MATH 136
The Inverse Function Theorem
Exercises
1. Let f (x) = 9 − 4 15 − 2x , for x ≤ 7.5 . Use the Inverse Function Theorem to compute
( f −1)′ (−11) . Then find a function formula for f −1(x) and use it to compute ( f −1)′ (−11) .
12 €
−10 , for x ≠ 10 . Use the Inverse Function Theorem to compute
6− x
( f −1 )′(−16) . Then find a function formula for f −1(x) and use it to compute ( f −1 )′(−16) .
2.
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Let f (x) =
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3. Let f (x) = cos x for 0 ≤ x ≤ π . Compute ( f −1)′ ( 2 / 2) and ( f −1)′ (− 3 / 2) .
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4. Let f (x) = sin x for −π /2 ≤ x ≤ π /2 . Compute ( f −1)′ ( 3 / 2) and ( f −1)′ (−1/ 2) .
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5. Let f (x) = tan x for –π/2 < x < π/2. Compute ( f −1)′ (1) and ( f −1)′ (−1/ 3) .
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Solutions
1. Solve 9 − 4 15 − 2x = –11 to get 20 = 4 15 − 2x , then 25 = 15 − 2x , and x = –5.
1
4
4
5
and f ′(–5) = ; thus, ( f −1)′ (−11) =
= .
f ′(–5)
15 − 2x
5
€ 4
€
€
2
 − ( x − 9) 
− 15

(x − 9)
( x − 9 )2 15
4 
−1
Also f (x) =
=−
, so ( f −1)′ (x ) = −
and ( f −1)′ (−11)
+
16
−2
32
2
( −20) 5
=−
= .
4
16
------------------------------------------------------------------------------------------------------------------12
12
2. First solve
−10 = –16, which gives
= −6 , then −2 = 6 − x , and x = 8 .
6− x
6− x
1
12
Next, f ′(x) =
and f ′(8) = 3 . Thus, ( f −1 )′(−16) = .
2
3
(6 − x)
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12
12 1
12
−1 ′
Also f −1(x) = 6 −€
. So ( f −1)′€(x ) =
,
and
=
= .
(
f
)
(−16)
2
x + 10
36
3
(x
+
10)
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------------------------------------------------------------------------------------------------------------------3.
First solve cos x =
Then f ′(x) = −sin x and
2 / 2 , to get x = π /4 .
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1
−2
2
€ ( f −1)′ ( 2 / 2) =
. Thus,
=€
.
f ′(π / 4)€= −sin(π / 4) = −
f ′(π / 4)
2
2
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Next solve cos x = – 3 / 2 , to get x = 5π /6 .
Then f ′(x) = −sin x and
1
1
= –2.
f ′(5π / 6) = −sin(5π / 6) = − . Thus, ( f −1)′ (− 3 / 2) =
f ′(5π / 6)
2
------------------------------------------------------------------------------------------------------------------€
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Then f ′(x) =
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4.
First
solve
sin x =
Then
3 / 2 , to get x = π /3.
1
1
= 2.
f ′(π / 3) = cos(π / 3) = . Thus, ( f −1)′ ( 3 / 2) =
f ′(π / 3)
2
and
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= –1/2, to get x = −π /6 .
Then f ′(x) = cosx and
1
2
3
. Thus, ( f −1)′ (−1/ 2) =
=
.
f ′(−π / 6) = cos(–π / 6) =
f ′(−π / 6)
3
2
------------------------------------------------------------------------------------------------------------------€
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1
5. First solve tan x = 1, to get x = π /4 . Then f ′(x) = sec 2 x and ( f −1)′ (1) =
=
f ′(π / 4)
1
1
= cos 2 (π / 4) = .
2
2
sec (π / 4)
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1
Next solve tan x = –1/ 3 , to get x = −π /6 . Then ( f −1)′ (−1/ 3) =
=
f ′(−π / 6)
1
3
= cos 2 (−π / 6) = .
2
4
sec (− π / 6)
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Next
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solve
f ′(x) = cosx
sin x