Derivative of Inverse Function Problem

Question

For each of the given functions f(x), find the derivative (f^-1)'(c) at the given point c, first finding a=(f^-1)(c)

1. f(x) = 4x+6x^13 c = -10
2. a= ?
3. (f^-1)'(c) = ?
4. f(x) = x^2-13x+65 on the interval [6.5,inf); c=25
5. a= ?
6. (f^-1)'(c) = ?

I am having trouble with the inverse with multiple powers. How is this problem solved?

Answer 1

f(x) = 4x + 6x¹³

By trial and error, we see that f(-1) = -10. So, f^-1(-10) = -1 = a

f'(x) = 4 + 78x¹² So, f'(-1) = 82

(f^-1(-10))' = 1 / f'(a) = 1 / f'(-1) = 1/82

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f(x) = x² - 13x + 65

x² - 13x + 65 = 25 So, x² - 13x + 40 = 0

(x - 5)(x - 8) = 0 So, x = 5 or 8. Since x ≥ 6.5 by hypothesis, x = 5 must be disregarded.

x = 8 = a

f'(x) = 2x - 13

So, (f^-1(25))' = 1 / f'(a) = 1 / f'(8) = 1/3

Answer 2

1. f(x) = 4x + 6x¹³

a. a = f⁻¹(–10)

find f(a) = –10

f(a) = 4x + 6x¹³ = –10 which means 6x¹³ + 4x + 10 = 0

6x¹³ + 4x + 10 = 0 at x = –1 which means f(–1) = –10

If f(–1) = –10 then = f⁻¹(–10) = f⁻¹(f(–1)) = –1

Answer (1, part a)

a = f⁻¹(–10) = –1

b. (f⁻¹)’(–10)

f'(x) = 4 + 78x¹²

f'(–1) = 4 + 78(–1)¹² = 82

(f⁻¹)’(–10) = 1/ f'(–1) = 1/82

Answer (1, part b)

(f⁻¹)’(–10) = 1/82

2. with the correct answers

f(x) = x² – 13x +65 [6.5, ∞)

a. a = f⁻¹(25)

find f(x) = 25 = x² – 13x + 65

x² – 13x + 40 = (x – 5) (x – 8) = 0

x = 5 and x = 8 are both solutions, but

only x = 8 is in the domain of f(x) [6.5, ∞)

which means f(8) = 25

If f(8) = 25 then = f⁻¹(25) = 8

Answer (2, part a)

a = f⁻¹(25) = 8

b. (f⁻¹)’(25)

f’(x) = 2x – 13

f(8) = 2(8) – 13 = 3

(f⁻¹)’(25) = 1/ f'(8) = 1/3

Answer (2, part b)

(f⁻¹)’(25) = 1/3