1. ## Critical point and Saddle Point Question

Given f(x,y) = (y - x^2)(y - 4x^2) show the origin is a critical point for f which is a saddle point but that on any line through the origin f has a local min @ (0,0).

2. Originally Posted by katielaw
Given f(x,y) = (y - x^2)(y - 4x^2) show the origin is a critical point for f which is a saddle point but that on any line through the origin f has a local min @ (0,0).
Okay, what have YOU done? Since this function is continuous for all x and y, a "critical point" is any point where the partial derivatives are 0. What are the partial derivatives of f?

"On any line through the origin", y= mx for some number m. Replace y with mx in f(x,y) so you have f(x,mx), a function of the single variable x. Show that that function has a local minumum at (0,0).

(Well, almost "any line through the origin". The vertical line through the origin is x=0 and cannot be written that way. Look at f(0,y)= y^2.)

3. Originally Posted by HallsofIvy
Okay, what have YOU done? Since this function is continuous for all x and y, a "critical point" is any point where the partial derivatives are 0. What are the partial derivatives of f?

"On any line through the origin", y= mx for some number m. Replace y with mx in f(x,y) so you have f(x,mx), a function of the single variable x. Show that that function has a local minumum at (0,0).

(Well, almost "any line through the origin". The vertical line through the origin is x=0 and cannot be written that way. Look at f(0,y)= y^2.)
Edit: I re-wrote the partials, hopefully this is better.

f_x = -10xy - 16x^3

f_y = 2y - 4x^3 - x^2

For f(x, mx),

h(x) =f(x, mx) where m is any real number.

For this, I just need to find f'(0) and f''(0) to show there is a critical point and a local minimum approaching the origin, right? What about the saddle point part?