Notes by Lex Toumbourou

2.101 Introduction

• Variable
  • In everyday life, many quantities depend on change in variables:
    • Plant's growth depends on sunlight and rainfall.
    • A runner's speed == how long it takes to run a distance.

2.102 The definition of a function

• Function
  • A function is a rule that relates how one quantity depends on another.
  • It's central to programming and Computer Science.
  • It's a relationship between a set of inputs and a set of outputs, where inputs map to exactly one output.
  • Function is a "well-behaved relation"
    • Given a starting point, we have just one ending point.
  • A function $f$ from a set of $A$ to a set of $B$ is an assignment of exactly one element of $B$ to each element of $A$.
    • If $f$ is the function from A to B, we write: $f: A \rightarrow B$
    • We can read this as f maps A to B:
      • $x \in A: \ \ x \rightarrow f(x) = y \ \ \ (y \in B)$
  • Domain of a Function
    • Given the function above, $A$ is the set of all inputs and called the "domain" of $f$.
      • Written as $D_f = A$

  • Co-Domain of Function

    • $B$ is the set containing the outputs and called the co-domain of $f$.
      • Written as $\text{co-}D_f = B$
    • The set of all outputs is called the range of f and is written as $R_f$.
    • $y$ is called the image of $x$.

    • $x$ is called the pre-image of $y$.

      

    • Example: a set mapping characters to a length.

      • $f(\text{Sea}) \rightarrow 3$ (contains 3 characters)
      • $f(Land) \rightarrow 4$ (contains 4 characters)
      • $f(on) \rightarrow 2$
        • 2 is the image of "on"
        • "on" is the pre-image of 2.
    • Conditions under which a relation is not a function:
      • Some inputs do not have an image.
      • Some inputs have more than one image.
    • Exercise 1
      • Given the following function: $f: Z \rightarrow Z$ with $f(x) = |x|$, what is domain, co-domain and range for function $f$?
    • Domain: $Z$
    • Co-domain: $Z$
    • Range: $Z^{+}$
    • Exercise 2
      • Given the following function: $g: R \rightarrow R$ with $g(x) = x^2 + 1$
    • Domain: R
    • Co-domain: R
    • Range: $\{1, 5, 9 ...\}$
    • Pre-images(5) = {-2, 2}

2.104 Plotting functions

• Linear Function

  • Linear function is of form: $f(x) = ax + b$

    • Where $a$ and $b$ are real numbers.
    • Straight-line function that passes through point (0, b).
    • $a$ is the gradient of the function. Where $a > 0$ the function is increasing.
      • That is: $x_1 \leq x_2$ then $f(x_1) \leq f(x_2)$.
    • Example of increasing linear function:

    

    • When the gradient is < 0, the function is decreasing.
      • $f: R \rightarrow R$
    • $f(x) = ax +b$
      • If $a > 0$ then function is increasing.

      • If $x_1 \leq x_2$ then $f(x_1) \leq f(x_2)$

        

• Quadratric Functions

  • Quadratic functions: $f(x) = ax^2 + bx + c$

    • Where $a$, $b$ and $c$ are real numbers and $a \ne 0$.

      

    • Domain of function f(x) is set of real numbers.

    • Range of function is set of positive numbers.
    • Exponential Functions
      • If base $b$ in $f(x) = b^x$, $b > 1$ then function is increasing and represents growth shown in this graph:

  

  * Graph also shows that the point $(0,1)$ is a "common point".
  * Domain is equal to set of all real numbers.
  * Range is equal to set of all real positive numbers.
  * X-axis is horizontal asymtot to curve of function.
  

  • If base 0 < b < 1, then function is decreasing:

    

    • Domain and range are the same as previous function.
    • Laws Of Exponential Functions
      • $b^xb^y = b^{x + y}$
      • $\frac{b^x}{b^y} = b^{x-y}$
      • $(b^x)^y = b^{xy}$
      • $(ab)^x = a^xb^x$
      • $(\frac{a}{b})^x = \frac{a^x}{b^x}$
      • $b^{-x} = \frac{1}{b^x}$

2.106 Injective and surjective functions

• Injective Function

  • A function is considered injective or one-to-one if and only if:

    • any 2 distinct inputs will lead to 2 distinct outputs.
    • In other words:
      • for all $a, b \in A, \text{ if } a \ne b \text{ then } f(a) \ne f(b)$
      • same as saying: $a, b \in A, \text{ if } f(a) = f(b) \text{ then } a = b$
    • Example on the left is an injective function, as every element of $A$ has a unique image in B.
    • Example on the right is not injective. 2 or 4 in A have the same image 0. 1 and 3 have the same image 1.

    

  • You can show a function is not injective by finding two different inputs $a$ and $b$ with the same Function Image.

  • An example with a linear function:
    • To show function $f: R -> R$ with $f(x) = 2x + 3$ is an injective function, we must show that $\text{ if } f(a) = f(b) \text{ then } a = b$
      • $f(a) = f(b)$ => $2a + 3 = 2b + 3$ => $2a = 2b$ => $a = b$ => f is injective.
    • Proof 2:
      • Let $a, b \in R$, show that $\text{ if } a \ne b \text{ then } f(a) \ne f(b)$
        • $a \ne b$ => $2a \ne 2b$ => $2a + 3 \ne 2b+3$ => $f(a) \ne f(b)$ => f is injective
  • An example quadratic function that is not injection.
    • Show function $f: R -> R$ with $f(x) = x^2$ is not an injective function
    • Example with 2 counter examples that have the same image.
      • One example is 5 and -5 have the same image.
        • $f(5) = (5)^2 = (-5)^2 = f(-5)$
          • Since: $-5 \ne 5$ it's not injective.
          • If we change domain to $R^{+}$, the function becomes injective.
      • Proof 1:
        • Let $a, b \in R^{+}$ show that if $f(a) = f(b)$ then $a = b$.
          • Let $a, b \in R^{+}$ show that if $f(a) = f(b)$ then $a = b$
      • Proof 2:
        • Let $a, b \in R^{+}$ show that if $a \ne b$ then $f(a) \ne f(b)$
        • $a \ne b => a^2 \ne b^2$ as $a, b \in R+ => f(a) \ne f(b) => f$ is injective.
  • Surjective Function
  • A function is said to be a surjective (onto) function if and only if every element of the co-domain of $f$, $B$, has at least one pre-image in the domain of $f, A$.
    • In other words, every element in the output domain has some input that will return it.

  • for all $y \in B$ there exists $x \in A$ such that $y = f(x)$

    • Equivalent to saying range and co-domain of surjective function are the same.
      • $\text{ CO}-D_f = R_f$

    • Examples:

      

  • An example Linear Function

    • Show that the function $f: R -> R$ with $f(x) = 2x+3$ is a surjective (onto) function.
    • Need to show that for any element $y \in R$, there exists $x \in \mathbb{R}$ such that $f(x) = y$
  • Proof:
    • $f(x) = y$ => $2x + 3 = y$ => $2x = y - 3$ => $x = \frac{y-3}{2} \in R$
    • Hence, for all $y \in R$, there exists $x = \frac{y-3}{2} \in R$ such that $f(x) = y$
  • An example quadratic function that is not surjective
    • Show that function $f: R-> R$ with $f(x) = x^2$ not a surjective (onto) functions
    • Proof: * Let $y \in R$, show that there exists $x \in R$ such that $f(x) = y$ * $R_f (\text{ set images }) = [0, + \infty [\ne R(co-D_f) = R$ * We know the range of $Rf$ is positive integers only: all negative images have no pre-images.
  • Examples:
  • Injective, not surjective
    • Injective because each element in the domain has a unique image.
    • Not surjective because the element 2 in the co-domain has no pre-image.
  • Surjective but not injective
    • Not injective because a and d are different but have the same image.
  • Injective and surjective
    • Each element has a unique image.
    • Each element in co-domain has at least one pre-image.
  • Neither injective nor surjective
    • Not injective because a and c have the same image.
    • Not surjective because the 4 element of co-domain has no pre-image.
  • Not a valid function
  • Input a has 2 outputs. In a function, an input can only have a single output.

2.109 Functions (Peer-graded Assignment)

Part 1

• *$f_1 : \mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = x^{2} + 1$
  • Claim: This function is not injective.
  • Proof:
    • Let $a = 2$, $b = -2$
    • $f(2) = (2)^2 + 1 = 5$
    • $f(-2) = (-2)^2 + 1 = 5$
    • $f(2) = f(-2)$ therefore the function is not injective.
  • Claim: This function is not surjective.
  • Proof:
    • $f(x) = y$
    • $x^2 + 1 = y$
    • $x^2 = y - 1$
    • $x = \sqrt{(y - 1)}$
    • $R \sqrt{(y - 1)} = [1, + \infty [$
    • $[1, + \infty [ \ \ne \mathbb{R}$ therefore, this function is not surjective.
• *$f_2 : \mathbb{R} \rightarrow [1, + \infty [ \text{ where } f(x) = x^{2} + 1$
  • Claim: This function is not injective.
  • Proof:
    • Let $a = 2$, $b = -2$
    • $f(2) = (2)^2 + 1 = 5$
    • $f(-2) = (-2)^2 + 1 = 5$
    • $f(2) = f(-2)$ therefore the function is not injective.
  • Claim: This function is surjective.
  • Proof:
    • $f(x) = y$
    • $x^2 + 1 = y$
    • $x^2 = y - 1$
    • $x = \sqrt{(y - 1)}$
    • $R _{\sqrt{(y - 1)}} = [1, + \infty [$ therefore, this function is surjective.
• $f_3: \mathbb{R} \rightarrow \mathbb{R} \text{ where } f(x) = x^3$
  • Claim: This function is injective.
  • Proof:
    • $f(a) = f(b)$
    • $f(a) = a^3$
    • $f(b) = b^3$
    • $a^3 = b^3$
    • $(a^3)^{1/3} = (b^3)^{1/3}$
    • $a = b$ for all $a, b \in \mathbb{R}$ there the function is injective.
  • Claim: This function is surjective
  • Proof:
    • $f(x) = y$
    • $x^3 = y$
    • $x = \sqrt[3]{y}$
    • $\sqrt[3]{y} \in \mathbb{R}$ therefore, the function is surjective.
• $f_4 : \mathbb{R} \rightarrow \mathbb{R} \text{ where } f(x) = 2x + 3$
  • Claim: This function is injective.
  • Proof:
    • $f(a) = f(b)$
    • $2a + 3 = 2b + 3$
    • $2a = 2b$
    • $a = b$ therefore f is injective.
  • Claim: This function is surjective.
  • Proof:
    • $f(x) = y$
    • $2x + 3 = y$
    • $2x = y - 3$
    • $x = \frac{y-3}{2} \in \mathbb{R}$ therefore the function is surjective.
• $f_5: \mathbb{Z} \rightarrow \mathbb{Z} \text{ where } f(x) = 2x + 3$
  • Claim: This function is injective.
  • Proof:
    • $f(a) = f(b)$
    • $2a + 3 = 2b + 3$
    • $2a = 2b$
    • $a = b$ therefore $f$ is injective.
  • Claim: This function is not surjective
  • Proof:
    • $f(a) = 2a + 3$
    • $f(x) = y$
    • $2x + 3 = y$
    • $2x = y - 3$
    • $x = \frac{y-3}{2} \notin \mathbb{Z}$ for all $x$ therefore this function is not surjective.

Part 2

Let $f : \mathbb{R} \rightarrow ]1, +\infty[$ with $f(x) = e^{x} + 1$

1. Show that $f(x)$ is bijection.

   • Claim: $f(x)$ is a bijective as it is both injective and surjective.
   • Proof of injective:
     • $f(a) = f(b)$
     • $f(a) = e^a + 1$
     • $f(b) = e^b + 1$
     • $e^a = e^b$
     • $a = b$ therefore the function is injective.
   • Proof of surjective:
     • $R \ {e^x} = [1, +\infty[$
     • $\text{Co-D } e_x = [1, +\infty[$
     • $R \ {e^x} = \text{Co-D } e_x$ therefore the function is surjective.

2. Find the inverse function $f^{-1}$.

   • $f(x) = y$
   • $e^x + 1 = y$
   • $e^x = y - 1$
   • $x = \log_e(y - 1)$
   • $f^{-1}(x) = log_e(x - 1)$

3. Plot the curve of $f$ and $f^{-1}$ in the same graph.

4. What can you say about these two curves?

The curves are symmetric with respect to the line $y = x$.