Question

Find (f^-1)'(a) of:
2x^3+3x^2+7x+4

Answer 1

Can you do the derivative of f(x)
or do the inverse first and figure the Derivative?

Answer 2

\[f'(x) = 6x ^{2}+6x+7\]

Answer 3

Would it be easier ("nicer") to do the inverse of the f'(x) or just the inverse of f(x)?
I think it will be ugly either way?

Answer 4

Hmm I actually haven't seen this type of problem before :D lol

Lemme think about it a few minutes...
My initial guess is that we probably have to take the inverse first, get it back in terms of x (somehow..) THEN differentiate.

Answer 5

Ok, I will keep trying it also. I have done just that:
\[y=2x ^{3}+3x ^{2}+7x+4\]
\[x=2y ^{3}+3y ^{2}+7y+4\]
Then solve. That is what i have been doing.

Answer 6

Also tried the same above but with the derivative.

Answer 7

My chapter is eventually working into Exponential,Logarithmic, and Inverse Trig Functions.

Answer 8

\[x=2y ^{3}+3y ^{2}+7y+4\]
\[x-4 = y(2y ^{2}+3y+7)\]
\[x-4=y(2y+1)(y+3)\]

Answer 9

?

Answer 10

\[\large \left[f^{-1}\right]'(a)=\frac{1}{f'(f^{-1}(a))}\]I keep coming across this equation.. still trying to make sense of it though :D heh

Answer 11

DOH
SORRY, totally forgot. a=4

Answer 12

Oh hmm maybe we can do something with that :)

Answer 13

Which you still need to find the inverse though ... right?

Answer 14

seems like it XD lol
that seems to be the tricky part, hmmm

Answer 15

Of course it will be tricky :-)
Calculus = 1% Calc and 99% Algebra. But that 1% Calculus make the Algebra "tricky".

Answer 16

Is it "legal" to do the following?
\[x=2y ^{3}+3y ^{2}+7y+4\]
\[x ^{1/3}=2(\sqrt[3]{3y ^{2}+7y+4}\]

Answer 17

Where did the y^3 go? You were attempting to take the cube root of that separately from the rest of the thing? Mmm yah that doesn't quite work the way you want it to :C

Answer 18

figured :-)

Answer 19

How about what I did up above, would that get us anywhere?

Answer 20

Maybe yes, let's think about this a second...\[\large x=2y^3+3y^2+7y+4\]Here is the inverse function of f, written in terms of Y.
If we were ABLE to write it in terms of x, we would be plugging in a=4 for the X value right?

Let's try plugging in 4 for the X value right now, maybe that will get us somewhere.\[\large 4=2y^3+3y^2+7y+4 \qquad \rightarrow \qquad 0=2y^3+3y^2+7y\]

Answer 21

hmmmmmm
\[f ^{-1}(y) = x <---> f(x)=y\]
Could we put a into f(x) and that would give us y?

Answer 22

Hmm that's an interesting idea :o

Answer 23

\[f(4)=208\]
\[f ^{-1}(208)=4\]
\[(f ^{-1})'(a)=\frac{ 1 }{ f'(f ^{-1}(4)) }\]
\[(f ^{-1})'(a) = \frac{ 1 }{ f'(208) }\]
\[(f ^{-1})'(a)=\frac{ 1 }{ 260839 }\]
But .... Not even close t the answer in the back.

Answer 24

What's the back say? :o

Answer 25

Heh
\[\frac{ 1 }{ 7 }\]

Answer 26

Close in some aspect :-)

Answer 27

lol :3

Answer 28

I typed this into Wolfram:
inverse of 2x^3+3x^2+7x+4
And got this:

Answer 29

I hope that this is not what I need to do.
That is ....... not pretty.

Answer 30

No I'm sure it's not what you need to do.
There is clearly some little trick that we're just not seeing. :\

Answer 31

And everyone who pops on to take a look runs screaming.

Answer 32

@Zarkon is a pretty smart dude :d Maybe he'll take a look at it if he has time. :D
I might have to give up on this one :C grr

Answer 33

That's ok. I appreciate all the help.
I may try to send a message and see what happens.
Thank you.

Answer 34

Well I did the little @ symbol thing. That is the best way to try and get someones attention. :D

It's like paging someone. It lets them know that you called, and they can click on the page to come directly to your thread.
So it's a handy little tool to use :D

Answer 35

Didn't know that.
Thanks for the little nugget of info.

Answer 36

Hello

Answer 37

@jim_thompson5910
Help?

Answer 38

@helder_edwin
Help?
Inverse Derivatives

Answer 39

Hello

Answer 40

Thank you for coming and taking a look.

Answer 41

a = 4 correct?

Answer 42

Correct.

Answer 43

when you plug in x = 0, you get f(x) = 4

so f^-1(4) = 0, agreed?

Answer 44

agreed .....

Answer 45

so...

[ f^-1]' (a) = 1/[ f' ( f^-1(a) ) ]

[ f^-1]' (4) = 1/[ f' ( f^-1(4) ) ]

[ f^-1]' (4) = 1/[ f' ( 0 ) ]

[ f^-1]' (4) = 1/7

Answer 46

so that would make:
\[\frac{ 1 }{ f'(0) }\]

Answer 47

f(x) = 2x^3+3x^2+7x+4

f ' (x) = 6x^2+6x+7

f ' (0) = 6(0)^2+6(0)+7

f ' (0) = 7

Answer 48

yes

Answer 49

yes.
So completly wrong plugging 4 in and solving the equation.

Answer 50

what do you mean?

Answer 51

Posted this earlier:

hmmmmmm
f−1(y)=x<−−−>f(x)=y

Could we put a into f(x) and that would give us y?

Answer 52

it's the other way around, you plug in y = a and solve for x

Answer 53

to find f^-1(a)

Answer 54

So I did f(x) = 2(4)^3+3(4)^2+7(4)+4
When I should of done:
4=.............

Answer 55

if you plugged in x = 4 into f(x), you'd get 208, but that's way off of what we want

Answer 56

yeah exactly

Answer 57

luckily 4 is the y-intercept, so it's kinda easy to solve for (if you know what to look for)

Answer 58

ohhh man
way to easy, thank you form coming to help.

Answer 59

yw

Answer 60

Thank you for coming to take a look.
Made it way to complecated.