Review - Inverse Functions
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f and g are inverse functions if for all x in the domains of f and g ,
f g (x) = x
and g f (x) = x
That is, f and g are inverses if g undoes f and f undoes g .
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Example: e x and ln(x) are inverse functions:
For all x, e ln(x) = x
ln(e x ) = x
For instance,
f g (1) = e ln(1) = e 0 = 1
Math 104-Calculus 2 (Sklensky)
g f (1) = ln(e 1 ) = 1
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Review - Inverse Functions
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The graphs of inverse functions are reflections across the line y = x,
as illustrated with the graphs of e x , ln(x), and y = x shown below:
3
2
1
0
−3
−2
−1
0
1
2
3
x
−1
y
−2
−3
Math 104-Calculus 2 (Sklensky)
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Review - Inverse Functions
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Recall: Not every function has an inverse.
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Example: f (x) = x 2
The reflection of y = x 2 across
the line y = x results in a graph
which is not a function (there
are two outputs for most inputs).
y = x 2 is not one-to-one – For
most outputs (y values), you
could pick either one of two inputs (x-values) to give it.
Math 104-Calculus 2 (Sklensky)
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Review - Inverse Functions
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If we restrict the domain of f (x) = x 2 to x ≥ 0, then it is invertible
√
and g (x) = x is the inverse.
Math 104-Calculus 2 (Sklensky)
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Definition:
For any x in [−1, 1], define arcsin(x) to be the unique y -value in the
π π
interval [− , ] where x = sin(y ). arcsin(x) = 2 ⇔ sin(2) = x.
2 2
Math 104-Calculus 2 (Sklensky)
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Definition:
For any x in [−1, 1], define arccos(x) to be the unique y -value in the
interval [0, π] where x = cos(y ). arccos(x) = 2 ⇔ cos(2) = x.
Math 104-Calculus 2 (Sklensky)
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Definition:
For any x in (−∞, ∞), define arctan(x) to be the unique y -value in the
π π
interval (− , ) where x = tan(y ). arctan(x) = 2 ⇔ tan(2) = x.
2 2
Math 104-Calculus 2 (Sklensky)
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Definition:
For any x in (−∞, ∞), define arcsec(x) to be the unique y -value in the
interval (0, π) where x = sec(y ). arcsec(x) = 2 ⇔ sec(2) = x.
Math 104-Calculus 2 (Sklensky)
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In Class Work
1. Find the following derivatives. Don’t worry about algebraic
simplifications.
d
arcsin(x 2 )
dx
d x
(b)
e arctan(4x)
dx d
arctan ln(x)
(c)
dx
(a)
2. For each of the following, find the signed area given by the integral
shown.
Z
1
1
√
dx
4
1
− x2
0√
Z 3
1
(b)
dx
9
+
9x 2
0
(a)
Math 104-Calculus 2 (Sklensky)
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Solutions
1. Find the following derivatives. Don’t worry about algebraic
simplifications.
1
2x
d
arcsin(x 2 ) = p
· 2x = √
dx
1 − x4
1 − (x 2 )2
1
d x
(b)
e arctan(4x) = e x
·
4
+ e x arctan(4x)
dx
1 + (4x)2
(a)
=
d
(c)
dx
arctan ln(x)
Math 104-Calculus 2 (Sklensky)
4e x
+ e x arctan(4x)
1 + 16x 2
=
1
1 + ln(x)
2 ·
1
1
=
2
x
x(1 + ln(x) )
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Solutions
2. For each of the following, find the signed area given by the integral
shown.
Z
1
(a)
0
√
Z
(b)
0
1
1
1
1
√
dx = arcsin(x) =
arcsin(1) − arcsin(0)
2
4
4
4 1−x
0
π
1 π
=
−0 =
4 2
8
3
√3
√
1
1
1
dx = arctan(x) =
arctan( 3) − arctan(0)
2
9 + 9x
9
9
0
π
1 π
=
−0 =
9 3
27
Math 104-Calculus 2 (Sklensky)
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