Graphing Functions. 0, < x < 0 1, 0 x < is defined everywhere on R but has a jump discontinuity at x = 0. h(x) =

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1 Graphing Functions Section. of your tetbook is devoted to reviewing a series of steps that you can use to develop a reasonable graph of a function. Here is my version of a list of things to check. You can use either of these lists, but you must understand, first, that not every step is useful in each case, and, second, that some of the calculations my be so difficult that to do by hand that it is wiser to use some other information about the function instead. For eample, the graph may never cross the -ais so in looking for - intercepts, that is, points for which f() = 0, it may be that the equation has no solution. Or, it may be that the equation is quite difficult or even impossible to solve by hand. The point is that all lists like these must be used with intellegence and insight!. Determine the domain of the function. Look for points where the function is either not defined at all or for points where the function has a discontinuity. (a) The function f() = is not defined at the point =. It is defined + everywhere else on R. (b) The function h() = { 0, < < 0, 0 < is defined everywhere on R but has a jump discontinuity at = 0.. Find the y-intercept if there is one. That is, evaluate the function at = 0. The y-intercept is the point of the form (0, f(0)). (a) The function f() = + has y-intercept (0, ). (b) The function g() = + does not have a y-intercept. The function is not defined at = 0!

2 0 8 y 0 Figure : A function with a y- intercept Figure : The function +. Find the -intercepts if any. These are the points of the form (ˆ, 0) where f(ˆ) = 0. To find the -intercepts, it is necessary to solve the equation f() = 0. For simple functions, this may be possible. For others, it may be quite difficult. Sometimes, we must use approimate solutions. One method to find these is Newton s Method which we will discuss later (see.9). (a) If p() = +, then the graph has -intercepts =, and =. (b) If q() = +, then the graph has no -intercepts. (Try solving the quadratic equation!) (c) If c() = cos(), then it will be impossible to solve this equation by hand, even though it is clear that there is such a root.. Look at the asymptotic behavior of the function f, that is, compute the limits lim f(), and lim f(). (a) If f() = +, then lim f() =, and lim f() =. The line y represents a horizontal asymptote for the graph.

3 Figure : (a):a graph with - intercepts Figure : (b):a function without any -intercepts Figure : (c): Graph of cos ()

4 (b) If j() = +, then lim f() =, and lim f() =. y Figure : (a): A curve with a single asymptote Figure 7: (b): A curve with two asymptotes (c) If v() = +, then as ±, the graph of v approaches the + graph of the straight line y =. This line is called an oblique asymptote. Note that an equivalent form of the function is obtained by dividing both the numerator and the denominator by : + + = Find the critical points of the function if any. That is, find all points at which f () = 0 or f does not eist. (a) If f() =, then f () = = ( ). So the critical points of f are = 0, and =. (b) If g() = + +, then g () = + +. There are no critical points for this function.

5 Figure 8: (c): An oblique asymptote (c) If v() = ( ), then v () = ( ). However, v is not differentiable at =. Notice that ( ) is not defined at =. It is important to go back to the definition of the derivative and use a difference quotient: v( + h) v() h = ( ) h 0 h = ( ) [ The left and right-hand limits of this latter epression are, respectively and. However, the limit itself does not eist. It is now a matter of looking at a geometric interpretation in terms of secant lines, say, to the left of =. The slopes are given by the difference quotients; the limit is the limiting slope. In such a case we say that the continuous function has a vertical tangent at =. The resulting shape is called a cusp which is shown in the net figure. h ].. Intervals of Increase or Decrease. Having found the critical points, one looks at intervals between the -coordinates of the critical points. On the one hand, if the function is increasing to the left and decreasing to the right of a critical point, then that point represents a relative maimum, whereas if the function is decreasing to the left and increasing to the right, it is clearly a point of relative minimum. (a) In the case of the function whose graph is the cusp, namely v() = ( ), the derivative, v () = ( ) is clearly negative to the left and positive to the right of =. Hence the cusp is a point of relative minimum.

6 Figure 9: (c): A cusp (b) For q() = +, the derivative is q () = so that the only critical point is =. Since the graph of is linear and therefore changes sign only once at the critical point and, moreover, is negative at = 0 which is to the left of the critical point, the function q is decreasing to the left and intcreasing to the right of =. point giving q a relative mimimum value. Therefore the critical point represents a (c) For P () = + 7), the derivative is P () = = ( )( + ). Since the first factor changes sign from negative to positive at = and the second at =, the factors are both negative to the left of and both positive to the right of and of different signs in between. So to the left of the derivative of P is positive so that P is increasing, that derivative is negative on (, ) so that P is decreasing on that interval, while, P must increase to the right of since P is negative there. Hence the critical point is a point of relative maimum, while the critical point is a point of relative minimum. 7. Concavity and Points of Inflection. If possible, we compute the second derivative f. To be useful, the result needs to be simplified to the stage that it can be analyzed, particularly with respect to its sign. We say that the graph of f is concave up on an interval I if the graph lies above all its tangent lines, while we say it is concave down of it lies below its tangent lines. We illustrate this behaviour with the function P () = + which

7 0 0 Figure 0: The function P of (c) we used immediately above in (c). The net figures show various tangent lines at points of the graph, and then look at various regions. Notice that it looks as if to the left of 0, the graph lies below its tangent lines while to the right, it lies above its tangent lines. Any point at which the concavity of a graph changes is called a point of inflection of the graph. In this case, there is a change of inflection at the point = 0. Cautionary note: we have noticed, for eample in the case of the function f() = that not every horizontal tangent represents either a relative maimum or a relative minimum. In that eample, the graph has a horizontal tangent at = 0 and that represents what we have called an inflection point. A simple sketch shows you that, indeed, = 0 is a point where the graph changes concavity, in this case from concave down to concave up. However not every point of inflection occurs at a point where the graph has an horizontal tangent! The function P is an eample. In this latter case, the inflection point occurs at =. We can use the second derivative to look for possible points of inflection, and to test concavity. The test is the following: if f is positive on and interval (a, b), then the graph of the function f is concave up, while if f is negative on (a, b), the graph of the function is concave down on (a, b). If the second derivative is in fact continuous on (a, b), and takes on both signs on that interval, then the Intermediate Value Theorem tells us that for some c (a, b), we must have f (c) = 0. Such a point, where the second derivative actually changes sign, is a point of inflection. (a) For the function P () = +, the first and second derivatives are: 7

8 Figure : curve P() Tangents to the Figure : Tangents above the curve: concave down Figure : The presence of an inflection point Figure : Tangents below the curve: concave up 8

9 P () =, P () + Notice that P () = 0 at the points = and = but nowhere else. We can evaluate P () at the two critical points and, doing so, we find that P ( ) = 7 < 0, and P ( ) = 7 > 07 > 07 > 07 > 07 > 07 > 07 > 0 and so the graph is concave down at the first critical point and concave up at the second. We infer that the original function P has a relative maimum at = and a relative minimum at =. This agrees with our analysis in (c) above. Moreover, we note that the graph of P is linear, changes sign once and only once at = and the sign changes from negative to positive. Therefore this value of indeed represents a point of inflection. (b) It is not always true that, at a zero of f the graph has an inflection point. Notice that if we take Q() = Then Q () = which has a zero at = 0. However, this function is always positive so the graph is concave up on all of R as illustrated in the net figure Figure : (7c): The graph is always concave up: f() = 9