MATH 151 Engineering Math I, Spring 2014
JD Kim
Week11 Section 4.6, 4.8, 5.1
Section 4.6 Inverse Trigonometric Functions
We have a difficulty to find a inverse trigonometric function, because the trigonometric functions are not one-to-one, they do not have a inverse functions. This difficulty is overcome by restricting the domain of these functions so that they become
one-to-one.
The inverse function of this restricted sine function f exists is denoted by sin−1
or arcsin. It is called the inverse sine function or the arcsine function.
1
Inverse Sine
π
π
If − ≤ x ≤ , then f (x) = sin x is one-to-one, thus the inverse exists, denoted
2
2
by sin−1 x or arcsin x.
Domain of arcsin x = range of sin x = [−1, 1],
π π
Range of arcsin x = domain of sin x = [− , ].
2 2
sin−1 x = y ⇔ sin y = x and −
and
π
π
≤y≤
2
2
π
π
≤x≤
2
2
sin(arcsin x) = x if − 1 ≤ x ≤ 1.
arcsin(sin x) = x if −
Ex1) Compute the following:
1-1) arcsin(0)
1-2) sin−1 (1)
1-3) arcsin(−1)
1
1-4) arcsin( )
2
2
−1
1-5) sin (−
√
3
)
2
2
1-6) tan(arcsin )
3
1-7) sin(arcsin
3
)
10
1-8) arcsin(sin
5π
)
4
π
1-9) arcsin(sin(− ))
6
1-10) arcsin(sin
π
)
120
3
Inverse Cosine
If 0 ≤ x ≤ π, the f (x) = cos x is one-to-one, thus the inverse exists, denoted by
cos−1 x or arccos x.
Domain of arccos x = range of cos x = [−1, 1],
Range of arccos x = domain of cos x = [0, π].
cos−1 x = y ⇔ cos y = x and 0 ≤ y ≤ π
and
arccos(cos x) = x if 0 ≤ x ≤ π
cos(arccos x) = x if − 1 ≤ x ≤ 1.
Ex2) Compute the following:
2-1) arccos(0)
2-2) cos−1 (1)
2-3) arccos(−1)
1
2-4) arccos( )
2
4
2-5) cos (−
−1
√
3
)
2
4
2-6) sin(2 arccos(− ))
5
π
2-7) arccos(cos )
6
2-8) arccos(cos
7π
)
6
π
2-9) arccos(cos(− ))
3
5
2-10) cos(cos−1 (2))
Inverse Tangent
If −π/2 < x < π/2, then f (x) = tan x is one-to-one, thus the inverse exists,
denoted by tan−1 x or arctan x.
Domain of arctan x = range of tan x = (−∞, ∞),
Range of arctan x = range of tan x = (−π/2, π/2).
tan−1 x = y ⇔ tan y = x and − π/2 < y < π/2
and
arctan(tan x) = x if − π/2 < x < π/2
tan(arctan x) = x for all x.
Ex3) Compute the following;
3-1) arctan(0)
3-2) tan−1 (1)
6
3-3) arctan(−1)
√
3-4) arctan(− 3)
3-5) tan(arcsin x)
3-6) arctan(tan(
5π
))
3
3-7) limx→∞ arctan x
3-8) limx→−∞ arctan x
7
Derivatives of Inverse Trigonometric Functions
1
d
arcsin x = √
dx
1 − x2
d
1
arccos x = − √
dx
1 − x2
d
1
arctan x =
dx
1 + x2
Ex4) Find the derivative of f (x) = arccos(2x − 1).
8
Ex5) Find the derivative of f (x) = tan−1 (arcsin x)
Ex6) What is the domain of arcsin(3x + 1).
Ex7) What is the domain of arctan(3x + 1).
9
Section 4.8 Indeterminate forms and L’Hospital’s rule
Indeterminate form
If we have a limit of the form
0
∞
f (x)
= or
,
x→a g(x)
0
∞
lim
then this limit may or may not exist and is called an indeterminate form.
L’Hospital’s Rule
Suppose f and g are differentiable and g ′ (x) 6= 0 on an open interval I that
contains a (except possibly at a). Suppose that
lim f (x) = 0 and lim g(x) = 0
x→a
x→a
or that
lim f (x) = ±∞ and lim g(x) = ±∞,
x→a
x→a
(In other words, we have an indeterminate form of type
∞
0
or
) then
0
∞
f (x)
f ′ (x)
= lim ′
x→a g(x)
x→a g (x)
lim
if the limit on the right side exists (or is ∞ or −∞).
f (x)
0
∞
=
or
, the limit is NOT indeterminate. CANNOT
g(x)
∞
0
use L’Hospital’s Rule.
NOTE If limx→a
Ex8) Find limx→1
ln x
x−1
10
Ex9) Find limx→0
2x − 1
x
Ex10) Calculate limx→0
Ex11) Find limx→0
sin x − x
x3
sin mx
sin nx
Ex12) Find limx→∞
(ln x)3
x2
ln x
Ex13) Find limx→2− √
2−x
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Indeterminate Products
If limx→a f (x)g(x) = 0 · ∞, this limit is an indeterminate product.
We can deal with it by writing the product f g as a quotient:
fg =
g
f
or f g =
.
1/g
1/f
0
∞
This converts the given limit into an indeterminate form of type or
so that we
0
∞
can use L’Hosptal’s Rule.
Ex14) Evaluate limx→0+ x ln x
Ex15) Find limx→0+
√
x sec x
Ex16) Find limx→1+ (x − 1) tan(πx/2)
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Indeterminate Difference
If limx→a (f (x) − g(x)) = ∞ − ∞, this limit is an indeterminate difference.
Ex17) Compute limx→(π/2)− (sec x − tan x)
Ex18) Find limx→1
1
1
−
ln x x − 1
Ex19) Find limx→0+
2x + 1 1
−
sin x
x
Ex20) Find limx→π/2− (sec x − tan x)
13
Indeterminate Power
If limx→a f (x)g(x) is of the form 00 , ∞0 , or 1∞ , then the limit is an indeterminate
power. To solve such a limit, take the natural logarithm.
Ex21) Calculate limx→0+ (1 + sin 4x)cot x
Ex22) Find limx→∞ x3/x
Ex23) limx→0+ (sin x)tan x
14
Ex24) limx→∞
2x + 3
2x + 5
2x+1
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Chapter 5. Applications of Differentiation
Section 5.1 What does f ′ say about?
1. If f ′ (x) > 0 on an interval, then f is increasing on that interval.
2. If f ′ (x) < 0 on an interval, then f is decreasing on that interval.
3. If f ′ goes from positive to negative at x = a, and x = a is in the domain of f ,
then f has a local maximum at x = a.
4. If f ′ goes from negative to positive at x = a, and x = a is in the domain of f ,
then f has a local minimum at x = a.
16
Ex25) Graph of the f , from the f ′ given.
17
Ex26) From the given graph of the derivative f , answer the following questions.
26-1) On what intervals is f increasing?
26-2) On what intervals is f decreasing?
26-3) At what x values does f have a local maximum or minimum?
18
Definitions Concave Upward, Concave Downward
If the slopes of a curve become progresively larger as x increases, then we say
f is Concave upward. If the slopes of a curve become progressively smaller as x
increases, then we say f is Concave downward.
What does f ′′ say about f ?
1. If f ′′ > 0 on an interval, then f is concave upward on that interval.
2. If f ′′ < 0 on an interval, then f is concave downward on that interval.
3. If f changes concavity at x = a, and x = a is in the domain of f , then x = a
is an inflection point of f .
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Ex27) The graph of the derivative f ′ of a function f is shown.
1. On what intervals is f increasing or decreasing?
2. At what values of x does f have a local maximum or minimum?
3. On what interval is f concave upward or downward?
4. State the x-coordinates of the points of inflection.
5. Sketch a graph of f .
20
Ex28) If f ′ (4) = 0 and f ′′ (4) = 5, what can be said about f ?
2
Ex29) If f ′ (x) = e−x , what can be said about f ?
21
Ex30) Sketch a graph of f satisfying the following conditions.
1. f ′ (x) > 0 on the interval (−∞, 1) and f ′ (x) < 0 on the interval (1, ∞).
2. f ′′ (x) > 0 on the interval (−∞, −2) and (2, ∞).
3. f ′′ (x) < 0 on the interval (−2, 2).
4. limx→−∞ f (x) = −2 and limx→∞ f (x) = 0.
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