Name: PAULIN T.

Assignment 1

1. Let[pic], where [pic] is the set of integers and[pic].

F(x) is a one-to-one function if we can show that:
For [pic] and[pic], [pic]=[pic]===> [pic]=[pic]
Let’s find out: [pic]==> [pic]=[pic], and [pic]=[pic]
So, [pic]=[pic]==> [pic]=[pic].
Subtracting 101 on both sides gives [pic]==> [pic]=[pic].
Since we’re able to show that [pic]=[pic] ==> [pic]=[pic], we then conclude that [pic] is a one-to-one function.

2. Let’s prove that [pic] is a one-to-one function.
To prove that, we have to show that
For [pic] and[pic], [pic]=[pic]===> [pic]=[pic]
So, [pic]==> [pic] and[pic].
[pic]=[pic]==> [pic]=[pic]
Subtracting -3 on both sides gives [pic]=[pic]
By simplifying 4, we easily have [pic]=[pic], hence the result!!!
Since we proved that [pic]=[pic]===> [pic]=[pic], we easily conclude that
[pic] is a one-to-one function.
3. The ceiling function maps every real number to the smallest integer greater than or equal to that number,[pic], where [pic] is the smallest integer greater than or equal to [pic].

a) Is the ceiling function a one-to-one function?
Let’s find out: The ceiling function is a one-to-one function if and only if for [pic] and[pic], [pic]=[pic]===> [pic]=[pic]

By considering these real numbers, [pic]=5.60 and [pic]=5.88, if we apply the ceiling function to both real numbers; we’ll have the following output: [pic]=6 and [pic]=6.
We notice here that [pic]=[pic]=6, but[pic][pic][pic].
We see that using the ceiling function, the image 6 can have more than one preimage. We can then conclude that the ceiling function is not one-to-one function.

b) Five numbers that have the same image:
The following numbers: 5.60; 5.75; 5.88; 5.92; 5.99 all have the same image through the ceiling function. Their image is 6.

4) Find [pic] when [pic]

When [pic], [pic] grows as fast as [pic].
When [pic], [pic] grows as fast as [pic].

As [pic], the fraction behaves like [pic]
So [pic]= n=[pic] when [pic]

5) Find [pic] as [pic]

As [pic], [pic] grows as fast as [pic].
As [pic], [pic] grows as fast as [pic].

As [pic], the fraction behaves as [pic].
So [pic]= [pic] as [pic]

6) Find [pic], as [pic]

As [pic], [pic] grows as fast as [pic].
As [pic], [pic] grows as fast as [pic].

So as [pic], the fraction behaves like [pic]

So [pic]=[pic], as [pic] (Because lim[pic] as [pic])

7) Find [pic], as [pic]

Let [pic]

As [pic], [pic]

So [pic][pic], as [pic]

8) Let’s Prove that [pic] is a continuous function…

We have to prove that[pic], as [pic].

[pic] =[pic]
Subtracting [pic] and -4 on both sides give us:[pic]
As [pic], so is the product [pic].
We then deduce that [pic] is a continuous function…

9) Let’s find the derivatives of the following functions, using the differentiation rules:

a) [pic]
Let’s consider[pic], where [pic]and [pic]
So, from [pic] above, C=constant=9 and [pic].
By applying the differentiation RULE 1 on[pic], we have:
[pic]then [pic].
Since [pic], applying the differentiation RULE 2, we have:
[pic]. Since [pic], and [pic]
[pic]
We therefore deduce that the derivative of [pic] is:[pic]

b) [pic]
Let’s assume that [pic], where C=const=3 and [pic]
By applying the second differentiation Rule (Rule 2) to [pic], we have:
[pic]. Since [pic]. By replacing C by its value, we can easily say that the derivative of [pic]is:[pic]

c) [pic]
Just like equation a) above, let’s say that [pic], where [pic], [pic] and [pic]. In this case, C=3 and [pic].
-Rule 1 of differentiation: [pic] then [pic].
- Rule 2 of differentiation: [pic] then [pic].
Since [pic], we will apply the Rule 4 of differentiation to find the derivative of [pic].
-Rule 4 of differentiation: [pic], then [pic].
Also, [pic]==>[pic]. With C=3, we can easily conclude that the derivative of [pic]is: [pic].
d) [pic]
Just like equation a) above, let us assume that [pic]
Where [pic] and [pic]
-Rule 1 of differentiation: [pic] then [pic].
-Rule 1 of differentiation on [pic]:[pic] then [pic].
- Rule 2 and 4 of differentiation on [pic]: [pic]
- Rule 1 and 2 of differentiation on [pic]:
[pic] then [pic]
So, we deduce that the derivative of [pic] is: [pic]

e) [pic]
Let [pic], where [pic] and [pic].
-Rule 3 of differentiation on [pic]:
[pic]
[pic], [pic]
By applying Rule 3 of differentiation on [pic], we have:
[pic]
We then conclude that the derivative of [pic] is:
[pic]
f) [pic]
Similar to equation e) above, let assume that [pic], with
[pic], [pic]and [pic]
-Rule 3 of differentiation on [pic]:
[pic]
[pic]
[pic]. With [pic]
Applying Rule 3 of differentiation gives us:
[pic]So, we can say that the derivative of [pic] is: [pic]

g) [pic]
Just like in equation d) above, let us assume that [pic]
Where [pic] and [pic]
-Rule 1 of differentiation: [pic] then [pic].
-Rule 1 of differentiation on [pic]:[pic] then [pic]
- Rule 2 and 4 of differentiation on [pic]: [pic]
- Rule 1 and 2 of differentiation on [pic]:
[pic] then [pic]
So, we deduce that the derivative of [pic] is: [pic]

10) Find the derivative of [pic] at [pic].
By using the definition of the derivative of a function: [pic](if it exists as [pic]) is called a derivative of [pic]at the point [pic].
So for [pic], let’s find the value of its derivative [pic] at the point [pic], that’s [pic].
[pic]=[pic]=[pic], as [pic]
By expanding and simplifying[pic], we have: [pic]=lim[pic] as [pic]
So, [pic]=lim[pic] as [pic]. We deduce that [pic]=[pic]
At the point [pic], [pic]
We then conclude that the derivative of [pic] at [pic] is: [pic].

11. Find the differentials of the functions:
a) [pic]
Differential of [pic] is by definition:[pic] Applying it for the above function, we’ll have: [pic]
But [pic] and [pic].
This gives us: [pic]
Simplifying by dx on both sides, we finally have: [pic]=[pic]

b) [pic]
Differential of [pic] is by definition:[pic] Applying it for the above function, we’ll have:
[pic], applying the rules [pic] and [pic]
We then have: [pic]
Simplifying by dx on both sides, we finally have: [pic]

c)[pic]
Differential of [pic] is by definition:[pic] Applying it for the above function, we’ll have:
[pic]Applying the rules [pic], [pic]and [pic],
We have: [pic]
Simplifying by dx on both sides, finally gives us: [pic]
Complementary Problems:

1) Let [pic] and [pic], where[pic]. Find [pic] and [pic]as well as the domain and range of these functions.

* [pic], Simply replace [pic] by [pic].
Here, the domain of [pic] is R. The Range for [pic] is the set of all positive numbers greater than or equal to zero.

*[pic]
So, Here, the domain of [pic] is R. The Range for [pic] is the set of all positive numbers greater than or equal to zero.

2) Let’s prove using the definition of a limit that [pic], as [pic]
Let’s consider the sequence[pic]. We say that a number [pic]is a limit of the sequence of numbers [pic]if for each very small number [pic], an infinite number of elements of that sequence will be located inside the ε -neighborhood of a and only a finite number of them will be located outside of the ε -neighborhood.

As [pic], [pic] grows bigger and bigger to infinity. But on the other hand, [pic], hence we have the result that [pic], as [pic].

3)Let’s find the derivative of [pic]by using the definition of the derivative:
By using the definition of the derivative of a function: [pic](if it exists as [pic]) is called a derivative of [pic].
So [pic]=[pic], as [pic]. Simplifying the top and bottom of the fraction with [pic], we have:

[pic]=[pic], as [pic]. We finally have: [pic]

We therefore deduce that the derivative of [pic] is: [pic]