The problem involves finding and classifying critical points for the function r=(1-x)(1+x), which is expressed in terms of one variable. Participants are exploring the application of the second derivative test and discussing the nature of the function.
The discussion is active, with participants providing insights into the function's behavior and questioning the application of certain tests. There is recognition of the function's maximum and the role of the vertex in determining critical points, though no consensus has been reached on the application of the second derivative test.
Participants note that the function is a quadratic and discuss its properties, including its maximum point and the implications of its decreasing and increasing intervals. There is also mention of difficulties in applying the second derivative test due to the function being in one variable.
This is a function of one variable, not two. The "second derivative test", for functions of one variable, says that at a critical point, x0, (f'(x0)= 0), if f''(x0)< 0, it is a maximum, if f''(x0)> 0, it is a minimum, if f''(x0)= 0, it is neither.Unemployed said:Homework Statement
r=(1-x)(1+x) Find and classify all of the critical points
Homework Equations
D=fxx-fyy-(fxfy)^2 (can't read my notes to well or find it on the internet)
The Attempt at a Solution
r=-x^+1
f'= -2X
f"=-2
Because -2<0 There is a maximum(1), I don't know the full argument for this. There is also no y so I feel I am at a dead end.
Can the d-test be applied to this equation?
Surely, you didn't mean to say this. This is parabola that opens downward. It is decreasing for x> 1 and increasing for x< 1. And, because it is a quadratic, you can find its maximum by finding its vertex (trivial in this case). Which, of course, is what you mean by sayingChantry said:It's D=fxx(fyy)-(fxfy)^2
If d > 0,
and fxx > 0, it's a local minimum or
fxx < 0, it's a local maximum
if d < 0,
then it's a saddle point.
However, you can't use this for this function as it's a function in terms of one variable.
So,
r=(1-x)(1+x)
= 1 - x^2
Which is a decreasing function.
It should be simple to get the result just by looking
but take the derivative, set it equal to zero and figure out the critical point. The function is decreasing, so whether it's a maximum or minimum should be obvious.