If $f(x,y) = x^2 + xy + y^2 - 3x + 4y - 5$. I know the domain is $\mathbb R^2$. How to determine the image of f is my issue.
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5$\begingroup$ What are your thoughts on the problem? What have you tried? Where are you running into trouble? $\endgroup$– Ben GrossmannCommented Jul 9, 2020 at 17:54
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3$\begingroup$ Also, why have you chosen the tag "complex analysis"? Your question doesn't seem to have anything to do with complex analysis. $\endgroup$– Ben GrossmannCommented Jul 9, 2020 at 17:55
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$\begingroup$ Welcome to MSE. Please read about how to ask a good question. $\endgroup$– Mohsen ShahriariCommented Jul 9, 2020 at 18:11
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$\begingroup$ Thanks. I will do that after you. $\endgroup$– QuadriCommented Jul 11, 2020 at 5:53
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3 Answers
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Have a look at my answer here: https://math.stackexchange.com/a/3619647/399263
You can always translate conics to cancel terms in $x,y$.
$f(x+a,y+b)=x^2+(-3+b+2a)x+xy+y^2+(2b+4+a)y+[\cdots]$
Solve $\begin{cases}2a+b-3=0\\2b+4+a=0\end{cases}\iff \begin{cases}a=\frac{10}3\\b=-\frac {11}3\end{cases}$
$$f(x+a,y+b)=\overbrace{(x^2+xy+y^2)}^{\ge 0}-\dfrac{52}3$$
In this case $x^2+xy+y^2\ge 0$ is always positive (discriminant $-3y^2<0$) with equality for $x=y=0$.
Thus $f$ has a minimum for $(x+a,y+b)=(0,0)\iff (x=-a,y=-b)$ of value $-\dfrac{52}3$.
Also $f$ is unbounded above since for instance $f(x,0)=x^2-3x-5$ is unbounded above.
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$\begingroup$ This is great! Now I see where to continue with the problem. $\endgroup$– QuadriCommented Jul 11, 2020 at 5:52
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Hint: This function is differentiable and has a unique critical point. It is helpful to determine whether this critical point is a saddle point, a maximum, or a minimum.
For instance, if the critical point is a local maximum and if that local maximum is also a global maximum, then the function attains a unique maximum but it unbounded below, which means that the range is of the form $(-\infty,c]$ for some $c \in \Bbb R$.
Alternatively, it would suffice to show that this function is convex.
answered Jul 9, 2020 at 17:57
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$\begingroup$ As this is true in dimension $1$ it's wrong in dimension $2$. Take $x^3+y^3-xy$. It has only one minimum in $(1/6,1/12)$, but $\lim_{x\to-\infty}x^3+y^3-xy=-\infty$ for any fixed $y$. $\endgroup$ Commented Jul 9, 2020 at 18:07
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1$\begingroup$ @MichaelHoppe one minimum, but two critical points $\endgroup$ Commented Jul 9, 2020 at 18:14
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$\begingroup$ @MichaelHoppe Also I believe the (local) minimum is at $(1/3,1/3)$. However, I realize now that I'm not sure about my statement, even if we assume that the critical point is unique. $\endgroup$ Commented Jul 9, 2020 at 18:20
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$\begingroup$ @MichaelHoppe In fact, my suspicion was incorrect after all. $\endgroup$ Commented Jul 9, 2020 at 18:24
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Hint Consider the qadratic form $$Q(x,y,z)= x^2 + xy + y^2 - 3xz + 4yz - 5z^2$$
Note that $$f(x,y)=Q(x,y,1)$$
$Q$ corresponds to the symmetric matrix $$A= \begin{bmatrix} 1 & \frac{1}{2} &\frac{-3}{2} \\ \frac{1}{2} & 1 & 2\\ \frac{-3}{2}&2 & -5 \end{bmatrix}$$
Now, orthogonally diagonalize $A$, and do the corresponding change of variable $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = P\begin{bmatrix} x \\ y \\ z\end{bmatrix} $ to transform $Q$ to the form $$Q(a,b,c)=\alpha a^2 + \beta b^2+ \gamma c^2$$
Swicth $a,b,c$ back to $x,y,z$ and set $z=1$.
