# Using the approximation, explain why the second derivative test works.

1. Jul 18, 2011

### jumboopizza

1. The problem statement, all variables and given/known data[/
Using the approximation, explain why the second derivative test works

approximation=f(x0+delta x, y0+delta y)

delta x and delta y are small...

2. Relevant equations

f(x0+delta x,y0+delta y)

3. The attempt at a solution

ok so i know the first derivative of it is:

fx(x0+y0)*delta x+fy(x0+y0)*delta y

and second derivative is:

fxx(x0,y0)*delta x^2+fxy(x0,y0)*delta x*delta y+fyy(x0,y0)*delta y^2

it justs seems like since delta x and delta y are getting smaller,that everything will be going towards 0....so why does it work?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 18, 2011

### SammyS

Staff Emeritus
What is the problem as written?

f appears to be a function of two variables. What is the second derivative test for a function of two variables?

3. Jul 18, 2011

### jumboopizza

Using the approximation, explain why the second derivative works. Give three exam-
ples for each scenario of the second derivative test.

isnt that what the approximation is? the f(x+delta x,y +delta y)?

its asking about finding local mins,local max and saddle points....now i can show those examples,but how can i explain how it works?its like a proof or something

4. Jul 18, 2011

### Ray Vickson

You don't say what you are testing for. If you mean the second-order test for a maximum or minimum, you still need to be more specific: the necessary conditions and the (most common) sufficient conditions are a bit different. You need to tell us which ones you want.

RGV

5. Jul 18, 2011

### SammyS

Staff Emeritus
Mathematical language is very exacting. You need to say what you mean & mean what you say.

f(x0+Δx, y0+Δy) is the (exact) value of the function, f, at the point (x0+Δx, y0+Δy).

If the first derivatives of f are zero at the point, (x0, y0), then the following is an approximation to f at the point (x0+Δx, y0+Δy).

f(x0+Δx, y0+Δy) ≈ f(x0, y0) + (1/2)[ fxx(x0, y0)(Δx)2 +2 fxy(x0, y0)(Δx)(Δy) + fyy(x0, y0)(Δy)2 ]