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.
The 2nd derivative test is a method used to determine the nature of critical points in Calculus 3. It involves taking the second derivative of a function and evaluating it at the critical points to determine whether they are local maxima, local minima, or saddle points.
To find critical points using the 2nd derivative test, you first need to take the derivative of the function and set it equal to 0 to find the critical points. Then, take the second derivative and evaluate it at each critical point. If the second derivative is positive, the critical point is a local minimum. If it is negative, the critical point is a local maximum. If it is 0, the test is inconclusive and further analysis is needed.
The 2nd derivative is significant in the 2nd derivative test because it helps determine the concavity of a function at a critical point. A positive 2nd derivative means the function is concave up, while a negative 2nd derivative means the function is concave down.
No, the 2nd derivative test can only be applied to twice-differentiable functions. This means that the function must have a continuous first and second derivative in the given interval.
If the 2nd derivative is 0 at a critical point, the 2nd derivative test is inconclusive and further analysis is needed. This could mean that the critical point is a point of inflection or that higher-order derivatives need to be evaluated to determine the nature of the critical point.