Download MATH 136 The Inverse Function Theorem

MATH 136
The Inverse Function Theorem
A function y = f (x) is one–to–one if it is always the case that different x values are
assigned to different y values. A continuous strictly increasing (or strictly decreasing)
function is one–to–one.
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For example, y = x 3 + 4 is one–to–one, but y = sin x is not one–to–one. The sine
function assigns many different x values to the same y . For instance 0, π, 2π, 3π, etc. all
have a sine of 0.
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€
€
3
y = sin x
y = x +4
A one–to–one function f (x) always has an inverse denoted by f −1(x) . The inverse
“undoes” the steps applied to x by the function f . For instance with f (x) = x 3 + 4 , the
function f first cubes, then adds 4. To undo the process, we first subtract 4, then take
the cube root. Thus f −1(x) = (x − 4)1/ 3 .
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One property of f and f −1 is that
Domain f = Range f −1
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and
Domain f −1 = Range f .
This property essentially means that coordinates are “paired up.” Consider f (x) = x 3
and f −1(x) = x1/ 3. Since f (2) = 8, then f −1(8) = 2. The pair (2, 8) is created by this
inverse relationship. The point 2 is in the domain of f and in the range of f −1 .
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Likewise, 8 is in range of f and in the domain of f −1 .
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The Derivative of the Inverse
Suppose we have a one–to–one function for which we have a specific known rule, such
as f (x) = x 3 + 4 . Then f −1(x) exists. We wish to find the derivative of f −1(x) at some
point b in the domain of f −1 . That is, we wish to find ( f −1)′ (b) .
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€ Note that b is also in the range of f and is paired up with a point a in the domain
of f such that f (a) = b .
The Inverse
Function Theorem: We find ( f −1)′ (b) in terms of f€′(a) by
€
€
1
, where f (a) = b .
( f −1)′ (b) =
f ′(a)
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Example 1. Let f (x) = x 3 + 4 . Find ( f −1)′ (220) .
Solution. The Inverse Function Theorem implies that we can compute ( f −1)′ (220)
−1
without actually
€ finding f or its derivative.
Step 1: Solve f (x) = 220: x 3 + 4 = 220 gives x 3 = 216 and x = 216 3 = 6. Thus f (6) =
220 and the pair (6, 220) is created.
2
Step 2: f ′(x) = 3x €and f ′(6) = 108. €
Step 3: ( f€−1)′ €
(220) =
1
1
=
.
f ′(6) 108
In this case, we actually can verify the answer by finding f −1 and its derivative.
Here f −1(x) = (x − 4)1/ 3 .
( f −1)′ (220) =
Thus, ( f −1)′ (x ) =
1
1
=
.
2
/
3
108
3(220 − 4)
1
1
.
(x − 4) −2/ 3 =
3
3(x − 4)2 / 3
Thus,
3x + 15 − 21
. Find ( f −1)′ (−9) using the Inverse Function
2
Theorem. Verify the result by finding f −1(x) and its derivative.
Example 2. Let f (x) =
3x + 15 − 21
= –9 gives 3x + 15 = 3 , which gives
2
3x + 15 = 9 and x = –2. Thus f (−2) = –9 and the pair (–2, 9) is created.
Solution. We first solve f (x) = –9:
3
1
21
1
Re-writing f (x) =
3x + 15 − , we have f ′(x) = (3x + 15)−1/ 2 × 3 =
4 3x + 15
2
2
4
€
1
1
−1
and f ′(−2) = .
Thus, ( f )′ (−9) =
= 4.
f ′(−2)
4
To find f −1(x) , we first observe the steps used to create f (x) : multiply by 3, add
15, take square root, subtract 21, divide by 2. To create f −1(x) , we undo these steps in
reverse order: Given x , multiply by 2, add 21, square, subtract 15, divide by 3. Thus,
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Then ( f −1)′ (x ) =
f −1(x) =
(2x + 21)2 − 15
3
=
1
( 2x + 21)2 − 5 .
3
2
4
4
( 2x + 21) × 2 = (2x + 21) and ( f −1)′ (−9) = (−18 + 21) = 4.
3
3
3
The Inverse Trig Functions
The trigonometric functions cos x , sin x , tan x are not one-to-one; thus they do not have
inverses. However we can make them one-to-one by restricting the domains to only
two quadrants so that we only have one segment of each graph.
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€the domain
€
With cos x , we restrict
to [0, π] (the 1st and 2nd quadrants). Then the
−1
“inverse cosine” exists: cos x is the angle (in radians) whose cosine is x , where the
angle must be in either the 1st or 2nd quadrant.
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€
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f (x) = cos x
Domain: 0 ≤ x ≤ π
Range: –1 ≤ y ≤ 1
x€
€
0
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
−1
f (x) = cos x
Domain: –1 ≤ x ≤ 1
Range: 0 ≤ y ≤ π
cos x
1
3
2
2
2
1
2
0
1
–
2
2
–
2
3
–
2
–1
x €
€
cos −1 x
1
3
2
2
2
1
2
0
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
0
1
–
2
2
–
2
3
–
2
–1
With sin x , we restrict the domain to [–π/2, π/2] (the 1st and 4th quadrants). Then
the “inverse sine” exists: sin−1 x is the angle (on radians) whose sine is x , where the
angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then
we
€ write it as a negative angle.
€
€
f (x) = sin x
Domain: –π/2 ≤ x ≤ π/2
Range: –1 ≤ y ≤ 1
x €
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
sin x
−1
f (x) = sin x
Domain: –1 ≤ x ≤ 1
Range: –π/2 ≤ y ≤ π/2
x €
–1
3
–
2
2
–
2
1
–
2
–1
3
–
2
2
–
2
1
–
2
0
1
2
2
2
3
2
0
1
2
2
2
3
2
1
1
sin −1 x
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
With tan x , we restrict the domain to (–π/2, π/2) (the 1st and 4th quadrants). Then
the “inverse tangent” exists: tan−1 x is the angle (in radians) whose tangent is x , where
the angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant,
then
€ we write it as a negative angle.
€
€
f (x) = tan x
Domain: –π/2 < x < π/2
Range: –∞ < y < ∞
€
−1
f (x) = tan x
Domain: –∞ < x < ∞
Range: –π/2 < y < π/2
€
x
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
tan x
x
–∞
–∞
– 3
– 3
–1
1
–
3
–1
1
–
3
0
1
3
0
1
3
1
1
3
3
∞
∞
tan −1 x
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
Example 3. (a) Let f (x) = cos x for 0 ≤ x ≤ π . Find ( f −1)′ (1 / 2) .
(b) Let f (x) = sin x for −π /2 ≤ x ≤ π /2 . Find ( f −1)′ (− 2 / 2) .
€
€
We first solve cos x = 1/2, to get x = π /3. Then f ′(x) = −sin x and
€
€
1
−2
3
. Thus, ( f −1)′ (1 / 2) =
=
.
f ′(π / 3) = − sin(π / 3) = −
f ′(π / 3)
3
2
€
€
(b)
We solve sin x = − 2 / 2 , to get x = −π /4 .
Then f ′(x) = cosx and
1
2
2
. Thus, ( f −1)′ (− 2 / 2) =
=
.
f ′(−π / 4) = cos(−π / 4) =
f ′(−π / 4)
2
2
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€
Solution.
(a)