Thread: Function within a function problem

Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

(A) g(8) = 1/3 and g'(8) = -1/9
(B) g(8) = 1/3 and g'(8) = 1/9
(C) g(8) = 3 and g'(8) = -9
(D) g(8) = 3 and g'(8) = -1/9
(E) g(8) = 3 and g'(8) = 1/9

f(g(x)) = x and g(f(x)) = x

This means that these functions are inverses.

Can you take it from there?

Originally Posted by mamori

Hello again! I am again lost on another problem. I wasn't sure what to call this, so hopefully it's still somewhat understandable. Thank you very much in advance!

The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?

(A) g(8) = 1/3 and g'(8) = -1/9
(B) g(8) = 1/3 and g'(8) = 1/9
(C) g(8) = 3 and g'(8) = -9
(D) g(8) = 3 and g'(8) = -1/9
(E) g(8) = 3 and g'(8) = 1/9

First from sub. x=3. Next differentiate wrt x so , then subs x= 3 again.

You're thinking way too much. =p

1.

since g(f(x)) = x

g(f(3)) = 3

But f(3) = 8

so g(8) = 3

2. f(g(x)) = x


f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

In general g ' (c) = 1/ f ' (a) where f(a) = c

Or From first principles:

[f(g(x)] ' = 1 for all x

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

at x = 3 f ' (g(8)) g ' (8) = 1

f ' (3) *g ' (8) =1

9* g ' (8) =1

g' (8) =1/9

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

Originally Posted by derfleurer

You're thinking way too much. =p

Originally Posted by Calculus26

1.

since g(f(x)) = x

g(f(3)) = 3

But f(3) = 8

so g(8) = 3

2. f(g(x)) = x


f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

In general g ' (c) = 1/ f ' (a) where f(a) = c

Or From first principles:

[f(g(x)] ' = 1 for all x

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

at x = 3 f ' (g(8)) g ' (8) = 1

f ' (3) *g ' (8) =1

9* g ' (8) =1

g' (8) =1/9

f and g are inverses so you have a theorem which tells you

g ' (8) = 1/ f'(3)

Apparently me and Calc26

Hi everyone! Thank you so much for all your help! I did understand that the functions are inverses, but I forgot the theorem for the derivatives. So thanks for your explanation, Calculus26! I do have one question. Where does

But [f(g(x)] ' also equals f ' (g(x)) g ' (x)

come from? Another theorem? I never learned this theorem/rule, so it's a bit tough on understanding this.

That's just chain rule.

sin(x^2)' = cos(x^2)(2x)

And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.

Danny-- I thought we were paid by the number of words

Originally Posted by derfleurer

That's just chain rule.

sin(x^2)' = cos(x^2)(2x)

And if you understood that they were inverses, you didn't really need to remember any calculus rules. Just know that the slope of the inverse is 1/m and that your domain/range swap. So f(3) = 8 yields f^-1(8) = 3.

Thanks. =] I just understood everything on this page. Somehow that inverse slope rule slipped my mind. Thanks a million everyone!

Originally Posted by Calculus26

Danny-- I thought we were paid by the number of words

Don't we wish