Help with inverse of derivative function

[hazy001]

Please help my homework is due tomorrow

Find (f −1)'(a).

1. f(x) = 9 + x² + tan(πx/2), −1 < x < 1, at a = 9

2. f(x) = $\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }$, at a = 3

I can figure out simpler ones but these are out of my way. thank you so much

the formula is $\displaystyle \left(f^{-1}\right)'(x)\, =\, \dfrac{1}{f'\left(f^{-1}(x)\right)}$

 

[stapel]

What relationship have they given you between a function, its inverse, its derivative, and the derivative of the inverse? How far have you gotten in applying this relationship?

 

[hazy001]

Hey

\(\displaystyle f(x)\, =\, \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}\, \mbox{ at }\, a\, =\, 3\)

the formula is $\displaystyle \left(f^{-1}\right)'(x)\, =\, \dfrac{1}{f'\left(f^{-1}(x)\right)}$

So I set $\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }$

then I plug x into that and put it 1 over

 

[HallsofIvy]

If you were able to do the "simpler ones" then you should have learned that $\displaystyle \frac{d f^{-1}(x)}{dx}= \frac{1}{\frac{df}{dx}}$.

 

[hazy001]

\(\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}\, \mbox{ at }\, a\, =\, 3\)

So I set:

$\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }\, =\, 3$

Solve for x which I can't do.

Find the derivative of $\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}$

Plug x into the derivative, then that goes under 1.

The formula is $\displaystyle \left(f^{-1}\right)'(x)\, =\, \dfrac{1}{f'\left(f^{-1}(x)\right)}$

 

[stapel]

Find (f −1)'(a).

2. f(x) = $\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }$, at a = 3

\(\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}\, \mbox{ at }\, a\, =\, 3\)

So I set:

$\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }\, =\, 3$

Solve for x which I can't do.

Find the derivative of $\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}$

Plug x into the derivative, then that goes under 1.

The formula is $\displaystyle \left(f^{-1}\right)'(x)\, =\, \dfrac{1}{f'\left(f^{-1}(x)\right)}$

I think you're maybe trying to find the value of x so that f(x) = a = 3, because then you'd know the value of f^-1(a). (I'm guessing. It really does help clearly to state your reasoning and show all of your work, so we can know, rather than merely guess, what you're doing and why you're getting stuck.) So let's try finding the value of x that gives f(x) = 3:

$\displaystyle \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, }\, =\, 3$

Square both sides:

$\displaystyle 3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\, =\, 9$

$\displaystyle 3x^3\, +\, 3x^2\, +\, 2x\, -\, 8\, =\, 0$

Look at the graph in your graphing calculator to find the one real root, and factor. (Or else use the old-style method to find the zeroes.) The poly factors as (x - 1)(3x² + 6x + 8) so the only real-number solution is x = 1. Then f(1) = 3 so f^-1(3) = 1.

What is the derivative of f(x)? Once you find this, you plug in the value of f^-1(3) and simplify. Then find the reciprocal, and you're done.

 

[hazy001]

inverse of derivative function help

f(x) =

3x3 + 3x2 + 2x + 1

, a = 3

formal is

Homework is due tonight and this is the only problem i cant solve

Your suppose to
3=

3x3 + 3x2 + 2x + 1

,
then plug x into that and put it under 1.

 

[Ishuda]

I'm not sure what you want but just to make it more clear for me the definition of an inverse function is given by something like
"g(x) is the inverse function of f(x) if and only if g(f(x)) = x".
That being the case, if f(x) is differentiable, then by the chain rule
$\displaystyle \frac{dg}{dx} = g' f' = 1$
or
$\displaystyle g'(x) = \frac{1}{f'(x)}$

So suppose f(2) = 7 . If g is the inverse of f then g(f(2)) = g(7) = 2 and g'(7) = 1/f'(7).

Note that if x=1 then $\displaystyle \sqrt{3 x^3 + 3 x^2 + 2 x + 1} = 3$

 

[stapel]

\(\displaystyle f(x)\, =\, \sqrt{3x^3\, +\, 3x^2\, +\, 2x\, +\, 1\,}\, \mbox{ at }\, a\, =\, 3\)

The formula for this is $\displaystyle \left(f^{-1}\right)(x)\, =\, \dfrac{1}{f'\left(f^{-1}(x)\right)}$

Homework is due tonight and this is the only problem i cant solve

This is at least your third posting of this question. (The various threads have been merged into one.) Is there any particular reason you're not following any of the suggestions, advice, or helps you've already received?