Considering s is implicit to function of p, given by s 6 − p 4...

anudoneddbv

anudoneddbv

Answered

2022-07-16

Considering s is implicit to function of p, given by s 6 p 4 = 1. For what s is it increasing and decreasing?
Well, I answeblack first like following: Calculating the first derivative using the implicit differentiation is s = 2 p 3 3 s 5 . s is increasing where s > 0 and decreasing where s < 0.
Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

Answer & Explanation

slapadabassyc

slapadabassyc

Expert

2022-07-17Added 21 answers

We are talking here about the solution set
S := { ( p , s ) | F ( s , p ) = 0 }   ,
where
F ( p , s ) = s 6 p 4 1   .
The implicit function theorem says the following: When ( p 0 , s 0 ) S and the "technical condition"
(1) F s | ( p 0 , s 0 ) = 6 s 0 5 0
is satisfied then there is a window
W :=   ] p 0 h , p 0 + h [   ×   ] s 0 h , s 0 + h [  
and a C 1 -function
ψ : ] p 0 h , p 0 + h [     ] s 0 h , s 0 + h [   ,
such that S W equals the graph of ψ. Furthermore one has
ψ ( p 0 ) = F p ( p 0 , s 0 ) F s ( p 0 , s 0 ) = 2 p 0 3 3 s 0 5   .
Now S contains no points with s = 0; therefore (1) is satisfied at all points of SS.

Whether the local function ψ is increasing or decreasing in its window depends on the signs of p 0 and s 0 : When p 0 and s 0 have the same sign ψ is increasing (for small enough h > 0), otherwise ψ is decreasing. The points ( 0 , ± 1 ) S are special, since ψ ( 0 ) = 0 there. Here further analysis is necessary, which I leave to you.

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