Question

g(t)=t^3-3+5
what is (g^-1)'(t)

Answer 1

is your g(t) written correctly?

Answer 2

just asknig because as is it is t^3+2

Answer 3

my bad i meant t^3-3t+5

Answer 4

HI
you are not going to find the inverse

Answer 5

what is the t value here

Answer 6

\[(g^{-1})'(t)=\frac{1}{g'(g^{-1}(t))}\]
at what number t

Answer 7

so i take the inverse and plug in t?

Answer 8

well you aren't going to be able to find the inverse relation as my dear friend @misty1212 said

Answer 9

why not

Answer 10

i bet it is \[(g^{-1})'(2)=\frac{1}{g'(g^{-1}(2))}\] or sommat

Answer 11

because you do not know how to solve a cubic equation

Answer 12

can you solve this for t: y=t^3-3t+5
because I can't

Answer 13

whered u get 2

Answer 14

can i just set t^3-3t+5=5

Answer 15

i made it up

Answer 16

so t is 5

Answer 17

and that t value is g^-1(5)

Answer 18

\[t^3-3t+5=5\]
observe which t satisfies this

Answer 19

use \(t=0\)

Answer 20

t= 0 and t = sqrt(3)

Answer 21

which one should i use

Answer 22

i don't think the latter works

Answer 23

but t^3-3t=0

Answer 24

and take out t and its t^2-3=0

Answer 25

and t=0

Answer 26

oops I was wrong
lol
this function doesn't seem to be one to one

Answer 27

do you have any restrictions on g

Answer 28

the question before says find the largest interval on which g has an inverse

Answer 29

that actually gives you 3 values
i bet you are going to use the t=0

Answer 30

containing 0

Answer 31

then 0!

Answer 32

lol

Answer 33

makes no difference, the derivative still exists locally

Answer 34

so -1/3

Answer 35

\[(g^{-1})'(0)=\frac{1}{g'(0)}\]

Answer 36

sounds great