i cant remember how to do these... The given function f is differentiable. Verify that f has an inverse and find (f^-1)'(c) f(x)= x^3 + 1 ; c=9
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i think you have to show that f is bijective.
what do you mean ?
Originally Posted by runner07 find (f^-1)'(c) f(x)= x^3 + 1 ; c=9 Use the formula: $\displaystyle \left( f^{-1} \right)'(c) = \frac {1}{f' \left( f^{-1}(c) \right)}$
Originally Posted by runner07 Verify that f has an inverse f(x)= x^3 + 1 ; c=9 I think we should just find the inverse. do you know how to do that?
yes.. so then we dont use that formula you posted earlier? we just find the inverse? i got- inverse= (x-3)^(1/3) so according to that formula you posted earlier, i would plug in 9 where x is?
Originally Posted by runner07 yes.. so then we dont use that formula you posted earlier? we just find the inverse? i got- inverse= (x-3)^(1/3) so according to that formula you posted earlier, i would plug in 9 where x is? the question has two parts. for the second part, you could use the formula i gave, but if you want to differentiate that, i guess that's fine too it should be $\displaystyle f^{-1}(x) = (x - 1)^{1/3}$
so then... (F^-1)'(c)= 1 / f'(f^-1(c)) (F^-1)'(9)= 1 / f'(f^-1(9)) ---> f^-1(9)= (9-1)^(1/3) am i on the right track?
Originally Posted by runner07 so then... (F^-1)'(c)= 1 / f'(f^-1(c)) (F^-1)'(9)= 1 / f'(f^-1(9)) ---> f^-1(9)= (9-1)^(1/3) am i on the right track? yeah, you're ok. but instead of (9 - 1)^(1/3), just write 2 so then $\displaystyle \left( f^{-1} \right)'(9) = \frac {1}{f'(2)}$ you can do it the way you suggested to verify your answer
Originally Posted by Jhevon find all the pieces and then plug them into the formula. what is the formula for $\displaystyle f'(x)$ in this case? what is $\displaystyle f^{-1}(9)$ ? $\displaystyle f^{-1}(x) = (x - 1)^{1/3} $ and the derivative of the inverse is... f'(x)= 1/3(x-1)^(-2/3) right? and then i plug in 9 for the x's and thats it?
Originally Posted by runner07 $\displaystyle f^{-1}(x) = (x - 1)^{1/3} $ and the derivative of the inverse is... f'(x)= 1/3(x-1)^(-2/3) right? and then i plug in 9 for the x's and thats it? yes
Originally Posted by runner07 $\displaystyle f^{-1}(x) = (x - 1)^{1/3} $ and the derivative of the inverse is... f'(x)= 1/3(x-1)^(-2/3) right? and then i plug in 9 for the x's and thats it? If you are going to use ASCII to write mathematical expressions try to use more brackets to make your meaning clear and unambiguous. So write: f'(x)= (1/3) (x-1)^(-2/3) say. RonL
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