Fantastic Function
f − 1 = f ′
The inverse of function f is equal to its first derivative as shown above. If f ( 2 ) = 2 , what is the value of f ′ ′ ( 2 ) ?
The answer is 0.5.
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3 solutions
The derivative of an inverse function can be formulated as: [ f − 1 ]' × [f'( f − 1 )] = 1.
Since f − 1 = f', then f'' × f − 1 ( f − 1 ) = 1.
f(2) = 2 = f − 1 (2).
Substitute the value of 2 for f − 1 ,
f''(2)× f − 1 ( f − 1 (2)) = 1.
f''(2)×2 = 1; f"(2) = 0.5
Is there a function that satisfies this condition?
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I think so. The simplest form of polynomial f would have a degree of a golden ratio though I haven't found which would have f(2) = 2. (I caught a cold lately.) :(
A function which satisfies this condition could have the form A*x^B where B is the golden ratio (phi) and A = 2^(1-phi).
By inverse function theorem: f − 1 ( x ) = f ′ ( x ) → f ′ ( f − 1 ( x ) ) 1 = f ′ ′ ( x ) → f ′ ( f ′ ( x ) ) 1 = f ′ ′ ( x ) Also we are given that f ( 2 ) = 2 → f − 1 ( 2 ) = 2 → f ′ ( 2 ) = 2 → f ′ ′ ( 2 ) = f ′ ( f ′ ( 2 ) ) 1 = f ′ ( 2 ) 1 = 0 . 5
Because f ′ = f − 1 we can get f ′ ( f ( x ) ) = f − 1 ( f ( x ) ) = x
so f ′ ( f ( x ) ) = x so we can get f ′ ( f ( 2 ) ) = 2 or f ′ ( 2 ) = 2
we derivate each side so we can get
f " ( f ( x ) ) f ′ ( x ) = 1 substitute the value of 2 we can get
f " ( f ( 2 ) f ′ ( 2 ) ) = 1 ⇔ f " ( 2 ) = f ′ ( 2 ) 1 = 2 1 = 0 . 5