What to expect when graphing transcendental
(non-algebraic) functions. For
graphing purposes, transcendental functions can be broken down
into two categories: those that involve one or more of the six
trigonometric functions and those that do not. In effect, the
ones that do not involve trigonometric functions can be graphed
completely with all their mathematical features included: all x-
and y-intercepts, all maxima and minima, all points of
inflections, all holes, all jump discontinuities, all vertical
asymptotes, all horizontal asymptotes, all vertical tangents,
all vertical cusps, all corners, and all isolated points. On
the other hand, if the function includes any trigonometric function,
the graph may have infinitely many such mathematical features.
For example, What is the best approach to graph transcendental functions? Graphing transcendental functions requires that you first learn to recognize their basic characteristics and graphs. We have found it useful to learn about these graphs in conjunction with their inverses. This focuses on creating a mental image of the complete graph as well as its inverse. We rely on the graphical interpretation of inverse as the reflection of the graph across the line y=x. While plotting certain points is indispensable, developing a mental image of the complete graph is vital to successful graphing. To introduce you to transcendental functions, the following collection of examples has been selected to help "model" the transcendental functions met most commonly. It is not intended to be exhaustive. Examples of transcendental functions that exhibit various characteristics. Exponential functions. Any
number b>0, b≠1, gives rise to an exponential
function Natural base. There is a special number, denoted by the letter e, which occurs frequently in mathematics. It is used as a base for the exponential function y=e^{x} and is referred to as the natural base. Its value can be computed, for instance, as the limiting value of the expression (1+1/n)^{n} as n→∞, which is approximately 2.718. Like the special number π, the decimal expansion of e never repeats. The exponential function y=e^{x} has the unusual property that for each number x, the slope of the tangent line to its curve at (x,e^{x}) equals its y-value, e^{x}. Logarithmic functions. The
inverse of the exponential function y=b^{x} (not
the reciprocal, which is the inverse with respect to multiplication,
but the inverse with respect to composition) is called the
logarithmic function, The base e≈2.718>1 The base 1/e≈0.368<1 The sine function,
Observe that the reflection across the line y=x produces a sine wave along the y-axis, and as such it is not a function (it does not pass the vertical line test). To obtain an inverse, we need to restrict the sine curve to where it is one to one (a restricted domain where it takes on all its y-values exactly once). While there are many ways this could be done, it is standard to use the smallest interval that contains the origin, [–π/2,π/2]. Here is the graph of the sine curve and its inverse, arcsin(x). The cosine function, y=cos(x),
and its inverse, y=cos^{-1}(x)=arccos(x). The
graph of the basic cosine function is similar to that of the
sine function. It is a "wave-like" pattern that repeats
on every interval of length 2π. It is actually a translation
of the sine curve (shift the sine curve left by π/2 units
and you get the cosine curve). Like the sine, the cosine has
an amplitude of 1 and is defined everywhere. In this case, to
obtain the inverse, the standard restricted domain where The tangent function, y=tan(x),
and its inverse, y=tan^{-1}(x)=arctan(x). Your
understanding of the graph of the basic tangent function can
be anchored by interpreting the identity: Graphing
a Transcendental Function Over the Interval [0,2π]: Since the sine function is periodic, it repeats every time its argument, in this case 3x+π/2, takes on all values through an interval of length 2π: 0≤3x+π/2≤2π. We can solve this inequality for x in two steps. First, we subtract π/2 (equivalently, add –π/2) to each term, –π/2≤(3x+π/2)–π/2≤2π–π/2, which simplifies to –π/2≤3x≤3π/2. Second, we divide each term by 3 (equivalently, multiply each term by 1/3), –π/6≤x≤π/2. Here, the period is the length of the resulting interval, π/2–(–π/6)=2π/3. At this point, we can sketch one complete period of the sine wave over this interval. Note that this is a sketch, so there is no need to plot more than the beginning and ending points. The factor of 2 in the expression Noting that the next complete wave starts at π/2
and ends at
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