Question

@zepdrix

Answer 1

I'm stuck on the f'(g(0)) part

Answer 2

It's a toughie isn't it zep ?

Answer 3

I know g equals the inverse, which is x = y^3 - y - 6, but how to solve for y to find f'(g(0)), idk

Answer 4

\[\large\rm f(x)=x^3-x-6\]g is the inverse function,
it takes values from the range, and maps them back to the domain.\[\large\rm g(y)=f^{-1}(y)\]So if we have something like this,\[\large\rm g(0)=f^{-1}(0)\]it's telling us that our y-coordinate is 0,
and we want the x-coordinate which corresponds to that.\[\large\rm 0=x^3-x-6\]Factoring! :)

Answer 5

right. that's the part I'm stuck on...how do I find x?

Answer 6

Rational Root Theorem tells us that we can simply look at the factors of our -6:
1, -1, 2, -2, 3, -3, 6, -6

One of those values has to work.
If factoring is too hard, we can just plug and chug :)
See which of those values gives us 0.

For x=1,\[\large\rm (1)^3-(1)-6\ne0\]Ok so x=1 is NOT a solution.

Answer 7

Ah sorry, I should've said,
`if this polynomial has a rational root`, then rational root theorem guarantees one of those roots :P

Answer 8

Luckily it does :)

Answer 9

Oh! 2 works

Answer 10

Ooo nice.
You could do polynomial long division or synthetic division,
dividing out the x-2 factor, to get a remaining 2nd degree polynomial.
But it turns out the roots of that polynomial involve imaginary numbers.
So this is our only real solution.

Good good good,
so we've determined that: \(\large\rm g(0)=2\)

Answer 11

Okay, so then we find teh derivative of f and plug in 2! Thanks!

Answer 12

yay \c:/

Answer 13

Oh oh ummm

Answer 14

what?

Answer 15

I think I did the g'(0) part wrong, so I'm gonna redo that part too

Answer 16

no, I didn't do it wrong. Still got -1, lol

Answer 17

for g'(0) I mean

Answer 18

Oh it's g'(0) on the outside +_+ woopsy my bad

Answer 19

I think I did something wrong.. I got a final answer of -11....and that's not an answer choice

Answer 20

Ahh sorry getting distracted XD

Answer 21

Hmm this one has me pretty confused XD

Answer 22

Ya we might need somebody smart for this one :D
My brain isn't working right