechowski Batory A & A HL December 7, 2021 19 / 20

In case of any questions you can email me at t.j.lechowski@gmail.com.

Tomasz Lechowski Batory A & A HL December 7, 2021 20 / 20