f(g(x)) = x and g(f(x)) = x
This means that these functions are inverses.
Can you take it from there?
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
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)
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.