Question

Let f be a differentiable function such that f(-4)=12, f(9)=-4, f'(4)=-6, and f'(9)=3. The function g is differentiable and g(x)=finverse(x). What is the value of g'(-4)?

Answer 1

g(x) = f^-1(x) ?

Answer 2

Yes! that's what I meant

Answer 3

g'(4) = 1/f' (f^-1(-4)) = 1/f'( g(-4) )

Answer 4

so you need to know what g(-4) is, say c. I.e g(-4) = c

and then you do 1/f'(c)

Answer 5

you can find c based on the givens: f(-4)=12, f(9)=-4, f'(4)=-6, and f'(9)=3

Answer 6

you get it?

Answer 7

I follow that you're doing the derivative of an inverse formula where you have g'(x) =[ f-1(x) ], but when you have
g'(4) = 1/f' (f^-1(-4)) = 1/f'( g(-4) )
I don't get why you put (f^-1(-4) in the denominator's argument for f'(x), or why that became the same thing as g(4) in the next step

Answer 8

sorry that was a typo, it should have been
g'(-4) = 1/f' (f^-1(-4)) = 1/f'( g(-4) )

Answer 9

the formula for derivative of inverse function is:
\[\frac{d}{dx} f^{-1} (x) = \frac{1}{f'(f^{-1} (x))}\]

Answer 10

I let g(x) = f^-1(x)

Answer 11

so if you replace g(x) with f^-1(x), you have:
\[\frac{d}{dx} g(x) = \frac{1}{f'(g(x))}\]

Answer 12

and x = -4, hence
\[\frac{d}{dx} g(4) = \frac{1}{f'(g(4))}\]

Answer 13

urg.. i meant -4
-.-

Answer 14

so you need to know what g(-4) is. Like I said earlier, let g(-4) = c,

find c, then do 1/f'(c)

Answer 15

here is a hint: since g and f are inverse of each other, (-4,c) is g means (c,-4) is on f.
In other words f(c) = -4

Answer 16

I got g(-4) = 9, and then f'(9)=3, so doing 1/f'(c), the final answer would be 1/3? Is that right?

Answer 17

so f(9) = -4, hence c = 9

1/f'(9) = 1/3.

Yes, 1/3 is correct

Answer 18

okay thanks! but why is the formula 1/f'(f-1x) in the first place? how did you get that?

Answer 19

it's just a formula that you have to memorize to be able to do this problem