2nd derivative test-critical points calc 3

[Unemployed]

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?

 

[Chantry]

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.

 

[HallsofIvy]

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)

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.

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?

 

[HallsofIvy]

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.

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 saying

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.

 

[Unemployed]

Thank you.