Stationary points of a function

[BigBeachBanana]

12?

Let's try this again.
$$f_{xy}=\frac{d}{dy}(6x^2-96)=\frac{d}{dy}(\text{Constant)}=?$$Think back to single variable, what's the derivative of a constant?

 

[Bener123]

-96

 

[Bener123]

Oh sorry derivative of a constant is 0

 

[Bener123]

So fxy is 0?

 

[BigBeachBanana]

Oh sorry derivative of a constant is 0

Yes. Now try the other way.
If $f_y=6y+12$, then $\red{f_{yx}=\frac{d}{dx}(6y+12)}=?$.

Use a similar idea.

 

[Bener123]

Yes. Now try the other way.
If $f_y=6y+12$, then $\red{f_{yx}=\frac{d}{dx}(6y+12)}=?$.

Use a similar idea.

So it's 0 again?

 

[BigBeachBanana]

So it's 0 again?

Yes, correct. In fact, this always holds true for continuous functions. It's called Clariot's Theorem.
$$f_{xy}=f_{yx}$$So if you know one, you'll know the other.

 

[Bener123]

Hi guys,

Cheers for the help. Can you see I'm I'm correct on this please?

 

[BigBeachBanana]

Hi guys,

Cheers for the help. Can you see I'm I'm correct on this please?

View attachment 32153

Those are correct.

 

[Bener123]

Awesome! I had doubts about 288 as I thought it was 12 initially. So you can clarify 288 is correct? Just thought it was quite a high number that's all.

 

[BigBeachBanana]

Awesome! I had doubts about 288 as I thought it was 12 initially. So you can clarify 288 is correct? Just thought it was quite a high number that's all.

You flipped the order $\triangle= f_{xx}f_{yy}-f_{xy}^2$.
You got lucky that $f_{xy}^2=0$, otherwise your answer could be wrong. Other than that, your answer is correct.