MHB Determining the Maximum and Minimum of a Discontinuous Function $f(x,y)$

[mathmari]

Hey!

We have the function $f: (\mathbb{R}^2,\|\cdot\|_2)\rightarrow (\mathbb{R},|\cdot |)$ with \begin{equation*}f(x,y)=\begin{cases}y-x & y\geq x^2 \\ 0 & y<x^2\end{cases}\end{equation*}

I have shown that $f(x,y)$ is discontinuous at the points $(x,x^2)$ with $x\neq 0,1$ and continuous at the other points, i.e. at the points $(x,y)$ with $y<x^2$ and $y>x^2$, and also at the points $(0,0)$ and $(1,1)$.

Now I want to determine the maximum and minimum of $f$, if they exist. When $y\geq x^2$ we have that $f(x,y)=y-x\geq x^2-x\geq -\frac{1}{4}$.
($-\frac{1}{4}$ is the minimum of $x^2-x$ at $x=\frac{1}{2}$)
Since $-\frac{1}{4}$ is smaller than $0$ (the value of the function when $y<x^2$) it follows that the function $f(x,y)$ has a minimum at $(x,y)=(x,x^2)=\left (\frac{1}{2}, \frac{1}{4}\right )$ which is equal to $-\frac{1}{4}$.

Is this correct? If yes, could we improve the justification? (Wondering) About the maximum: In the first case, $y$ is greater than $x^2$, so I think that the value of the function can grow infinitely, i.e. it has no maximum. Is this correct? But how could we justify that formally? (Wondering)

 

[Opalg]

Hey!

We have the function $f: (\mathbb{R}^2,\|\cdot\|_2)\rightarrow (\mathbb{R},|\cdot |)$ with \begin{equation*}f(x,y)=\begin{cases}y-x & y\geq x^2 \\ 0 & y<x^2\end{cases}\end{equation*}

I have shown that $f(x,y)$ is discontinuous at the points $(x,x^2)$ with $x\neq 0,1$ and continuous at the other points, i.e. at the points $(x,y)$ with $y<x^2$ and $y>x^2$, and also at the points $(0,0)$ and $(1,1)$.

Now I want to determine the maximum and minimum of $f$, if they exist. When $y\geq x^2$ we have that $f(x,y)=y-x\geq x^2-x\geq -\frac{1}{4}$.
($-\frac{1}{4}$ is the minimum of $x^2-x$ at $x=\frac{1}{2}$)
Since $-\frac{1}{4}$ is smaller than $0$ (the value of the function when $y<x^2$) it follows that the function $f(x,y)$ has a minimum at $(x,y)=(x,x^2)=\left (\frac{1}{2}, \frac{1}{4}\right )$ which is equal to $-\frac{1}{4}$.

Is this correct? If yes, could we improve the justification? (Wondering) About the maximum: In the first case, $y$ is greater than $x^2$, so I think that the value of the function can grow infinitely, i.e. it has no maximum. Is this correct? But how could we justify that formally? (Wondering)

Your answers are correct. The minimum is $-\frac14$, at the point $\left (\frac{1}{2}, \frac{1}{4}\right )$. To show that there is no maximum, notice that $f(0,y) = y$ for all $y\geqslant0$.

 

[mathmari]

Your answers are correct. The minimum is $-\frac14$, at the point $\left (\frac{1}{2}, \frac{1}{4}\right )$. To show that there is no maximum, notice that $f(0,y) = y$ for all $y\geqslant0$.

Ah ok! Thank you very much! (Yes)