Question

g(x) = x^3+2x-1

Answer 1

@umerlodhi ca you help with this?

Answer 2

ye i can but not rn lol sorry i have prior engagements starting in exactly 1 minute

Answer 3

Okay if you know anyone who can could you tag them real fast please

Answer 4

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

Answer 5

so we want to evaluate this at x=2

Answer 6

Yep

Answer 7

g^(-1)(2) is given

Answer 8

so the next step is to find g'(x)
and plug in the result of g^(-1)(2) into it

Answer 9

Okay I will find g'(x)

Answer 10

\[3x ^{2}+2\]

Answer 11

ok you still need to plug in the result of g^(-1)(2) into that

Answer 12

and then put 1/that result and you are done

this is what you have done so far
\[(g^{-1})'(2)=\frac{1}{g'(g^{-1}(2))}=\frac{1}{3(g^{-1}(2))^2+2}\]

Answer 13

not g' i meat g^-1

Answer 14

what

Answer 15

you are already given g^(-1)(2)
you just need to plug in its value

Answer 16

you don't have to any math to figure out g^(-1)(2)

Answer 17

oh okay

Answer 18

g^(-1)(x) at the points (2,1)
means we are given g^(-1)(2)=1

Answer 19

of course you can see that g(1)=2 holds which it should since g^(-1) is suppose to be g 's inverse relation

Answer 20

Yeah so for the equation you gave me \[(g^{-1})'(2)=\frac{1}{g'(g^{-1}(2))}=\frac{1}{3(g^{-1}(2))^2+2}\] I just use 1 to represent g^-1(x)?

Answer 21

you only need to replace g^(-1)(2) with 1
since g^(-1)(2)=1 is given by the problem

Answer 22

we know this is given because it tells (2,1) is a point on g^(-1)(x)

Answer 23

i have to go
but the rest of this problem is just order of operations

Answer 24

Do i just plug in 2 into the differentiated equation giving me 14?

Answer 25

how do you get 14 from 1/(3*1+2) ?

Answer 26

because at (2,1) the x value is 2

Answer 27

oh i just plugged 2 into g'

Answer 28

oh it looks like you avoided that there was a fraction altogether and you used that g^(-1)(2) is 2 even though they told us this was 1

Answer 29

yeah

Answer 30

so based off of the expression you gave then the answer is 1/5?

Answer 31

\[(g^{-1})'(2)=\frac{1}{g'(g^{-1}(2))}=\frac{1}{3(\color{red}{g^{-1}(2)})^2+2}\]
the value in red is given as 1 since (2,1) is a point on g^(-1)(x)

Answer 32

yes!

Answer 33

Okay!! Thanks!!

Answer 34

I hope you understand that (a,b) being on h means h(a)=b ...

Answer 35

Thank you for our time!

Answer 36

anyways i have to go put food in my tummy super hungry

Answer 37

Yeah that is in my notes