Second derivative test Definition and 24 Threads

Hello, In second-order derivative test, the test is inconclusive when ##f''(c)=0##, so we had to generalize to higher-order derivative test. I was wondering how such tests can be generalized and derived? For example, how can I prove that ##f(x)=x^4## have minimum at 0? Bagas

According to this link: http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx The second derivative test can only be applied if ##f''## is continuous in a region around ##c##. But according to this link...

Homework Statement So the test is to take the determinant (D) of the Hessian matrix of your multivar function. Then if D>0 & fxx>0 it's a min point, if D>0 & fxx<0 it's a max point. For D<0 it's a saddle point, and D=0 gives no information. My question is, what happens if fxx=0? Is that...

Homework Statement Let A be a set of critical points of the function f(x). Let B be a set of roots of the equation f''(x)=0. Let C be a set of points where f''(x) does not exist. It follows that B∪C=D is a set of potential inflection points of f(x). Q 1: Can there exist any inflection points...

I am doing critical points and using the second derivative test (multivariable version) Homework Statement f(x,y) = (x^2+y^2)e^{x^2-y^2} Issue I am having is with the system of equations to get the critical points from partial wrt x, wrt y The Attempt at a Solution f_{x} =...

This is a problem from My book which I have hard understanding what they are asking for, I am pretty confused on the question would like to have help! Second derivate test works as follows: If f (c) = 0 and f'' (c)> 0 Then c is a local min point for function f (a) Show that c need not be a...

Suppose ##f^{\prime\prime}## is continuous on an open interval that contains x = c 1. If ##f^{\prime}(c)=0## and ##f^{\prime\prime}(c)<0##, then ##f## has local maximum at x = c. 2. If ##f^{\prime}(c)=0## and ##f^{\prime\prime}(c)>0##, then ##f## has local minimum at x = c. 3. If...

The graph of a differentiable function y=f(x) is 1. concave up on an interval I if f' is increasing on I. 2. concave down on an interval I if f' is decreasing on I. Let y=f(x) is twice differentiable on an interval I 1. If f'' > 0 on I, the graph of f over I is concave up. 2. If f'' <...

How does one derive the second derivative test for three variables? It's clear that D(a,b) = fxx * fyy - (fxy)^2 AND fxx(a,b) Tells us almost all we need to know about local maxima and local minima for a function of 2 variables x and y, but how do I make sense of the second directional...

Hey guys. I am having some trouble visualizing one aspect of the Second derivative test in the 2 variable case (related to #3 below). Essentially, what does the curve look like when f_{xx}f_{yy} > 0, BUT f_{xx}f_{yy} < [f_{xy}]^{2}? To be more detailed, if the function is f(x,y), H(x,y) is the...

Under what conditions on the constants a and b does the second derivative test guarantee that the function g(x,y,z)=ax^2+2axz+by^2-2byz+z^2 has a local maximum at (0,0,0)? a local minimum at (0,0,0)? well, i used the Hessian matrix to compute the eigenvalues to set them above zero. but...

Homework Statement So fx is how much f changes when you change x. Thus fxx is the rate of change of fx, or geometrically how fast the functions slope is changing. The same can be said for fy and fyy. But what about fxy and fyx? Could someone please explain to me what they mean? I want...

Homework Statement I'd always used the 2nd derivative test for the nature of stationary points. But I came across this equation in one of my exercises, for which the test doesn't seem to work at all. Find the stationary points of: y=(x^2-1)4, stating the nature of each. Homework...

Hi there, just wanted to make a clarification before my final exam. The second derivative test for partial derivatives (or at least part of it) states if D = ∂2f/∂x2 * ∂2f/∂y2 - (∂2f/∂x∂y)2 and (a,b) is a critical point of f, then a) if D(a,b) > 0 and ∂2f/∂x2 < 0, then there is a local...

Homework Statement \begin{pmatrix} -2 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & -2 \end{pmatrix} If I evaluate with eigenvalues, I get: det\begin{pmatrix} -2-\lambda & 0 & 0\\ 0 & -2-\lambda & 0\\ 0 & 0 & -2-\lambda \end{pmatrix}=0 (-2-\lambda{)}((-2-\lambda{)}(-2-\lambda{)})=0 and thus...

Hello! I am wondering if someone could let me know if my understanding is right or wrong. The Taylor series gives the function in the form of a sum of an infinite series. From this an approximation of the change in the function can be derived: f_{a} and f_{a,a} are the first and second...

urgent! second derivative test for functions of 2 variables Homework Statement f(x,y)=x^4 - y^2 - 2x^2 + 2y - 7 Homework Equations classify points (0,1) and (-1,1) as local maximum, local minimum or inclusive The Attempt at a Solution f(x,0)=4x^3 - 0 - 4x + 0 - 0 = 4x^3-4x...

[b]1. Homework Statement [/ 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... Homework Equations f(x0+delta x,y0+delta y) The Attempt at a Solution ok so i know the first derivative...

Homework Statement I am doing the various ins and outs of curve sketching and the mean value theorem and all that jazz with this function: f(x)=sec(x)+tan(x) Homework Equations The Attempt at a Solution I took the first derivative to be: f'(x)=sec(x)tan(x)+sec^{2}(x) I am having...

Hi, I am having a little trouble understanding something my lecturer said about using the table of signs to check whether there exists a point of inflection when y'' = 0. I understand that in order for there to be a point of inflection at x0 say, I require to check the value of y'' at either...

Can someone tell me what this actually is. So, in the case when the Hessian is positive (or negative) semidefinite, the second derivative test is inconclusive. However, I think I've read that even in the case where the Hessian is positive semidefinite at a stationary point x, we can still...

Homework Statement For example with f(x,y) = x2y + xy2 Homework Equations The Attempt at a Solution Well I know there is a critical point at (0,0). So I calculated the second derivatives but they are all 0 here so that doesn't help. I also tried using the Taylor expansion to...

Does the second derivative test fail for x3 at x=0: f'(x)=3x2 f''(x)=6x , for x=0, f'(0)=0 & f''(0)=+ve , so it should be a point of local maxima , but it is not!

Homework Statement Find all relative extrema using the second derivative test for H(x) = x * lnx Homework Equations The Attempt at a Solution H'(x) = (1 * ln x) + (x * 1/x) = lnx + 1 H''(x) = 1/x + 0 Is H''(x) right? Then I am having trouble finding the relative extrema from...