# finding inverse

• Nov 29th 2009, 01:36 PM
johnnyc418
finding inverse
f(x)=x³+3x+6

i need to find f^-1'(2)

do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?
• Nov 29th 2009, 01:50 PM
tonio
Quote:

Originally Posted by johnnyc418
f(x)=x³+3x+6

i need to find f^-1'(2)

do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?

I suppose you already know that $\displaystyle f^{-1}{}'=\frac{1}{f'(x)}$ , when the derivative in the denominator is expressed in the same variable as the one of $\displaystyle f^{-1}$ , of course.

So in this case $\displaystyle f^{-1}{}'(y)=\frac{1}{3x^2+3}$ , when x is expressed as function of y, and we don't have the faintest idea what is this...but we don't need to have: just check that as $\displaystyle f(x)=2\Longleftrightarrow\,x=-1$ , then $\displaystyle f^{-1}(2)=-1\Longrightarrow\,f^{-1}{}'(2)=\frac{1}{3(-1)^2+3}=\frac{1}{6}$

Tonio
• Nov 29th 2009, 01:51 PM
skeeter
Quote:

Originally Posted by johnnyc418
f(x)=x³+3x+6

i need to find f^-1'(2)

do i need to find the inverse of f(x) before i derive it. or do i derive it first, then find the inverse, and then plug 2 in for x?

$\displaystyle f[f^{-1}(x)] = x$

$\displaystyle f'[f^{-1}(x)] \cdot {f^{-1}}'(x) = 1$

$\displaystyle {f^{-1}}'(x) = \frac{1}{f'[f^{-1}(x)]}$

now, use the fact that $\displaystyle f(-1) = 2$ and $\displaystyle f^{-1}(2) = -1$ to find $\displaystyle {f^{-1}}'(2)$
• Nov 29th 2009, 02:04 PM
johnnyc418
Quote:

Originally Posted by skeeter
$\displaystyle f[f^{-1}(x)] = x$

$\displaystyle f'[f^{-1}(x)] \cdot {f^{-1}}'(x) = 1$

$\displaystyle {f^{-1}}'(x) = \frac{1}{f'[f^{-1}(x)]}$

now, use the fact that $\displaystyle f(-1) = 2$ and $\displaystyle f^{-1}(2) = -1$ to find $\displaystyle {f^{-1}}'(2)$

the second post confused me. is f^-1'(2)=1/6. i didnt quite understand the last part
• Nov 29th 2009, 02:12 PM
skeeter
$\displaystyle {f^{-1}}'(2) = \frac{1}{f'[f^{-1}(2)]}$

$\displaystyle {f^{-1}}'(2) = \frac{1}{f'(-1)}$

since $\displaystyle f(x) = x^3+3x+6$

$\displaystyle f'(x) = 3x^2 + 3$

$\displaystyle f'(-1) = 3(-1)^2 + 3 = 6$

so ...

$\displaystyle {f^{-1}}'(2) = \frac{1}{f'(-1)} = \frac{1}{6}$

unconfused now?
• Nov 29th 2009, 02:14 PM
johnnyc418
yes thank you