1. ## 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?

2. 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 $f^{-1}{}'=\frac{1}{f'(x)}$ , when the derivative in the denominator is expressed in the same variable as the one of $f^{-1}$ , of course.

So in this case $f^{-1}{}'(y)=\frac{1}{3x^2+3}$ , 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 $f(x)=2\Longleftrightarrow\,x=-1$ , then $f^{-1}(2)=-1\Longrightarrow\,f^{-1}{}'(2)=\frac{1}{3(-1)^2+3}=\frac{1}{6}$

Tonio

3. 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?
$f[f^{-1}(x)] = x$

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

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

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

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

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

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

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

the second post confused me. is f^-1'(2)=1/6. i didnt quite understand the last part

5. ${f^{-1}}'(2) = \frac{1}{f'[f^{-1}(2)]}$

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

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

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

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

so ...

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

unconfused now?

6. yes thank you