Setting the gradient of f (a two variable function) to (0,0)

[RM5152]

Hello. I have been asked to solve the gradient of f, denoted by the upside down triangle before f and I have done this as can be seen in my work. I’m now asked to solve the gradient of f=(0,0). However, when I set the 2 functions to zero they just leave me with either x^2=x or x=-y.. what I mean can be seen from my work below. Can anyone help please? Thanks

 

[blamocur]

You have 2 equations with 2 unknowns. You have almost solved them by isolating an equation for [imath]x[/imath] -- can you figure out the solutions for it?

 

[RM5152]

You have 2 equations with 2 unknowns. You have almost solved them by isolating an equation for [imath]x[/imath] -- can you figure out the solutions for it

I’m getting x=1, y= -1 ?

 

[blamocur]

I’m getting x=1, y= -1 ?

Only one solution for [imath]x[/imath]?

 

[topsquark]

I’m getting x=1, y= -1 ?

Yes, but you still aren't quite done yet. What are the possible solutions to [imath]y^2 + y = 0[/imath]? There are two of them...

FYI: The upside-down triangle operator has the name "del" (or occasionally "nabla" but del is much more common.)

-Dan

 

[RM5152]

I have confused myself even further… I got 2 critical points (1,-1) and (0,0) but then I also got another 2 critical points by using the b^2-4ac formula after inserting y=-1 into the formula. I presume this is not correct as I can’t have 4 critical points.. can you tell me where I have gone wrong please and thank you

 

[RM5152]

This is another example of simultaneous equations with variables xy that I am having trouble with solving … is there any rules or tricks to solving simultaneous equations like this?

 

[topsquark]

I have confused myself even further… I got 2 critical points (1,-1) and (0,0) but then I also got another 2 critical points by using the b^2-4ac formula after inserting y=-1 into the formula. I presume this is not correct as I can’t have 4 critical points.. can you tell me where I have gone wrong please and thank you

Note that [imath]\sqrt{18^2 - 4*24*(-6)} = 30[/imath], not [imath]\sqrt{30}[/imath]. That's a tough mistake to catch.

-Dan

 

[topsquark]

This is another example of simultaneous equations with variables xy that I am having trouble with solving … is there any rules or tricks to solving simultaneous equations like this?

So far as I know you did it right. Are you expected to be able to solve these numerically? Otherwise you have to use Cardano's method. (Look under the section "depressed cubic."

You really need to start factoring out constants. It makes everything simpler to analyze.

-Dan

 

[RM5152]

but how can I solve 3x^3 -12x - 3 = 0 to find the critical points? i presume I do need to solve x numerically so that I can find y and find critical point (x,y)? thanks for the help

 

[blamocur]

I have confused myself even further… I got 2 critical points (1,-1) and (0,0) but then I also got another 2 critical points by using the b^2-4ac formula after inserting y=-1 into the formula. I presume this is not correct as I can’t have 4 critical points.. can you tell me where I have gone wrong please and thank you

You cannot just say "if y=-1 then x = ...". You have 2 equations, and both have to be satisfied. One of the equations is [imath]x+y=0[/imath], so if [imath]y=-1[/imath] then you can only have [imath]x=1[/imath].

 

[RM5152]

Sorry I realised I am meant to start a new thread for a new question .. I’ll do this now. If you could tell me how to get any further than x^3-4x-1=0 to solve critical points that would be much appreciated .. thanks ?

So far as I know you did it right. Are you expected to be able to solve these numerically? Otherwise you have to use Cardano's method. (Look under the section "depressed cubic."

You really need to start factoring out constants. It makes everything simpler to analyze.

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