Question

If f(x) = x^3+2x+1 and g is the inverse function of f, so, g'(1) is equal to:
-1/4
-1/2
0
1/4
1/2

Someone know how to do?

Answer 1

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

Answer 2

Do you know how to find \[f^{-1}(1)=? \]

Answer 3

I've found that f inverse is equals to x-1/(y^2+2). Is this correct?

Answer 4

no you can't easily find the general form of the inverse
answer my question :)

Answer 5

\[f^{-1}(0)=? \text{ implies } f(?)=0\]
What can you plug into your function that will give you 0 as output?

Answer 6

oops that 0 is suppose to be a 1 by the way

Answer 7

We are looking for \[f^{-1}(1)=? \text{ \implies } f(?)=1 \]

Answer 8

Sorry, I went dinner. ^^. It makes sense! Then x^2 = -1/3 right?

Answer 9

answer my question first
I don't know what you mean
what can you plug into your function so that the output is 1

Answer 10

what does \[f(0)=?\]

Answer 11

I don't know ^^. This is the enuncioation of the exercis!

Answer 12

\[f(x)=x^3+2x+1\]
To evaluate f(0) just replace all the x's with 0

Answer 13

Tell what is \[f(0)=0^3+2(0)+1\]

Answer 14

But the derivative of f(x) that equal to 1 right?

Answer 15

?
Right now we are trying to find the ? in
f^{-1}(1)=? \text{ \implies } f(?)=1

Answer 16

\[f^{-1}(1)=? \text{ \implies } f(?)=1\]

Answer 17

Aha right

Answer 18

Can you evalue f(0) for me ?

Answer 19

Like I wrote what f(0) was all you have to do is simplify

Answer 20

f(0) = 1

Answer 21

great so remember we are trying to evaluate
\[(f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))}=\frac{1}{f'(0)}\]
Now you need find f'
and evaluate f'(0)

Answer 22

Is equal to 2 right?

Answer 23

first what did you get for f'

Answer 24

3x^2 +2

Answer 25

Ok great

Answer 26

The answer is 1/2 right?

Answer 27

yes

Answer 28

Thanks @myininaya !!