We can clearly see that f(0) = 7, therefore f⁻¹(7) = 0
f(x) = 2x³ + 3x² + 3x + 7
f'(x) = 3x² + 6x + 3
f'(0) = 3
By definition of inverse functions:
f(f⁻¹(x)) = x
Differentiate both sides, using chain rule on left side
f'(f⁻¹(x)) * (f⁻¹)'(x) = 1
(f⁻¹)'(x) = 1/f'(f⁻¹(x))
(f⁻¹)'(7) = 1/f'(f⁻¹(7))
........... = 1/f'(0)
........... = 1/3
------------------------------
General rule: f(b) = a ----> (f⁻¹)'(a) = 1/f'(b)

Positive: 47 %

Answer #3 | 09/02 2014 07:05

f(g(x)=x f(0)=7=>g(7)=0
f'(g(x)g'(x)=1
g'(x)=(f^-1)'(x)=1/f'(g(x))
= 1/f'(0)
f'(x)=6x^2+6x+3
f '(0)=3
so (f^-1)'(7)=1/3

Positive: 27 %

Answer #4 | 08/02 2014 13:23

Let f^-1(7) = c
f(c) = 7
2c^3+3c^2+3c+7 = 7
2c^3+3c^2+3c=0
c(2c^2+3c+3) = 0
c=0
2c^2+3c+3 has complex roots
Therefore, f^-1(7) = 0
(f^-1)'(7) = 1 / f' (f^-1(7))
= 1 / f' (0)
f'(x) = 6x^2+6x+3
f'(0) = 3
1/f'(0) = 1/3

Positive: 10 %

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Answer #1| 08/02 2014 12:1253 %