Math 1 (MA111) Lecture 2 PDF

## Math 1 (MA111) ### Lecture # 2 ### Function - **Definition:** Let X, Y be non-empty sets, Let f: X →Y be a relation from X into Y. - **How to show that f(x) is a function:** - **Algebraically:** Show that: for all x∈X there exists a unique y∈Y : y = f(x) - **Graphically:** Use the follo...

## Math 1 (MA111) ### Lecture # 2 ### Function - **Definition:** Let X, Y be non-empty sets, Let f: X →Y be a relation from X into Y. - **How to show that f(x) is a function:** - **Algebraically:** Show that: for all x∈X there exists a unique y∈Y : y = f(x) - **Graphically:** Use the following Vertical line test: A relation f is a function if and only if its graph intersects each vertical line at most once - X = D₁ =Domain, Y = C₁ =Co-domain - **Example1:** the relation f(x) = 2x+1 is a function using the two methods. - **Example2:** the relation x² + y² = 9 is _NOT_ a function using the two methods. ### The Concept of a Function - **DEFINITION Function:** A function from a set D to a set Y is a rule that assigns a unique (single) element f(x) ∈ Y to each element x ∈ D. - **We usually write y = f(x)** - **Input (domain)** _x_ - **Output (range)** _f(x)_ ### Domain and Range of a function - **D = domain set** - **Y = set containing the range** - A function from a set D to a set Y assigns a unique element of Y to each element in D. ### Examples of functions | Function | Domain (x) | Range (y) | |---|---|---| | y = x² | (-∞, ∞) | [0, ∞) | | y = 1/x | (-∞, 0) U (0, ∞) | (-∞, 0) U (0, ∞) | | y = √x | [0, ∞) | [0, ∞) | | y = √4 - x | (-∞, 4] | [0, ∞) | | y = √1 - x² | [-1, 1] | [0, 1] | ### One-to-One function - **DEFINITION One-to-One Function:** A function f(x) is one-to-one on a domain D if f(x1) ≠ f(x2) whenever x1 ≠ X2 in D. Equivalently, f(x1) = f(x2) implies x₁ = X2 - **How to show that f(x) is one-to-one:** - **Algebraically:** Show that: f(x₁) = f(x2) implies X₁ = X2 - **Graphically:** Use the following Horizontal line test: ### The Horizontal Line Test for One-to-One Functions - A function y = f(x) is one-to-one if and only if its graph intersects each horizontal line at most once. - **Exactly one point:** - One-to-one: Graph meets each horizontal line at most once. - **Exactly two points:** - Not one-to-one: Graph meets one or more horizontal lines more than once. - **Using the horizontal line test, we see that y = x³ and y = √x are one-to-one on their domains (-∞, ∞) and [0, ∞), but y = x² and y = sin x are not one-to-one on their domains (-∞, ∞).** - **More than two points** ### Exercises: **Show algebraically and graphically that the following functions are one-to-one:** 1. f(x) = 2x + 1 2. f(x) = x³ 3. f(x) = (3x + 5)1/3 + 3 (HW) ### Solution (algebraically) (1) Let f(x1) = f(x2), then 2x1 + 1 = 2x2 + 1, then x₁ = x2, hence f(x) is one –to – one. ### Inverse Function **Definition:** If f is a function with domain D and range R, then a function g with domain R and range D is the inverse of the function fiff: (1) The function f is a One-to-One function (2) g(f(x)) = x, for all x in D and f(g(y)) = y, for all y in R. **Notation:** Usually g is denoted by f-1. So we write f-1 (f(x)) = x and f(f-1 (y)) = y It is clear that: if x = f¯¹ (y), then y = f(x) ### How to show that a given function is the inverse of another function **Example:** Show that f(x) is the inverse of g(x) where, (1) f(x) = 2x+1, g(x) = (x-1)/2 (2) f(x) = x³, g(x) = x1/3 (3) f(x) = (3x + 5)1/3 + 3, g(x)=[(x-3)3-5]/3 **Solution:** We have to prove that: (i) The function f(x) is one - to – one (ii)g(f(x)) = x, for x in D and f(g(y)) = y, for y in R ### How to sketch the graph of f-1, if you know the graph of f - (1) Draw the line y = x - (2) Reflect the graph of f on this line to get the graph of f-1. ### How to find f-1(x) if you know f(x) **Find and graph the inverse function of each of the following functions:** 1. y = (1/2)x + 1 2. y = vx 3. y = (x + 5)1/3 + 3 (HW) 4. y = 2 + (2x - 4)-1 (what do you notice?) **First you have to show (from the graph) that the given functions are One-to-One** ### Solution: (1) Since y = (1/2)x + 1, (i) Put f(x) = (1/2)x + 1 (ii) Replace every x by f1(x) we get f(f1(x)) = (1/2) f1(x) +1 (iii) Replace f¹(f(x)) by x We get, x = (1/2) f1(x) +1 (iv) Solving for f1(x) we get 2x = f1(x) + 2 Hence, f1(x) = 2x - 2 **There is another practical method. Can you find it?** **f(x) = (1/2)x + 1 and f¯¹(x) = 2x - 2 together shows the graphs' symmetry with respect to the line y = x.** ### Use the first method to do the algebraic solution by yourself ### Solution: (the second method): (2) Let y = √x, (i) Solving for x we get x = y2 (ii) Interchange the x with the y we get y = x2 (iii) Replace y by f1(x) to get, f1(x) = x² ### The functions y = √x and y = x², x ≥ 0, are inverses of one another ### Thank you Dr. Ahmad Moursy