Addison Gross

2022-01-25

How do you find the critical point and determine whether it is a local maximum, local minimum, or neither for $f(x,y)={x}^{2}+4x+{y}^{2}$ ?

trasahed

Beginner2022-01-26Added 14 answers

The first-order partial derivatives of $z=f(x,y)={x}^{2}+4x+{y}^{2}$ are $\frac{\partial z}{\partial x}=2x+4$ and $\frac{\partial z}{\partial y}=2y$ .

Setting both of these equal to zero results in a system of equations whose unique solution is clearly$(x,y)=(-2,0)$ , so this is the unique critical point of $f$ .

The second-order partials are$\frac{{\partial}^{2}z}{\partial {x}^{2}}=2,\frac{{\partial}^{2}z}{\partial {y}^{2}}=2$ and $\frac{{\partial}^{2}z}{\partial x\partial y}=\frac{{\partial}^{2}z}{\partial y\partial x}=0$

This makes the discriminant for the (multivariable) Second Derivative Test equal to

$D=\frac{{\partial}^{2}z}{\partial {x}^{2}}\cdot \frac{{\partial}^{2}z}{\partial {y}^{2}}-{\left(\frac{{\partial}^{2}z}{\partial x\partial y}\right)}^{2}=2\cdot 2-{0}^{2}=4>0$

which means the critical point is either a local max or a local min (it's not a saddle point).

Since$\frac{{\partial}^{2}z}{\partial {x}^{2}}=2>0$ , the critical point is a local min.

Setting both of these equal to zero results in a system of equations whose unique solution is clearly

The second-order partials are

This makes the discriminant for the (multivariable) Second Derivative Test equal to

which means the critical point is either a local max or a local min (it's not a saddle point).

Since

lilwhitelieyc

Beginner2022-01-27Added 10 answers

The critical point makes both partial derivatives 0 (simultaneously).

For this function there is one critical point:

To determine whether f has a local minimum, maximum or neither at this point we apply the second derivative test for functions of two variables. (Well, we try to apply it. It does not always give an answer.)

Evaluate the second partials at the critical point (In this case they are all constant, but in general we cannot skip this step.)

At the critical point

Calculate

Apply the second derivative test:

Since D is positive, we look at A and with

f

f has a local minimum of−4 at

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A)Subtraction

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The significance of partial derivative notation

If some function like $f$ depends on just one variable like $x$, we denote its derivative with respect to the variable by:

$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$

Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:

$\frac{\mathrm{\partial}}{\mathrm{\partial}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Now my question is what is the significance of this notation? I mean what will be wrong if we show "Partial derivative" of $f$ with respect to ${x}_{i}$ like this? :

$\frac{\mathrm{d}}{\mathrm{d}{x}_{i}}f({x}_{1},\cdots ,{x}_{i},\cdots ,{x}_{n})$

Does the symbol $\mathrm{\partial}$ have a significant meaning?The function $f(x,y,z)$ is a differentiable function at $(0,0,0)$ such that ${f}_{y}(0,0,0)={f}_{x}(0,0,0)=0$ and $f({t}^{2},2{t}^{2},3{t}^{2})=4{t}^{2}$ for every $t>0$. Define $u=(6/11,2/11,9/11)$, with the given about. Is it possible to calculate ${f}_{u}(1,2,3)$ or ${f}_{u}(0,0,0)$, or ${f}_{z}(0,0,0)$?

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So, Derivative of $f(df)$ if exists, will satisfy $limH\to 0\frac{||f(A+H)-f(A)-df(H)||}{||H||}=0$.if $F(x,y)$ and $y=f(x)$,

$\frac{dy}{dx}=-\frac{\frac{\mathrm{\partial}}{\mathrm{\partial}x}\left(F\right)}{\frac{\mathrm{\partial}}{\mathrm{\partial}y}\left(F\right)}$

1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$

2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?