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Rational functions, and how to graph them
There is no single picture that can show a generic form for all rational functions,
f (x) = g(x)/h(x) where g(x) and h(x) are both polynoimials, since they can
look very different. Still, there are some pretty firm rules for determining how
to graph them. Essentially, there are five steps:
1. To determine the y-intercept, evaluate f (0) by ignoring everything except
the constant terms at the end of each polynomial. This gives you one
point to plot.
2. To determine the behavior as x → ±∞, ignore everything except the leading terms. Divide one by the other, and draw the resulting function as a
dotted line. If the resulting function takes the form xn , where n is positive, i.e., the numerator is higher order than the denominator, the rational
function behaves like this polynomial on both sides. If the numerator and
denominator are of the same order, the resulting value a gives you a horizontal asymptote at y = a, on both sides. If the order of the denominator
is greater than that of the numerator, the expression goes like x−n , and
you will find a horizontal asymptote at y = 0.
3. Factor the numerator. Where it equals zero, the function is either equal
to zero or undefined (see below).
4. Factor the denominator. Where it equals zero, we either have a vertical
asymptote or the function is undefined (see below).
5. If the numerator and denominator have the same root, the point will be
undefined. If both are roots appear the same number of times, we have
a removable discontinuity. The function can be graphed by cancelling
out those terms and graphing it as if they didn’t exist, except for the
missing point. If the root appears extra times in the numerator, it’s still a
removable discontinuity, with the limit of the function equal to zero there.
If the root appears more times in the denominator, we have a vertical
asymptote.
Some examples:
f (x) =
(x − 1)(x + 1)
x(x + 1)(x − 1)
x3 − x
x2 − 1
=
;
g(x)
=
=
(1)
2
2
x −4
(x − 2)(x + 2)
x − 3x + 2
(x − 2)(x − 2)
1
h(x) =
2
x(x + 1)2
x+1
x+1
x3 + 2x2 + x
(2)
=
; j(x) = 2
=
x3 − 4x
x(x − 2)(x + 2)
x − 2x + 1
(x − 1)2
More on inverse trig
Let’s consider the standard inverse trig functions: sin−1 (x), cos−1 (x), and
tan−1 (x):
We can construct a table of known values covering both trig functions and
their inverses:
sin θ
x
√1
√3/2
2/2
1/2
0
−1/2
√
−√2/2
− 3/2
-1
θ
sin−1 (x)
π/2
π/3
π/4
π/6
0
−π/6
−π/4
−π/3
−π/2
cos θ
x
√1
√3/2
2/2
1/2
0
−1/2
√
−√2/2
− 3/2
-1
θ
cos−1 (x)
0
π/6
π/4
π/3
π/2
2π/3
3π/4
5π/6
π
tan θ
x
∞
√
3
1
√
3/3
√0
− 3/3
-1
√
− 3
−∞
θ
tan−1 (x)
π/2
π/3
π/4
π/6
0
−π/6
−π/4
−π/3
−π/2
Notice that if x ≥ 0, then all of the inverse trig functions return values
between 0 an π/2. For the problems where we compose a trig function with
an inverse trig function, we can use triangle diagrams from the last handout to
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determine the result. For specific values, though, it’s important to learn how
to use them. If the argument we use is positive the result will be positive and
there’s not much to discuss. Let’s consider a negative case:
√
sin−1 (− 3/2) = −π/3
(3)
We’ve previously derived relations that said
tan(sin−1 (x)) = √
x
;
1 − x2
cos(sin−1 x) =
p
1 − x2
(4)
, and can no confirm that these are true everywhere, even if the original triangle
doesn’t apply because one side is negative. From the table above,
√
√
√
− 3/2
−1
√
(5)
tan(sin (− 3/2)) = tan(−π/3) = − 3 =
1 − (− 3/2)2
q
√
√
cos(tan−1 (− 3/2)) = cos(−π/3) = 1/2 = 1 − (− 3/2)2
(6)
We find that the rules we derived using triangles work, even when one side has
a negative value. Just remember to be careful with minus signs when doing the
algebra.
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More on the squeeze theorem
In class, we discussed the squeeze theorem, which states that if f (x) ≤ g(x) ≤
h(x) and
lim f (x) = lim h(x) = c
(7)
x→a
x→a
then limx→a g(x) = c. The typical use of this theorem involves sine and cosine functions for which we cannot determine a limit, typically because their
argument goes to infinity. Consider the case of g(x) = sin(1/x), for which
limx→0 sin(1/x) is undefined. If you are curious why, look in the textbook.
We first note that the sine of anything has a range of at most −1 ≤ sin(???) ≤
1. Thus, if we have F (x) = f (x) sin(g(x)), we can always show that −|f (x)| ≤
F (x) ≤ |f (x)|, i.e., the value of F always falls between −f and f (f is referred
to as an “envelope function”, since F oscillates within a region enveloped by
f ). If f (x) ≥ 0 everywhere, but has a zero somewhere, we can use the squeeze
theorem if the sine term doesn’t have a limit (note that if the sine term has a
well-defined limit, we can just use the product rule instead!).
Consider h(x) = x2 sin(1/x), for which we cannot use the product rule. We
know that −x2 ≤ h(x) ≤ x2 , and that
lim −x2 = lim x2 = 0
x→0
x→0
We conclude that limx→0 h(x) = 0.
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(8)
What about the case where the envelope function crosses zero? Consider
the case of j(x) = x sin(1/x). For x ≥ 0, we have −x ≤ j(x) ≤ x, and see that
lim −x = lim+ x = 0
x→0+
(9)
x→0
and use the one-sided squeeze theorem to conclude that limx→0+ j(x) = 0.For
x < 0, we have x ≤ j(x) ≤ −x, and see that
lim x = lim− −x = 0
x→0−
(10)
x→0
and use the one-sided squeeze theorem from the other side to conclude that
limx→0− j(x) = 0. Since j(x) has the same limit from both sides, we conclude that limx→0 j(x) = 0. Below, we show this in practice for a few functions:
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