jamesbond007 Messages 1 Reaction score 0 Thread starter Nov 30, 2009 #1 given that f(x)=x³+3x+6 i need to find f^-1'(2)
coki2000 Messages 91 Reaction score 0 Nov 30, 2009 #2 jamesbond007 said: given that f(x)=x³+3x+6 i need to find f^-1'(2) Hi, Derivatives of inverse functions for y=f(x) [tex]y=f(x)\Rightarrow f^{-1}(y)=x=f^{-1}(f(x))[/tex] if we take derivatives of both sides dependent to x, [tex]f'(x)[(f^{-1})'(f(x))]=1\Rightarrow(f^{-1})'(f(x))=\frac{1}{f'(x)}[/tex].
jamesbond007 said: given that f(x)=x³+3x+6 i need to find f^-1'(2) Hi, Derivatives of inverse functions for y=f(x) [tex]y=f(x)\Rightarrow f^{-1}(y)=x=f^{-1}(f(x))[/tex] if we take derivatives of both sides dependent to x, [tex]f'(x)[(f^{-1})'(f(x))]=1\Rightarrow(f^{-1})'(f(x))=\frac{1}{f'(x)}[/tex].
HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Dec 1, 2009 #3 Of course, f'(x) has to be evaluated at the x such that f(x)= 2, so you need to start by solving [itex]x^3+ 3x+ 6= 2[/itex]. Fortunately, that's very easy.
Of course, f'(x) has to be evaluated at the x such that f(x)= 2, so you need to start by solving [itex]x^3+ 3x+ 6= 2[/itex]. Fortunately, that's very easy.