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 ${s}^{\prime}=\frac{2{p}^{3}}{3{s}^{5}}$. $s$ is increasing where ${s}^{\prime}>0$ and decreasing where ${s}^{\prime}<0$.

Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

Well, I answeblack first like following: Calculating the first derivative using the implicit differentiation is ${s}^{\prime}=\frac{2{p}^{3}}{3{s}^{5}}$. $s$ is increasing where ${s}^{\prime}>0$ and decreasing where ${s}^{\prime}<0$.

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

Expert

2022-07-17Added 21 answers

We are talking here about the solution set

$S:={\textstyle \{}(p,s)\phantom{\rule{mediummathspace}{0ex}}{\textstyle |}\phantom{\rule{mediummathspace}{0ex}}F(s,p)=0{\textstyle \}}\text{},$

where

$F(p,s)={s}^{6}-{p}^{4}-1\text{}.$

The implicit function theorem says the following: When $({p}_{0},{s}_{0})\in S$ and the "technical condition"

$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial}F}{\mathrm{\partial}s}{{\textstyle |}}_{({p}_{0},{s}_{0})}=6{s}_{0}^{5}\ne 0\end{array}$

is satisfied then there is a window

$W:=\text{}]{p}_{0}-h,{p}_{0}+h[\text{}\times \text{}]{s}_{0}-{h}^{\prime},{s}_{0}+{h}^{\prime}[\text{}$

and a ${C}^{1}$-function

$\psi :\phantom{\rule{1em}{0ex}}]{p}_{0}-h,{p}_{0}+h[\text{}\to \text{}]{s}_{0}-{h}^{\prime},{s}_{0}+{h}^{\prime}[\text{},$

such that $S\cap W$ equals the graph of $\psi $. Furthermore one has

${\psi}^{\prime}({p}_{0})=-\frac{{F}_{p}({p}_{0},{s}_{0})}{{F}_{s}({p}_{0},{s}_{0})}=\frac{2{p}_{0}^{3}}{3{s}_{0}^{5}}\text{}.$

Now $S$ contains no points with $s=0$; therefore (1) is satisfied at all points of $S$S.

Whether the local function $\psi $ 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 $\psi $ is increasing (for small enough $h>0$), otherwise $\psi $ is decreasing. The points $(0,\pm 1)\in S$ are special, since ${\psi}^{\prime}(0)=0$ there. Here further analysis is necessary, which I leave to you.

$S:={\textstyle \{}(p,s)\phantom{\rule{mediummathspace}{0ex}}{\textstyle |}\phantom{\rule{mediummathspace}{0ex}}F(s,p)=0{\textstyle \}}\text{},$

where

$F(p,s)={s}^{6}-{p}^{4}-1\text{}.$

The implicit function theorem says the following: When $({p}_{0},{s}_{0})\in S$ and the "technical condition"

$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial}F}{\mathrm{\partial}s}{{\textstyle |}}_{({p}_{0},{s}_{0})}=6{s}_{0}^{5}\ne 0\end{array}$

is satisfied then there is a window

$W:=\text{}]{p}_{0}-h,{p}_{0}+h[\text{}\times \text{}]{s}_{0}-{h}^{\prime},{s}_{0}+{h}^{\prime}[\text{}$

and a ${C}^{1}$-function

$\psi :\phantom{\rule{1em}{0ex}}]{p}_{0}-h,{p}_{0}+h[\text{}\to \text{}]{s}_{0}-{h}^{\prime},{s}_{0}+{h}^{\prime}[\text{},$

such that $S\cap W$ equals the graph of $\psi $. Furthermore one has

${\psi}^{\prime}({p}_{0})=-\frac{{F}_{p}({p}_{0},{s}_{0})}{{F}_{s}({p}_{0},{s}_{0})}=\frac{2{p}_{0}^{3}}{3{s}_{0}^{5}}\text{}.$

Now $S$ contains no points with $s=0$; therefore (1) is satisfied at all points of $S$S.

Whether the local function $\psi $ 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 $\psi $ is increasing (for small enough $h>0$), otherwise $\psi $ is decreasing. The points $(0,\pm 1)\in S$ are special, since ${\psi}^{\prime}(0)=0$ there. Here further analysis is necessary, which I leave to you.

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