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1 2 Function 2.1 Function notation The equation f(x) = x defines a function f from the set R of real numbers to itself (written f : R R). This function accepts an input value x and returns an output value f(x), which, according to the equation, is obtained by squaring the input value and adding 3. For instance, when the input is 5, the output is f(5), which is 28. One way to study a given function is to make a table of input values x versus output values f(x), here illustrated for the function f(x) = x 2 + 3: f(x) x Input values can be chosen to be anything, but they are usually chosen to be convenient numbers (often integers) close to the region of interest (often not far from 0). A Venn diagram provides a useful way to think about a function. Pictured below is a Venn diagram representation of the function f(x) = x It shows a general input value x being mapped to the corresponding output value f(x) as well as how this mapping works for the chosen inputs. The bubble on the left represents the domain of f (here, R), which is the set of input values. The bubble on the right represents the codomain of f (here, R), which is a set containing the output values. 1

2 2 FUNCTION 2 The range of f is the set of all values that actually occur as outputs. Here, the range is the set [3, ) of all numbers greater than or equal to 3 (the smallest x 2 can be is 0). If a function f has domain A and codomain B we write f : A B and say that f is a function from A to B. Strictly speaking, in order to define a function, one must specify the domain and the codomain as we did above when we said that f was a function from the set R of real numbers to itself. However, if a function is given by means of a formula, like f(x) = x 2 + 3, we assume that the domain is taken to be the set of all real numbers x for which the expression on the right makes sense (usually meaning no division by 0 or even roots of negative numbers), and the codomain is taken to be the set of all real numbers. Instead of saying something like the function f given by f(x) = x 2 + 3, we will often say the function f(x) = x 2 + 3, or even just the function x when there is no need to give the function a name Example Let f(x) = 2x 3 x 1. (a) Find f(5). (b) Find f(x + h). (c) Find the domain of f. Solution 5: (a) We get f(5) by replacing every occurrence of x in the formula by f(5) = 2(5) = 7 2. (b) The expression f(x + h) will come up when we talk about the derivative of a function. It is computed just as before: Replace every occurrence of x in the formula by x + h. If this is found to be confusing, it might help to imagine the formula as being f( ) = 2( ) 3, 1 so that f(x + h) is obtained by putting x + h in every box: f(x + h) = The parentheses in 2(x + h) are essential. 2(x + h) 3 x + h 1. (c) Since the square root of a negative number is undefined, the formula requires x 1 0. But division by zero is also undefined, so there is the further requirement x 1 0. Putting these together, we get x 1 > 0, which is the

3 2 FUNCTION 3 same as saying x > 1. Therefore, the domain of f is (1, ), the set of all real numbers greater than 1. Incidentally, the codomain of f in the example is taken to be R as usual. It is usually not an easy task to find the range of a function, and that is the case in this example. An accurate graph of the function (see 2.2) usually helps considerably; we will find that calculus provides a means for producing accurate graphs. 2.2 Graph of function Let f be a function (by which we always mean a function from some subset of R to R). The graph of f is the set of all points (x, f(x)) with x in the domain of f Example Sketch the graph of the function f given by f(x) = x 2. Solution We are to depict all points in the plane of the form (x, f(x)) (that is, (x, x 2 )), with x in the domain of f, which is R. We make a table of inputs versus outputs for some conveniently chosen input values, plot those points, and then connect them to form the graph: x f(x) This method of plotting points and connecting them to form the graph brings up a question: How do we know that the graph proceeds smoothly between the points we plotted and does not have, say, ripples? It turns out that calculus gives the answer; it shows that the graph is in fact smooth. We will see why later, but for now we will take it on faith that this is the case. The graph of the function f(x) = x 2 is the same as the graph of the equation y = x 2 (so, replace f(x) by y). The graph of y = x 2 is the set of all points (x, y)

4 2 FUNCTION 4 that satisfy the equation. More generally, the graph of a function f is the graph of the equation y = f(x) Example Sketch the graph of the function f given by { f(x) = 1 x, x 0 2, x > 0. Solution This is what is called a piecewise-defined function. For x 0, the function is given by f(x) = 1 x, and for x > 0, the function is given by f(x) = 2. The graph is the same as that of the line y = 1 x all the way up to and including x = 0, and then it changes to the graph of the horizontal line y = 2 from that point on: It was mentioned earlier that having an accurate graph aids one in the determination of the range of a function. Since the graph of f consists of all points (x, f(x)), the range of f is the set of y-coordinates of the points on the graph. As a practical matter, one can quickly visualize the range by projecting the graph horizontally (from both directions) onto the y-axis. In the first example above, the range is [0, ) (which we could have determined without looking at the graph). In the second example, the range is [1, ). 2.3 Composition of functions The function f(x) = (2x + 3) 2 can be thought of as being built up from the two simpler functions g(x) = 2x + 3 and h(x) = x 2 : f(x) = (2x + 3) 2 = (g(x)) 2 = h(g(x)). For a given input x, the function g produces the output g(x). With this output used as input, the function h produces the output h(g(x)). According to the equation above, the function f has the same effect as this two-step process.

5 2 FUNCTION 5 The composition of the functions g and h is the function h g given by (h g)(x) = h(g(x)). Here is a Venn diagram depiction of the composition h g: For the particular functions defined above, we have the relationship f = h g (which means that f(x) = (h g)(x) for all x) Example Find (h g)(x), given that g(x) = x+1 and h(x) = x 2x. Solution We have (h g)(x) = h(g(x)) = g(x) 2g(x) = x + 1 2(x + 1). In the example, g has domain R, while h g has domain [ 1, ). This shows that the domain of a composition can be smaller than the domain of the first function applied. 2.4 Identity function The simplest function is the identity function ε defined by ε(x) = x. Given the input value x, it returns the output value x, so, in effect, it does nothing. The graph of this function is the graph of the equation y = x (the 45 line through the origin). For any function f, we have ε f = f and f ε = f, since (ε f)(x) = ε(f(x)) = f(x) and (f ε)(x) = f(ε(x)) = f(x). With composition viewed as a sort of multiplication, these equations show that ε acts like a multiplicative identity (like the number 1). This is why ε is called the identity function.

6 2 FUNCTION Inverse function Let f(x) = x 3 and g(x) = 3 x. These functions satisfy g(f(x)) = x and f(g(x)) = x, since g(f(x)) = 3 f(x) = 3 x 3 = x and f(g(x)) = (g(x)) 3 = ( 3 x) 3 = x. The equation g(f(x)) = x says that an input x, changed by f to f(x), gets changed back to x by g, that is, g undoes what f does. For example, 2 f 8 g 2. Similarly, the equation f(g(x)) = x says that f undoes what g does. We say that g is an inverse of f. Existence of inverse function. Not every function has an inverse. For instance, the squaring function f(x) = x 2 has no inverse. One might suppose that the square root function g(x) = x undoes what f does, since, for instance, 2 f 4 g 2. However, it does not work for every input value. For example, 2 f 4 g 2. The problem is that the squaring function f sends two inputs to the same output (both 2 and 2 get sent to 4). Put another way, the graph of f fails the horizontal line test. (A graph is said to pass the horizontal line test if each horizontal line intersects the graph in at most one point.) Indeed, the horizontal line y = 4 intersects the graph of f at both ( 2, 4) and (2, 4). The remedy is to use a smaller domain for f, that is, to allow fewer input values. If we declare the domain of f to be just the set [0, ) of nonnegative real numbers, then it is no longer the case that f sends two inputs to the same output (for example, 2 is no longer an allowable input). In this new situation, the left half of the graph of f disappears and we are left with a graph that passes the horizontal line test. With this adjustment, the square root function g(x) = x does indeed serve as an inverse for f Example (a) Find 4. (b) Solve the equation x 2 = 4. Solution (a) The square root function is the inverse of the squaring function with reduced domain [0, ), so 4 is the number in the interval [0, ) that you square to get 4. Therefore, 4 = 2. (b) Here, we are searching for all real numbers x satisfying the equation. The answer is x = ±2.

7 2 FUNCTION 7 This example is intended to serve as an illustration in a familiar setting of an analogous example given later in the less-familiar setting of inverse trigonometric functions (see Example 4.4.1). Also, it is intended to correct the common misconception that a is ± the number you square to get a. Rather, a is the number 0 that you square to get a. The issue of ± comes up only in solving equations like in (b). A function f is injective if it never sends two inputs to the same output (i.e., its graph passes the horizontal line test). If f is injective, then it has an inverse function that sends an output of f back to the (unique) input that produced the output. Definition of inverse function. Let f : A B be an injective function with range B. The function f 1 : B A defined by is called the inverse of the function f. f 1 (y) = x f(x) = y (1) Inverse function identities. With notation as above (i) f 1 (f(x)) = x for all x in A, (ii) f(f 1 (x)) = x for all x in B. If a function is not injective, then it does not have an inverse, but we can reduce its domain in order to achieve an injective function just as we did for the squaring function. This process is illustrated here using Venn diagrams: Example Let f(x) = x

8 2 FUNCTION 8 (a) Explain why f has no inverse as is. (b) From here on, assume that the domain of f has been reduced to [ 2, ). Then the function f : [ 2, ) [0, ) is injective and its range is [0, ) so that it has an inverse f 1 : [0, ) [ 2, ). Find a formula for the function f 1. (c) Verify that f and f 1 satisfy the inverse function identities (i) and (ii) (see 2.5). Solution (a) f( 3) = 1/3 = f( 1) so f is not injective (the two inputs 3 and 1 produce the same output 1/3). (b) For x in the restricted domain [ 2, ), the quantity x+2 is always greater than or equal to 0, so the absolute value sign in the formula can be dropped giving f(x) = (x+2)/3. By (1) in the definition of inverse function, f 1 (y) is the x satisfying f(x) = y, that is, the x satisfying (x+2)/3 = y. Solving for x, we get x = 3y 2, so f 1 (y) = 3y 2. Replacing y with the more traditional letter x, we get f 1 (x) = 3x 2. (c) For x [ 2, ), we have f 1 (f(x)) = f 1 ( x + 2 ) = 3( x + 2 ) 2 = (x + 2) 2 = x, 3 3 and for x [0, ), we have as desired. f(f 1 (x)) = f(3x 2) = (3x 2) = 3 x 3 = x = x, The inverse function identity (i) can be written (f 1 f)(x) = ε(x), which says that f 1 f = ε. Similarly, (ii) can be written f f 1 = ε. With composition viewed as multiplication and ε viewed as a multiplicative identity (as in 2.4), these equations say that f 1 is a multiplicative inverse of f. This is the reason for the notation f 1 as well as the terminology inverse function. The trigonometric functions (see 4.2) are not injective, but we will reduce their domains in order to achieve injective functions, just as we did for the squaring function, and this will enable us to define the inverse trigonometric functions (see 4.4). Graph of inverse function. The graph of the function f(x) = x 2, with domain reduced to [0, ), is shown below together with the graph of its inverse function f 1 (x) = x.

9 2 FUNCTION 9 We see that the graph of f 1 is the reflection of the graph of f across the 45 line y = x. If the graph of f is drawn, and the paper is folded along the dotted line before the ink dries, then the graph of f 1 is produced (albeit, in a different color here). This is a general property relating the graph of a function f to the graph of its inverse f 1 (if such exists). It amounts to saying that the point (x, y) is on the graph of f if and only if the point (y, x) is on the graph of f 1, and this is precisely what (1) says Example Let f be as in Example 2.5.2, with domain [ 2, ). Sketch the graph of f together with the graph of f 1. Solution As noted in the earlier example, for x in the domain [ 2, ), the function f is given by f(x) = (x + 2)/3. The graph of f is the same as the graph of the equation y = (x + 2)/3 = 1 3 x + 2 3, which is a line with slope 1 3 and y-intercept 2 3. The graph of the inverse f 1 is obtained by reflecting across the 45 line y = x: (By the earlier example, f 1 (x) = 3x 2, the graph of which is the same as that of y = 3x 2, a line with slope 3 and y-intercept 2. This agrees with the graph of f 1 as shown above.)

10 2 FUNCTION 10 2 Exercises 2 1 Let f be the function given by f(x) = x 2 (a) Find f(5). (b) Find f(x + h). (c) Find the domain of f x x Sketch the graph of the function f given by 1 x 2, x < 0, f(x) = 1 2x, 0 x 1, 1/x, x > Let f(x) = 2 x + x 2 and g(x) = x 1. (a) Find (f g)(x). (b) Find (g f)(3). 2 4 Let f(x) = (x 1) 2 1. (a) Explain why f has no inverse as is. (b) From here on, assume that the domain of f has been reduced to [1, ). Then the function f : [1, ) [ 1, ) is injective and its range is [ 1, ) so that it has an inverse f 1 : [ 1, ) [1, ). Sketch the graph of f together with the graph of f 1. (c) Find a formula for the function f 1. (d) Verify that f and f 1 satisfy the inverse function identities (i) and (ii) (see 2.5).