G. GRAPHING FUNCTIONS

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1 G. GRAPHING FUNCTIONS To get a quick insight int o how the graph of a function looks, it is very helpful to know how certain simple operations on the graph are related to the way the function epression looks. We consider these here.. Right-left translation. Let c > 0. Start with the graph of some function f(). Keep the -ais and y-ais fied, but move the graph c units to the right, or c units to the left. (See the pictures below.) You get the graphs of two new functions: { { right f( c) () Moving the f() graph c units to the gives the graph of. left f( + c) If f() is given by a formula in, then f( c) is the function obtained by replacing by c wherever it occurs in the formula. For instance, f() = 2 + f( ) = ( ) 2 + ( ) = 2, by algebra. Eample. Sketch the graph of f() = Solution. By algebra, f() = ( ) 2. Therefore by (), its graph is the one obtained by moving the graph of 2 one unit to the right, as shown. The result is a parabola touching the -ais at =. 2 ( ) 2 To see the reason for the rule (), suppose the graph of f() is moved c units to the right: it becomes then the graph of a new function g(), whose relation to f() is described by (see the picture): value of g() at 0 = value of f() at 0 c = f( 0 c). This shows that g() = f( c). The reasoning is similar if the graph is translated c units to the left. Try giving the argument yourself while referring to the picture. h() f() c c o c o g() The effect of up-down translation of the graph is much simpler to see. If c > 0, { { up f() + c (2) Moving the f() graph c units gives the graph of. down f() c since for eample moving the graph up by c units has the effect of adding c units to each function value, and therefore gives us the graph of the function f() + c Eample 2. Sketch the graph of +. Solution Combine rules () and (2). First sketch, then move its graph unit to the right to get the graph of, then unit up to get the graph of +, as shown. +

2 2 Eample 3. Sketch the curve y = Solution We complete the square : = ( ) 3 = ( + 2) 2 3, 2 so we move the graph of 2 to the left 2 units, then 3 units down, getting the graph shown Changing scale: stretching and shrinking. Let c >. To stretch the -ais by the factor c means to move the point to the position formerly occupied by c, and in general, the point 0 to the position formerly occupied by c 0. Similarly, to shrink the -ais by the factor c means to move 0 to the position previously taken by 0 /c. What happens to the graph of f() when we stretch or shrink the -ais? (3) { Stretching Shrinking the -ais by c changes the graph of f() into that of { f(/c) f(c). The picture eplains this rule; it illustrates stretching by the factor c >. The new function has the same value at 0 that f() has at 0 /c, so that it is given by the rule 0 f( 0 /c), which means that it is the function f(/c). f( ) f( c) 0 c 0 If the y-ais is stretched by the factor c >, each y-value is multiplied by c, so the new graph is that of the function cf(): (4) { Stretching Shrinking the y-ais by c changes the graph of f() into that of { c f() f()/c. Eample 4. Sketch the graph of 2. Solution. Start with the graph of /, move it unit to the right to get the graph of /( ), then shrink the -ais by the factor 2 to get the graph of the given function. See the picture. /2 3. Reflecting in the - and y-aes: even and odd functions. To reflect the graph of f() in the y-ais, just flip the plane over around the y-ais. This carries the point (, y) into the point (, y), and the graph of f() into the graph of f( ). Namely, the new function has the same y-value at 0 as f() has at 0, so it is given by the rule 0 f( 0 ) and is the function f( ). f (- ) f( ) Similarly, reflecting the y-plane in the -ais carries (, y) to the point (, y) and the graph of f() gets carried into that of f(). Finally, relecting first in the y-ais and then in the -ais carries the point (, y) into the point (, y). This is called a reflection through the origin. The graph of f() gets carried into the graph of f( ), by combining the above two results. Summarizing: - 0 f() -f(-) 0

3 G. GRAPHING FUNCTIONS 3 (5) Reflecting in the y-ais -ais origin moves the graph of f() into that of f( ) f() f( ). Of importance are those functions f() whose graphs are symmetric with respect to the y-ais that is, reflection in the y-ais doesn t change the graph; such functions are called even. Functions whose graphs are symmetric with respect to the origin are called odd. In terms of their epression in, (6) (7) f( ) = f() f( ) = f() definition of even function definition of odd function Eample 5. Show that a polynomial with only even powers, like , is an even function, and a polynomial with only odd powers, like , is an odd function this, by the way, eplains the terminology even and odd used for functions. Solution. We have to show (6) and (7) hold for polynomials with respectively only even or odd powers, but this follows immediately from the fact that for any non-negative integer n, we have { n ( ) n = ( ) n n, if n is even, = n, if n is odd. The following easily proved rules predict the odd- or even-ness of the product or quotient of two odd or even functions: (8) (9) even even = even odd odd = even odd even = odd even/even = even odd/odd = even odd/even = odd Eample is of the form odd/even, therefore it is odd; (3 + 4 ) /2 ( 3 ) has the form even odd, so it is odd. 4. The trigonometric functions. The trigonometric functions offer further illustrations of the ideas about translation, change of scale, and symmetry that we have been discussing. Your book reviews the standard facts about them in section 9., which you should refer to as needed. The graphs of sin and cos are crudely sketched below. (In calculus, the variable is always to be in radians; review radian measure in section 9. if you have forgotten it. Briefly, there are 2π radians in a 360 o angle, so that for eample a right angle is π/2 radians.) As the graphs suggest and the unit circle picture shows, (0) cos( ) = cos (even function) sin( ) = sin (odd function). From the standard triangle at the right, one sees that cos(π/2 ) = sin, π/2 -

4 4 and since cos is an even function, this shows that () cos( π/2) = sin. From (), we see that moving the graph of cos to the right by π/2 units turns it into the graph of sin. (See picture.) The trigonometric function cos cos - cos( - ) (2) tan = sin cos is also important; its graph is sketched at the right. It is an odd function, by (9) and (0), since it has the form odd/even. sin sin π - sin (- ) π 2π π Periodicity An important property of the trigonometric functions is that they repeat their values: (3) sin( + 2π) = sin, cos( + 2π) = cos. This is so because + 2π and represent in radians the same angle. From the graphical point of view, equations (3) say that if we move the graph of sin or cos to the left by 2π units, it will coincide with itself. From the function viewpoint, equations (3) say that sin and cos are periodic functions, with period 2π. In general, let c > 0; we say that f() is periodic, with period c, if (4) (4 ) f( + c) = f() c for all, and is the smallest positive number for which (4) is true. By rule (), the graph of a periodic function having period c coincides with itself when it is translated c units to the left. If we replace by c in (4), we see that the graph will also coincide with itself if it is moved to the right by c units. But beware: if a function is made by combining other periodic functions, you cannot always predict the period. For eample, although it is true that tan( + 2π) = tan and cos 2 ( + 2π) = cos 2, the period of both tan and cos 2 is actually π, as the above figure suggests for tan. The general sinusoidal wave. The graph of sin is referred to as a pure wave or a sinusoidal oscillation. We now consider to what etent we can change how it looks by applying the geometric operations of translation and scale change discussed earlier. a) Start with sin, which has period 2π and oscillates between ±. b) Stretch the y ais by the factor A > 0; by (4) this gives Asin, which has period 2π and oscillates between ±A. c) Shrink the -ais by the factor k > 0; by (3), this gives Asin k, which has period 2π/k, since Asin k( + 2π ) = Asin(k + 2π) = Asin k. k d) Move the graph φ units to the right; by (), this gives

5 G. GRAPHING FUNCTIONS 5 (5) A sin k( φ), A, k > 0, φ 0, general sinusoidal wave which has period 2π/k angular frequency k amplitude A phase angle φ (the wave repeats itself every 2π/k units); (has k complete cycles as goes from 0 to 2π); (the wave oscillates between A and A); (the midpoint of the wave is at = φ). Notice that the function (5) depends on three constants: k, A, and φ. We call such constants parameters; their value determines the shape and position of the wave. By using trigonometric identities, it is possible to write (5) in another form, which also has three parameters: A φ - A π/k (6) a sink + b cosk The relation between the parameters in the two forms is: (7) a = Acos kφ, b = Asin kφ; A = a 2 + b 2, tan kφ = b a. Proof of the equivalence of (5) and (6). (5) (6): from the identity sin(α + β) = sin α cosβ + cosαsin β, we get Asin(k( φ)) = Asin(k kφ) = Acoskφsin k Asin kφcosk which has the form of (6), with the values for a and b given in (7). (6) (5): square the two equations on the left of (7) and add them; this gives a 2 + b 2 = A 2 (cos 2 kφ + sin 2 kφ) = A 2, showing that A = a 2 + b 2. If instead we take the ratio of the two equations on the left of (7), we get b/a = tankφ, as promised. Eample 7. Find the period, frequency, amplitude, and phase angle of the wave represented by the functions a) 2 sin(3 π/6) b) 2 cos(2 π/2) Solution. a) Writing the function in the form (5), we get 2 sin3( π/8), which shows it has period 2π/3, frequency 3, amplitude 2, and phase angle π/8 (or 0 o ). b) We get rid of the sign by using cos = cos( π) translating the cosine curve π units to the right is the same as reflecting it in the -ais (this is the best way to remember such relations). We get then 2 cos(2 π/2) = 2 cos(2 π/2 π) = 2 sin(2 π), by (); = 2 sin2( π/2).

6 6 Thus the period is π, the frequency 2, the amplitude 2, and the phase angle π/2. (Note that the first three could have been read off immediately without making the above transformation.) Eample 8. Sketch the curve sin2 + cos2. Solution Transforming it into the form (5), we can get A and φ by using (9): A = 2; tan 2φ = 2φ = 35 o = 3π/4, φ = 3π/8. So the function is also representable as 2 sin2( 3π/8); it is a wave of amplitude 2, period π, frequncy 2, and phase angle 3π/8, and can be sketched using this data. 5. Reflection in the diagonal line; inverse functions. As our final geometric operation on graphs, we consider the effect of reflecting a graph in the diagonal line y =. This reflection can be carried out by flipping the plane over about the diagonal line. Each point of the diagonal stays fied; the -and y-aes are interchanged. The points (a, b) and (b, a) are interchanged, as the picture shows, because the two rectangles are interchanged. To see the effect of this on the function, let s consider first a simple eample. (b,a) y= Eample 9. If the graph of f() = 2, 0 is reflected in the diagonal, what function corresponds to the reflected graph? Solution. The original curve is the graph of the equation: y = 2, 0. Reflection corresponds to interchanging the two aes; thus the reflected curve is the graph of the equation: = y 2, y 0. To find the corresponding function, we have to epress y eplicitly in terms of, which we do by solving the equation for y: y =, 0 ; the restriction on follows because if = y 2 and y 0, then 0 also. y y y a,b) reflect no change 2 y=, >0 =y 2, y>0 y=, >0 Remarks.. When we flip the curve about the diagonal line, we do not interchange the labels on the - and y-aes. The coordinate aes remain the same it is only the curve that is moved (imagine it drawn on an overhead-projector transparency, and the transparency flipped over). This is analogous to our discussion in section of translation, where the curve was moved to the right, but the coordinate aes themselves remained unchanged. 2. It was necessary in the previous eample to restrict the domain of in the original function 2, so that after being flipped, its graph was still the graph of a function. If we

7 G. GRAPHING FUNCTIONS 7 hadn t, the flipped curve would have been a parabola lying on its side; this is not the graph of a function, since it has two y-values over each -value. The function having the reflected graph, y =, 0 is called the inverse function to the original function y = 2, 0. The general procedure may be represented schematically by: y = f() = f(y) y = g() original graph switch and y reflected graph solve for y reflected graph In this scheme, the equations = f(y) and y = g() have the same graph; all that has been done is to transform the equation algebraically, so that y appears as an eplicit function of. This function g() is called the inverse function to f() over the given interval; in general it will be necessary to restrict the domain of f() to an interval, so that the reflected graph will be the graph of a function. To summarize: f() and g() are inverse functions if (i) geometrically, the graphs of f() and g() are reflections of each other in the diagonal line y = ; (ii) analytically, = f(y) and y = g() are equivalent equations, either arising from the other by solving eplicitly for the relevant variable. Eample 0. Find the inverse function to, >. Solution. We introduce a dependent variable y, then interchange and y, getting We solve this algebraically for y, getting = y, y >. (20) y = +, > 0. (The domain is restricted because if y >, then equation (20) implies that > 0.) The right side of (20) is the desired inverse function. The graphs are sketched. It often happens that in determining the inverse to f(), the equation (2) = f(y) /(-) +/ cannot be solved eplicitly in terms of previously known functions. In that case, the corresponding equation (22) y = g() is viewed as defining the inverse function to f(), when taken with (2). Once again, care must be taken to restrict the domain of f() as necessary to ensure that the relected will indeed define a function g(), i.e., will not be multiple-valued. A typical eample is the following.

8 8 Eample. Find the inverse function to sin. Solution. Considering its graph, we see that for the reflected graph to define a function, we have to restrict the domain. The most natural choice is to consider the restricted function (23) y = sin, π/2 π/2. The inverse function is then denoted sin, or sometimes Arcsin ; it is defined by the pair of equivalent equations (24) = sin y, π/2 π/2 y = sin,. The domain [, ] of sin is evident from the picture it is the same as the range of sin over [ π/2, π/2]. As eamples of its values, sin = π/2, since sin π/2 = ; similarly, sin /2 = π/6. Care is needed in handling this function. For eample, substituting the left equation in (24) into the right equation says that (25) sin (sin y) = y, π/2 y π/2. It is common to see the restriction on y carelessly omitted, since the equation by itself seems obvious. But without the restriction, it is not even true; for eample if y = π, sin (sin π) = 0. π /2 - - y=sin π /2 Eercises: Section A

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