Question

f(g(x))=x, f(2)=7, and f'(2)=6 What is g'(7)?

Answer 1

I don't even know where to begin, we've never done a problem like this in class.

Answer 2

How about by differentiating both sides of
f(g(x)) = x ?

Answer 3

Okay, so I would get f'(g(x))=1?

Answer 4

Chain rule woes... you (should) know that there should also be a factor of g'(x) on the left-hand side ^_^

Answer 5

\[\Large f(g(x))= x\\\Large f'(g(x))g'(x) = 1\]
Chain rule... right?

Answer 6

yeah.
That makes sense.

Answer 7

i forgot to multiply by the derivative of the function inside the parentheses

Answer 8

Hopefully, it more than makes sense D:
Promise that you'll always remember rules of differentiation, ok?
Now, if we plug in x = 7, we get
\[\Large f'(g(7)) \color{blue}{g'(7)}=1\]

Bringing out an unknown value, g'(7) but what about f'(g(7)) ?
Do we know what that it? ;)

Answer 9

that is*

Answer 10

Pay attention to the first equation...
f(g(x)) = x...

What does that tell you about f and g?

Answer 11

It means that they're inverses right?

Answer 12

That is correct...
So, using that knowledge and the second equation, f(2) = 7
How do you manipulate that to something more...usable?

Answer 13

It becomes f(7)=2?

Answer 14

Hell no :P
What you did was here was apply f to both sides of the equation, and somehow getting rid of the f in f(2)... but f is not the inverse of f... is it?

IF no, then what is? :P

Answer 15

g(7)=2?

Answer 16

That is correct ^_^
Now, back to the equation....
\[\Large f'(\color{red}{g(7)}) \color{blue}{g'(7)}= 1\]
Can you wrap it up from here?

Answer 17

I think so, thanks so much. You're really smart.

Answer 18

:)
No problem.

Answer 19

and it's g'(7) =\[\frac{ 1 }{ 6 }\]

Answer 20

Brilliant ^_^
That wasn't so hard, now, was it?

Answer 21

Thanks again. I really appreciate it.

Answer 22

You're welcome.
Now I'm signing off, good luck with the rest of your maths stuff :D

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Terence out