Classifying Critical Points: Finding Local Extrema and Saddle Points

[kukumaluboy]

Homework Statement

Homework Equations

The Attempt at a Solution

1)
I found the asymptote as (+/- 1)

2)
Let f(x) = y;

dy/dx = -2x^2 / (x^4 - 2x^2 + 1) = 0

-2x^2 - 0
x = 0;

Since f() != 1, f(2) > 0 Increasing
Since f() != -1, f(-2) < 0 Decreasing

So i guess range is increasing or x >=2, decreasing for x<=-2

3)
Since x = 0, how find local max or min or what the toot is saddle point..

 

[Dick]

Homework Statement

Homework Equations

The Attempt at a Solution

1)
I found the asymptote as (+/- 1)

2)
Let f(x) = y;

dy/dx = -2x^2 / (x^4 - 2x^2 + 1) = 0

-2x^2 - 0
x = 0;

Since f() != 1, f(2) > 0 Increasing
Since f() != -1, f(-2) < 0 Decreasing

So i guess range is increasing or x >=2, decreasing for x<=-2

3)
Since x = 0, how find local max or min or what the toot is saddle point..

Your first derivative dy/dx is wrong. Use the quotient rule!

 

[Vagn]

Homework Statement

Homework Equations

The Attempt at a Solution

1)
I found the asymptote as (+/- 1)

2)
Let f(x) = y;

dy/dx = -2x^2 / (x^4 - 2x^2 + 1) = 0

-2x^2 - 0
x = 0;

Since f() != 1, f(2) > 0 Increasing
Since f() != -1, f(-2) < 0 Decreasing

So i guess range is increasing or x >=2, decreasing for x<=-2

3)
Since x = 0, how find local max or min or what the toot is saddle point..

I think you've differentiated that incorrectly, remember to use the product (or quotient) rule.