Section - 9 GRAPHS. (a) y f x (b) y f x. (c) y f x (d) y f x. (e) y f x (f) y f x k. (g) y f x k (h) y kf x. (i) y f kx. [a] y f x to y f x

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1 44 Section - 9 GRAPHS In this section, we will discuss graphs and graph-plotting in more detail. Detailed graph plotting also requires a knowledge of derivatives. Here, we will be discussing some general issues related to graphs. Graphs will be discussed again later in the section on differential calculus. Up till now, we have discussed some standard functions and their graphs. We have also plotted graphs for other functions that are some variants of the standard functions. Lets formalise all that discussion here. Given the graph of y = f (), how will you draw the graphs of (a) y f (b) y f (c) y f (d) y f (e) y f (f) y f k (g) y f k (h) y kf (i) y f k [a] y f to y f This part is the easiest. We just flip the graph about the ais so that positive values become negative and negative become positive.

2 45 [b] y f to y f To draw y = f () from y = f (), we flipped our original graph about the ais. What should we do to draw the graph of y = f ( )? Flip the graph about the y ais. This is obvious once you realise that any value of y that was initially associated with (i.e y = f ()) will now be associated with, (i.e y = f ( ( )). For eample consider y = 3. At =, y = 8. In y = ( ) 3, what value of gives y =8? Obviously, = Note that in this particular case, y = ³ can also be obtained as in part [a], that is, by flipping the graphs 3 3 about the ais. This is because ( ) and ( ) are the same. [c] y f to y f The modulus function gives the magnitude of its argument (and returns a positive output). In other words, wherever the graph of f() lies below the -ais (that is, wherever f() is negative), the modulus function will make it positive), of eactly the same magnitude, or put differently, will take a reflection of the negative part of f() into the upper half of the aes.

3 46 [d] y f to y f Consider the equation y f carefully. In this equation, the input to f is only positive.(even if you input negative values for, the modulus function reduces it to a positive value. Also, we see that the input () and ( ) give the same output. How should we obtain the graph? In the graph of y = f (), first discard the part that lies to the left of the y-ais. This part of the graph has no use now since the input to f (argument of f, or the independent variable) takes only positive values. After discarding this left part, take the reflection of the right side of the graph into the left side. (Because the output of f is same, whether the input is or ). The eamples below will make it clear: (i)

4 47 (ii) (iii) (iv) [e] y f to y f If you ve understood the case for y f, this should not be very hard. You can approach this problem in any of the following ways: * In drawing y f, we discarded the left part of the graph y = f () and took a reflection of the right part into the left part. Here, the modulus function is on y. What should we do? Discard the lower part and take the reflection of the upper part into the lower half of the aes. * In the equation y f, the LHS is non-negative, so f () necessarily needs to be non-negative. Hence, wherever f () becomes negative, we have to discard that part, that is, we have to discard that part of the graph that lies below the -ais. Also, if (, y ) satisfies the equation y f, (, y ) also satisfies this equation, or in other words, if (, y ) lies on the graph of y f, (, y ) will

5 48 also lie on this graph, or the graph will be symmetrical about the -ais. Therefore, to obtain the graph of y f from y = f (), we discard its lower part and take a reflection of the upper part into the lower half of the aes. Note that the equation y f does not represent a function since it becomes one-many. It represents a dependence between and y.

6 49 [f] y f to y f k Here, we are adding (or subtracting) k units to the value of y for every point, or in other words, we are incrementing or decrementing the value of y by the same amount everywhere. For a graph, we are shifting it upwards or downwards while eactly retaining its shape. Draw the graphs of,, and sin by yourself. [g] y f to y f k Recall that we ve discussed one such case earlier where we drew the graphs of y of y. This is elaborated again here. from the graph Consider the graph of y = f (). Assume a particular value for say 0 and for this value, suppose y takes the value y 0. Therefore y 0 = f ( 0 ). Now consider y = f ( + k). For what value of will I get the output as y 0? Obviously, = 0 k, because y 0 = f ( 0 ) = f (( 0 k) + k). Hence, the output y 0 will now come k units earlier on the graph. This argument is true for each value of (since 0 is arbitrary here). Hence the entire graph will appear k units earlier on the aes, or, as is standard terminology, advanced, k units to the left. Similarly, the graph of f ( k) will be delayed, k units to the right. Obviously, if k is negative, f ( + k ) is actually shifted to the right and f ( k) to the left. Some eamples will make this clear.

7 50 [h] y f to y k f Suppose y = is our original graph and we want to draw the graph for y =. What is happening? The value for y is increasing (being doubled every where). But this increase is not uniform. Hence there is no uniform upwards or downwards shift. Also, we can see that there is no left or right shift. We can infer that the graph does change but not in the form of a shift. It stays where it is. If y = 0 for some 0 on the graph y = f (), it is also 0 for the graph y = kf (). The zeroes of the function (or the graph) remain the same. Every where else, y increases or decreases in magnitude depending on whether k or k. We can say that the graph epands or compresses in the the y direction.

8 5 If it is helpful, think that some fictitious mathematical creature is stretching or compressing the graph of y = f () to obtain the graph of y = k f (), depending on the value of k (This creature will also be flipping the graph about the ais if k < 0.) Draw the graphs of, { } yourself.

9 5 [i] y f to y f k Lets consider a particular value of k to make things easier. Take k =. For our original function, suppose 0 gives the output y 0, that is, y0 f o. In the new function, y = f (), at what value of will the output be y 0? Obviously, at 0 / since y 0 = f ( 0 ) = f 0. Hence, as in the case y = f ( + k) where the same output came k units earlier, in y = f ( ), the same output will come at half the original value: our mathematical creature will compress the graph along the -ais by a factor of. What will happen if k = /? Our being will stretch the graph by a factor of. What will happen if k =? Our being will first compress the graph by a factor of and then flip it about the y ais (see part b). Or he can flip it first and then compress it. f ( ) f ( ) f ( ) or f ( ) f ( ) f ( Try drawing the graphs of cos,, log 3 5

10 53 S U M M A R Y () y = f () to y = f () : Flip about the ais () y = f () to y = f ( ) : Flip about the y ais (3) y = f () to y = f : Reflect the parts of the graph that lie in the lower half (negative parts) into the upper half of the aes. (4) y = f () to y f : Discard the left part of the graph (for < 0) and take a reflection of the right part of the graph into the left half of the aes. (5) y f to y f : Discard the lower part of the graph ( f () < 0 ) and take a reflection of the upper part of the graph into the lower half of the aes. (6) y = f () to y = f ()+k : Shift the graph k units upwards or downwards depending on whether k is positive or negative respectively. (7) y = f () to y = f ( + k) : Advance (shift left) or delay (shift right) the graph by k units depending on whether k is positive or negative respectively. (8) y = f () to y = kf () : Stretch or compress the graph along the y-ais depending on whether k or k respectively. Also, flip it about the -ais if k is negative (This latter statement follows from part ) (9) y = f () to y = f (k) : Stretch or compress the graph along the -ais depending on whether k or k respectively. Also flip it about the y ais if k is negative (The latter statement follows from part ) These rules summarize all that you need to know about plotting graphs for the time being. A knowledge of limits and derivatives will make graph plotting more accurate, but for now, lets plot some graphs using what we ve learnt upto this point.

11 54 Eample Plot the graphs for the following: (a) y (b) y (c) (e) y (d) y 3 (f) y 3 y 3 (g) y (h) y (i) y log (j) y log Solution: For the first three graphs, lets first draw the graph of y =. We can proceed in the following sequence. (a) We need to draw y y. In the last figure above, take the reflection of the negative parts into the upper half of the aes.

12 (b) We discard the lower part and take a reflection of the upper part of the graph into the lower half of the aes. 55 (c) We combine the methods of parts (a) and (b), in that sequence. For the net three parts, we first draw the graph of y = We can write 3 y. Hence we use the following sequence 4 Note that in the third graph, and are and respectively (the values of where y becomes 0). (d) We need to draw y y

13 56 (e) This is y y. We discard the lower part of y and take a reflection of the upper part into the lower half of the aes. (f) We apply the methods of parts (d) and (e) in that sequence (g) We can follow the following sequence. We could also have used the sequence y y y y (h) Here again we see that y can be written as y and hence we follow the sequence

14 57 (i) For y log, we follow the sequence (j) For y log, we follow the sequence y log y log y log Notice carefully the difference in the sequences we follow for log and log. For y log we cannot first draw y log and then shift units left. This is because a unit left shift means the argument is changing by, i.e., which would imply that y log becomes not y log but y log which is the second case.

15 58 TRY YOURSELF - IV Draw the graphs for the following functions, marking all the important points clearly: (a) y, y, y, y (b) y, y, y, y (c) y 7 6, y 7 6, y 7 6 (d) y e (e) y 0 0 (f) y, y (g) y, y (h) y, y, y, y * * * * * * * * *

16 59 SOLVED EXAMPLES The following pages contain some solved eamples that are a bit advanced than the questions we have discussed up till now. These will give you an idea on what approach you should follow: Eample Find the domain for f ( ) 5 4 Solution: The denominator is a bit complicated and we need to analyse it in detail to determine where it can become zero. The fastest and easiest way would be to visualise the graph. Draw the graphs for and 5, apply the greatest integer function on these graphs separately, then add them and find the values of for which this sum becomes 4. For these values of, the denominator of f() becomes 0.

17 60 Now add the graphs of A and B point by point We see that the value of (graphs A + graph B) is 4 for the following values of : 0,],{,3, 4},[5,6. Hence, D = \{(0, ],, 3, 4, [5, 6)} Eample Draw the graphs for : (a) y { }, y { } (b) y [ ] { } (c) y [ ] { } Solution: Let us first draw the graph for y { }. Now, {} is the same as for 0 < <. In this interval, { } will be the same as We can easily see that this same curve will be repeated in every previous and subsequent unit interval, since {} is the same in all such intervals. Hence we obtain the graph of y { }

18 6 (a) Now we can easily draw y { } by taking a reflection in the -ais (b) In any interval n < n + (where n is an integer), [] has the value n. In any such interval therefore, the graph of y [ ] { } will be the graph of { } + integer n. This will be the graph segment of { } lifted by n units. For eample. 0 < y { } The graph is drawn below. < y { } < 0 y { } Note that this graph has no holes (or breaks) now. We epress this fact by saying that this function is continuous. The function y { }, is, on the other hand, discontinuous. We will study continuity in more detail later.

19 6 (c) On following the same lines as part (b), we obtain the graph of y [ ] { }. This function, as we can see, is discontinuous. Based on the discussions above, draw the graphs for y { }, y [ ] { }, y [ ] { }, y [ ] { } y [ ] { } and y { } Eample 3 Find the roots of the function f ( ) a b c and plot its graph. Also solve the inequalities f ( ) 0 and f ( ) 0 Solution: The epression for f ( ) is quadratic, of which we have seen some eamples. Lets discuss a particular case for f ( ) and then go on to the general case above. Let f ( ) which can be written as f ( ) (3 )( 3). This is 0 when either 3 0 or 3 0 i.e. =, (i) Now f ( ) can be rearranged as f ( )

20 = We can obtain the graph for f() in the following steps: We see that the coordinates of the verte (the point v) are 5 6, 3 3, and the graph crosses the ais at (it has its roots at) 3, as we have determined in (i) above. The graph goes below the ais in 3 the interval 3, 3 so that in this interval f() < 0. In the intervals (, 3) and, 3 f () > 0. Now we know everything about this quadratic function Hence, f () < 0 between the roots and f() > 0 outside this interval. Now we apply this reasoning to f ( ) = a b c, f ( ) = b c a b c a a a = a b b b c a a a a = a b b 4ac a 4a = a b D a 4a

21 64 where D b 4ac is called the discriminant of the quadratic epression. f ( ) 0 when a b a = 4 D a b a = D b D a a The two roots of f() are b D, b D a a The graph is obtained in the following steps:...(ii) a a a b b D a a 4a We see that if a < 0, the graph will not open upwards but downwards b D Also, whether a > 0 or a < 0, the co-ordinates of the verte are, a 4a D < 0? The y-co-ordinate of the verte, 4 D a. What if a > 0 and is positive, and hence the verte lies above the -ais. f() in this case never cuts the ais and hence never becomes 0. We say that f () has no real roots. f () > 0 for all values of. Similarly, if a < 0 and D < 0, the y-co-ordinate of the verte is negative and the graph lies below the ais. Here also, f () has no real roots. We see that whether a > 0 or a < 0, if D < 0, f () will not have real roots.

22 65 This is also evident in the formula for the roots in (ii). If D < 0, then D is imaginary (non-real). What if a > 0 and D = 0? In this case, the y-coordinate of the verte is 0 or the verte lies on the -ais. Now we can easily determine the solutions to f () > 0 and f () < 0 All these results are summarized below. You are urged not to memorize the results but understand them, by verifying each of them on your own

23 66 Eample 4 For real, find the condition on a, b, c such that the function ( a)( b) f ( ) ( { c} ) ( c) is onto Solution: Let y f ( ) require the range to be ( a)( b). Since the co-domain for this function has been specified has, we ( c) Now y( c ) = ( a)( b) also if f() is to be onto. Hence we require y to take on all real values. ( a b y) ( ab cy ) = 0 We again follow the method discussed earlier. For to be real, the discriminant should be non-negative. ( a b y) 4( ab cy ) > 0...(*) For the function f to be onto, we require that each y have a real pre-image. This is only possible if that y satisfies the constraint (*). Hence, this constraint, or this inequality, should be true for all real y. Rearranging as a quadratic in y y ( a b c) y ( a b ) > 0...(**) As we saw in Q3, for the LHS of (**) to be always non-negative, we require its graph to lie above the -ais (or touching it, at the most). If it goes below the ais, the LHS will become negative. Hence we require the discriminant for (**) to be non-positive. i.e. D < 0 ( a b c) 4( a b ) < 0 ( a b) 4c 4( a b) c ( a b ) < 0 ab c ( a b) c < 0 As in Q3., we can treat the LHS above as a quadratic in c. LHS < 0 implies that c must lie within the ( a b) ( a b) 4 ab ( a b) ( a b) roots of this quadratic epression, which are There is no loss of generality in assuming that a > b since the epression for f () is symmetric about a and b. Hence, we get the constraint on a, b, c as a < c < b a, b

24 67 Eample 5 Plot the graphs for the following: (a) y sin sin (b) y sin (sin ) Solution: By observation, you will have a tendency to say that since y is a composition of two inverse functions, which should cancel out, the output should be y =, which is a straight line. But we have to be more cautious: (a) The inner function, sin, is defined only for [, ]. For these values of, sin(sin ) will give back again. Hence, the graph is the identity function but only in the interval [, ]. (b) The inner function sin, is defined for all and gives an output in [, ]. The outer function, sin ( ), will now give a value in,. For eample, if =, sin = 0, and sin (sin ) 0 and not. Similarly, if 3 We can state these things concisely as, sin = and sin (sin ) and not 3, and so on sin (sin ) = if if + ( ) if 3 if 3 3 and so on.

25 68 Convince yourself that the above formulation is correct. The graph is drawn below: Eample 6 Plot the curve y min,,. Solution: What does such a statement mean? It means that for each value of, we evaluate all the three quantities on the RHS, select the minimum of the three, and plot that value of y. What we can do is plot the separate graphs for these three quantities on the same aes and select those portions that lie lowermost, out of all the three. The darker line segments on the left hand side diagram show the minimum value out of all the three, considered at each point. This is therefore the graph of y. Eample 7 Solve the equation for : { } [ ] [ ] 3 Solution: A solution is not evident by mere observation. But it can be noted that the RHS is always positive: therefore, the LHS must be positive and hence > 0. Also, for the LHS to be defined, > and > > and hence [] >. Let I be the integral and f the fractional part of. I I [ f ] = f 3

26 69 We have retained [ f ] since f could be greater than and hence [ f ] is not necessarily 0. We will have to consider two different cases separately: (i) f I I = f 3 3 I = f 3 Now we can substitute different values of I; if these give valid values for f such that these solutions. f, we accept I = f 7 6 [not acceptable] I = f 5 [acceptable] I = 3 f 6 [acceptable] I = 4 f 4 [acceptable] I = 5 f 0 No more solutions will eist since f becomes < 0 for I > 5 [not acceptable] (ii) f I I = f 3 I = f = [not acceptable] I = f (f does not satisfy 30 f ) [not acceptable] I = 3 f 7 [not acceptable] I = 4 f = /36 [not acceptable] I = 5 f < 0 [not acceptable] Hence the valid solutions are I f ,, 6 4

27 70 Find Eample the range of 8 (a) y sin (b) y 3 4. (c) y 4 4 sin cos (d) y (e) y 3. Solution: (a) Let us first evaluate the range of z (say) z = (Notice the last steps carefully; 0 ) Now, z implies sin z sin. 6 sin z can take a maimum value of for z =, but we see that for no value of does z become (although z approaches or almost becomes as becomes larger and larger i.e. as, z, but z ). Hence sin = is not included in the range R =, 6 (b) Let us approach this problem in two different ways. (i) Define it piecewise (define it separately in each interval): < y = 3 4 = 0 4 < < y = 3 4 = 8 < < 3 y = 3 4 = 4 3 < < 4 y = 3 4 = > 4 y = 3 4 = 4 0 Now draw the graph:

28 7 The minimum is y = 4 for < < 3. The range is [4, ). 3 4 (ii) Symmetry suggests that we should get a minimum for, but it is not unique. The 4 function is minimum for the entire interval [,3]. This is due to the nature of this function, which contains the modulus functions. Consider the graphs of y Here, the minimum is y = 6 at = 3, , which is unique. Reread the above carefully to determine why the minimum is unique now (the function y = f() never becomes constant, as in the previous case) Now consider the graph of y 4

29 7 The minimum is at =, equal to y = 3 and not at Symmetry is wrong here! Actually, symmetry is not wrong, but this function does not have complete symmetry (this is obvious from the graph), while the previous two functions did. This is because in functions of the form a i, a turning point (a sharp point) will always come at one of the a ' s, and the etremum will always lie at such a point. In the first function, there is a minimum at =, 3 (which are sharp points) and hence at all the points in between. In the second function, there is a minimum at = 3 (a sharp point) which i happens to be the same as = 3). n a i (because the points,, 3, 4, 5 are symmetrically placed about In the third function, the minimum is at = but ai n lies somewhere else, because the points,, 4 are not distributed symmetrically about any of these sharp points So for eample f has 3 sharp point, {, 0, } distributed symmetrically about one of the sharp point {0}, and hence the minimum is at = 0 equal to y =, which is also obtained by symmetry. f has 3 sharp points {, 0, } which are not distributed symmetrically about any sharp point and hence, although 0, the minimum is at 0 and not / y has 4 sharp points {,,, } which are distributed symmetrically about and. The minimum is therefore at 0(and also at all values 4 between and since the function assumes a constant value in this interval (y = 6).) Therefore, we see that symmetry has to be used carefully. * For eample, a function of the form ai, could be non-symmetric but a minimum could still occur at ai n, when the number of ai ' s are even. You need not get confused with all these arguments above. Just remember to use symmetry with caution!) * One could argue that the minimum for ( a ) ( a ) ( a ) should come at one of the points a, a... a n and not at a a... an n n as we saw earlier. But this is false since this function does not have sharp points. It is hence not necessary that the etremum will be at one of the points a, a... a n. As we will see later on, this function is differentiable, meaning that it is smooth, while a is non differentiable, meaning it is not smooth). i

30 73 (c) y 4 4 sin cos = (sin cos ) sin cos sin cos (sin ) = Now sin 0 (sin ) 0 (sin ) (sin ) (sin ) Therefore, the range is [, ]. (d) y = = ( ) This is of the form a b, which we discussed earlier. The minimum for y comes at =, : y = 3 The maimum comes at The range is R = 3, 6 : y 6 (e) y 3 This function is no longer symmetric so we carry out a standard analysis. The domain is [, 3]. Also, y > 0. Now y ² = ( ) 4(3 ) 4 ( )(3 ) = 3 4 ( )(3 ) Rearranging and squaring gives ( y 3 ) 6( 4 3) = 0 4 y (3 ) y (3 ) 6( 4 3) = 0

31 74 For real, we require D > (6y 30) ( y y 69) 0 4 (3y 65) 5( y y 69) 0 y 4 6y 60y 0 0y 0 4 y 0 Also, at =, y = and at = 3, y =, and therefore, the minimum value of y is The range is, 0 (we will be able to evaluate the range much more easily using derivatives later on).

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