The discussion focuses on finding and classifying critical points for the function r=(1-x)(1+x), which simplifies to r=1-x². The second derivative test is applied, revealing that f''(x) = -2, indicating a local maximum at x=1. The conversation clarifies that the second derivative test is applicable only for functions of one variable, confirming that the function is a downward-opening parabola with a maximum at the vertex.
PREREQUISITESStudents studying calculus, particularly those focusing on critical points and the second derivative test, as well as educators teaching these concepts.
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.