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Find (f^-1)'(a) for f(x) = 2x^3+3x^2+3x+7, a=7?
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Find (f^-1)'(a) for f(x) = 2x^3+3x^2+3x+7, a=7?
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Answer #1
| 08/02 2014 12:12
(f^-1(x))' = 1/f'(f^-1(x) ) f'(x) = 6x^2 + 6x + 3 f^-1(7) = 0, because f(0) = 7 f'(0) = 3 so 1/f'(0) = 1/3
Answer #2
| 08/02 2014 12:48
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)
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
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
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Answer #1| 08/02 2014 12:12