Question

Determinati inversa functiei (f^-1)derivat(-2):

f:R->R
f(x)=x^3+3x+2

Answer 1

$f:\mathbb{R}\to\mathbb{R}$
$f(x) = x^3+3x+2 \\ (f^{-1})'(-2) = ? \\ \\ f(-1) = -2 \\ \Rightarrow f^{-1}(-2) = -1\\ \\f^{-1}\Big(f(x)\Big) = x \\f^{-1}(x^3+3x+2) = x \\ \Big[f^{-1}(x^3+3x+2)\Big]' = x' \\ \\ \text{Aplic\u{a}m regula lan\c{t}ului:}\\ \\ \Rightarrow (f^{-1})'(x^3+3x+2)\cdot (x^3+3x+2)' = 1 \quad (*)\\ \Rightarrow (f^{-1})(x^3+3x+2)\cdot(3x^2+3) = 1$

$\text{Trebuie s\u{a} rezolv\u{a}m ecuatia: } x^3+3x+2 = -2 \\ \\ \Rightarrow x^3+3x+4 = 0 \Rightarrow x^3-x+4x+4 = 0 \Rightarrow \\ \Rightarrow x(x^2-1)+4(x+1) = 0 \Rightarrow x(x-1)(x+1)+4(x+1) = 0 \Rightarrow \\ \Rightarrow (x+1)(x^2+3x+4) = 0 \Rightarrow x = -1 \text{ singura solu\c{t}ie real\u{a}.}$

$\text{Pentru }x = -1:}\\ \\ \Rightarrow (f^{-1})'\Big((-1)^3+3\cdot (-1)+2\Big)\cdot \Big(3\cdot (-1)^2+3\Big) = 1\\ \Rightarrow (f^{-1})'(-2) \cdot 6 = 1\\ \\ \Rightarrow \boxed{(f^{-1})'(-2) = \frac{1}{6}}$

$(*) \text{ De aici vine \c{s}i formula: } (f^{-1})'(y_0)\cdot f'(x_0) = 1 \Rightarrow \\ \\ \Rightarrow } (f^{-1})'(y_0) = \dfrac{1}{f'(x_0)}$



$A~NU~se~confunda~f'\Big(u(x)\Big)~ cu ~ \left[f\Big(u(x) \Big)\right]'.\\ \\ ~f'(x)=[f(x)]',\\ deoarece ~[f(x)]' = f'(x)\cdot x' = f'(x)\cdot 1 = f'(x)$