Calculus of Several Variables, Lecture 7: Functions

7. Functions Definition 7.1. A function f : A −→ B consists of: (1) the set A, which is called the domain of f , (2) the set B, which is called the range of f , (3) a rule which assigns to every element a ∈ A an element b ∈ B, which we denote by f (a). It is easy to write down examples of functions: (1) Let A be the set of all people and let B = [0, ∞). Let f (x) be the height of person x, to the nearest inch. (2) f : R −→ R given by f (x) = x2 . (3) f : R −→ [0, ∞) given by f (x) = x2 . (4) f : [0, ∞) −→ R given by f (x) = x2 . (5) f : [0, ∞) −→ [0, ∞) given by f (x) = x2 . (6) f : R −→ {0, 1} given by � 0 if x is rational f (x) = 1 if x is irrational. (7) f : N −→ {0, 1} given by � 0 if x is even f (x) = 1 if x is odd. Definition 7.2. If f : A −→ B and g : B −→ C are two functions, then we can compose f and g to get a new function g ◦ f : A −→ C. If a ∈ A, then (g ◦ f )(a) = g(f (a)). If f : R −→ R is the function f (x) = sin x and g : R −→ R is the function g(x) = 2x, then g ◦ f : R −→ R is the function (g ◦ f )(x) = 2 sin x, whilst f ◦ g : R −→ R is the function (f ◦ g)(x) = sin 2x. Definition 7.3. Let f : A −→ B be a function. We say that f is injective if whenever f (a1 ) = f (a2 ) for some pair a1 , a2 ∈ A, then in fact a1 = a2 . We say that f is surjective if for every b ∈ B, we may find an a ∈ A such that f (a) = b. We say that f is a bijection if f is both injective and surjective. It is interesting to go through the examples above. The function in (1) is neither injective or surjective. Indeed there are lots of people who are five foot nine, so just because there are two people with the same height, does not mean they are the same person. So f is not injective. On the other hand, there is no person who is a million inches high, so f is not surjective either. 1 The function in (2) is neither injective nor surjective as well. f (−1) = � −1. There is no real number whose square is −1, so 1 = f (1), but 1 = there is no real number a such that f (a) = −1. The function in (3) is not injective but it is surjective. f (−1) = f (1), and 1 �= −1. But if b ≥ 0 then there is always a real number a ≥ 0 such that f (a) = b (namely, the square root of b). The function in (4) is injective but not surjective. If f (a1 ) = f (a2 ), then a21 = a22 . As both a1 ≥ 0 and a2 ≥ 0, this implies a1 = a2 . On the other hand, there is still no number whose square is −1. The function in (5) is bijective. It is injective, as in (4) and it is surjective as in (3). The function in (6) is not injective but it is surjective. f (0) = 0 = � 1. On the other f (1), but 0 = √ hand, if b ∈ {0, 1} then either b = 0 or b = 1. But f (0) = 0 and f ( 2) = 1. The function in (7) is not injective but it is surjective. f (0) = f (2) = � 2. On the other hand, if b ∈ {0, 1} then either b = 0 or 0, but 0 = b = 1. But f (0) = 0 and f (1) = 1. (8) (9) (10) (11) f : R3 −→ R given by f (�v ) = �v�. f : R3 − {0} −→ R3 given by f (�v ) = ��vv� . r : R −→ R3 given by f (t) = (cos t, sin t, t). f : R −→ (0, ∞) is given by f (x) = ex . The function in (8) is neither injective nor surjective. There are plenty of unit vectors and there are no vectors of negative length. The function in (9) is neither injective nor surjective. There are plenty of vectors which point in the same direction and the image consists of vectors of unit length. The function in (10) is injective but not surjective. The function in (11) is bijective. If f : A −→ B is a bijective function, then f has an inverse function g : B −→ A. g ◦ f : A −→ A is the identity, it sends a to a. Similarly f ◦ g : B −→ B is the identity, it sends b to b. The inverse of the exponential function is the logarithm g : (0, ∞) −→ R given by g(x) = ln x. Definition 7.4. If A and B are two sets, the product of A and B, denoted A × B is the set of ordered pairs A × B{ (a, b) | a ∈ A, b ∈ B }. The graph of a function f : A −→ B is the subset of the product { (a, f (a) | a ∈ A } ⊂ A × B. The product R × R = R2 , R × R2 = R3 , and so on. 2 If f : R −→ R is the function f (x) = x2 , then the graph of f is the parabola, y = x2 , { (x, x2 ) | x ∈ R } ⊂ R2 . Now suppose that g : R2 −→ R is the function g(x, y) = x2 + y 2 . The graph of g is the set z = x2 + y 2 , { (x, y, x2 + y 2 ) | (x, y) ∈ R2 } ⊂ R3 . One way to picture g is to slice it using level curves and level sets. If g : R2 −→ R is any function, the level curve of g at height c is the set { (x, y) ∈ R2 | g(x, y) = c } ⊂ R2 . 2 2 In the example when √ f (x, y) = x + y , the level curves are concentric circles of radius c, centred at the origin. If c = 0 the level curve is the origin and if c < 0 the level curve is empty. Note that graph has a minimum at the origin. If g : R3 −→ R is any function, the level set of g at height c is the set { (x, y, z) ∈ R3 | g(x, y, z) = c } ⊂ R3 . If g(x, y, z) = z − x2 − y 2 then the level sets are parallel paraboloids. It is interesting to try to visualise various functions. If we consider f (x, y) = −x2 −y 2 , then this is an upside down paraboloid, which has a maximum at the origin. Note that the level curves are the same as the level curves for f (x, y) = x2 + y 2 (for different values of c, of course). Now consider the function f (x, y) = x2 − y 2 . If we consider fixing y = b and varying x, then we get a parabola z = x2 − b2 . If we consider fixing x = a and varying y, then we get an upside down parabola, z = a2 − y 2 . In fact the origin is a saddle point, a point which is neither a maximum nor a minimum. 3