Classifying Critical Points: Finding Local Extrema and Saddle Points

In summary, the conversation is about finding the asymptote, determining the range of a function, and finding local maxima and minima. The speaker incorrectly calculates the first derivative and is reminded to use the quotient rule.
  • #1
kukumaluboy
61
1

Homework Statement



v75ajc.png

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..
 
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  • #2
kukumaluboy said:

Homework Statement



v75ajc.png

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!
 
  • #3
kukumaluboy said:

Homework Statement



v75ajc.png

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.
 

Related to Classifying Critical Points: Finding Local Extrema and Saddle Points

1. What is a maximum/minimum derivative?

A maximum/minimum derivative is a point on a graph where the slope of the curve is either at its highest (maximum) or lowest (minimum) value. It is also known as a critical point.

2. How is a maximum/minimum derivative calculated?

The maximum/minimum derivative is calculated by taking the derivative of the original function and setting it equal to zero. This will give the x-value of the critical point. The y-value can then be found by plugging the x-value into the original function.

3. What does a maximum/minimum derivative represent in real-world applications?

In real-world applications, a maximum/minimum derivative represents a point of optimization or the highest/lowest value of a function. For example, in economics, it can represent the maximum profit or minimum cost for a business.

4. How do you determine if a critical point is a maximum or minimum derivative?

To determine if a critical point is a maximum or minimum derivative, you can use the first or second derivative test. If the first derivative is positive before the critical point and negative after, it is a maximum. If the first derivative is negative before the critical point and positive after, it is a minimum. The second derivative can also be used to confirm with a positive value indicating a minimum and a negative value indicating a maximum.

5. Can a maximum/minimum derivative exist at a point where the function is not defined?

Yes, a maximum/minimum derivative can exist at a point where the function is not defined. This is because the derivative only depends on the behavior of the function around that point, not necessarily at the point itself. However, it is important to check the original function to ensure that the critical point does not lie on a discontinuity or asymptote.

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