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July 15th, 2017, 11:02 AM   #1
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Second derivative test conclusions

Hello forum,

In multivariable calculus, when finding critical points and utilizing the second derivative test, if you are trying to determine if a critical point is a max, min, or saddle, you find D = D(a,b) = $\dfrac{\partial^2 f}{\partial x^2}$ $\dfrac{\partial^2 f}{\partial y^2}$ - $(\dfrac{\partial^2 f}{\partial xy})^2$

If D is any real value between negative and positive infinity, but $\dfrac{\partial^2 f}{\partial x^2}$ = 0, what could I conclude about the second derivative test here? My book only gives me cases for when $\dfrac{\partial^2 f}{\partial x^2}$ > 0 or $\dfrac{\partial^2 f}{\partial x^2}$ < 0.

Thank you for taking the time to read my post and help me out.

Jacob
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July 15th, 2017, 11:29 AM   #2
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The test in inconclusive. Compare with a real valued function $f:\mathbb{R}\rightarrow \mathbb{R}$ with $f'(a) = f''(a) = 0$, then it is also inconclusive. For example, take $f(x) = x^3$, then we have a saddle at $0$. While if $f(x) = x^4$, then we have a maximum at $0$. Note that in both cases, $f'(0) = f''(0) = 0$.

To conclude whether you have a saddle, minimum or maximum, you'll need to check higher derivatives.
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July 15th, 2017, 11:33 AM   #3
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Note that if $H:= f_{xx}f_{yy} - (f_{xy})^2 > 0$, then $f_{xx}$ has to be nonzero. If $f_{xx} = 0$, then $H = -(f_{xy})^2 < 0$.
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July 15th, 2017, 01:12 PM   #4
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Originally Posted by Micrm@ss View Post
The test in inconclusive. Compare with a real valued function $f:\mathbb{R}\rightarrow \mathbb{R}$ with $f'(a) = f''(a) = 0$, then it is also inconclusive. For example, take $f(x) = x^3$, then we have a saddle at $0$. While if $f(x) = x^4$, then we have a maximum at $0$. Note that in both cases, $f'(0) = f''(0) = 0$.

To conclude whether you have a saddle, minimum or maximum, you'll need to check higher derivatives.
Thanks a lot for helping me out.

What could I conclude from say $\dfrac{\partial^3 f}{\partial x^3}$ or $\dfrac{\partial^3 f}{\partial y^3}$ ?

For example... $f(x,y) = xy^2 + x^2 y + 8xy$

Graphically, I see there is a critical point at $f(x,y) = f(0,0)$

Also algebraically, $\dfrac{\partial f}{\partial x} = y^2 + 2xy+ 8y$

By factoring the $y$ out, I noticed that $\dfrac{\partial f}{\partial x} =y^2 + 2xy+ 8y = 0$ , when $y=0$

The same can be said about $\dfrac{\partial f}{\partial y}$ except this time $x$ will be factored out.

$\dfrac{\partial f}{\partial y} = x^2 + 2xy + 8x = 0$ when $x=0$.

$\dfrac{\partial f^2}{\partial y^2} = 2x$ and $\dfrac{\partial f^2}{\partial x^2}=2y$

So when $f(x,y) = f(0,0)$ the second derivative test is inconclusive for those values. Additionally, $\dfrac{\partial^3 f}{\partial x^3}$ and $\dfrac{\partial^3 f}{\partial y^3} = 0$

What would be an advisable approach at this point in the problem to help drive the point home that $(0,0)$ is a critical point on $f(x,y)$. Unless it is not a critical point and I made a mistake?

Thanks a ton,

Jacob
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July 15th, 2017, 01:29 PM   #5
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Quote:
Originally Posted by SenatorArmstrong View Post
Thanks a lot for helping me out.

What could I conclude from say $\dfrac{\partial^3 f}{\partial x^3}$ or $\dfrac{\partial^3 f}{\partial y^3}$ ?

For example... $f(x,y) = xy^2 + x^2 y + 8xy$

Graphically, I see there is a critical point at $f(x,y) = f(0,0)$

Also algebraically, $\dfrac{\partial f}{\partial x} = y^2 + 2xy+ 8y$

By factoring the $y$ out, I noticed that $\dfrac{\partial f}{\partial x} =y^2 + 2xy+ 8y = 0$ , when $y=0$

The same can be said about $\dfrac{\partial f}{\partial y}$ except this time $x$ will be factored out.

$\dfrac{\partial f}{\partial y} = x^2 + 2xy + 8x = 0$ when $x=0$.

$\dfrac{\partial f^2}{\partial y^2} = 2x$ and $\dfrac{\partial f^2}{\partial x^2}=2y$

So when $f(x,y) = f(0,0)$ the second derivative test is inconclusive for those values. Additionally, $\dfrac{\partial^3 f}{\partial x^3}$ and $\dfrac{\partial^3 f}{\partial y^3} = 0$

What would be an advisable approach at this point in the problem to help drive the point home that $(0,0)$ is a critical point on $f(x,y)$. Unless it is not a critical point and I made a mistake?

Thanks a ton,

Jacob
But $f_{xy}\neq 0$, so $H = f_{xx} f_{yy} - (f_{xy})^2 < 0$. So the second derivative test works. It only doesn't work when $H=0$.
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July 15th, 2017, 01:53 PM   #6
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But $f_{xy}\neq 0$, so $H = f_{xx} f_{yy} - (f_{xy})^2 < 0$. So the second derivative test works. It only doesn't work when $H=0$.
Okay thank you. And since in my case $H<0$ it is not a local maximum or minimum. That is strange though, since wolfram does consider it a critical point.
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July 15th, 2017, 01:54 PM   #7
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Okay thank you. And since in my case $H<0$ it is not a local maximum or minimum. That is strange though, since wolfram does consider it a critical point.
It's a saddle point. That's counted as a critical point.
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July 15th, 2017, 02:22 PM   #8
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It's a saddle point. That's counted as a critical point.
Thanks for all the help
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