Finding Stationary Points of a Function

  • Thread starter goldfish9776
  • Start date
  • Tags
    Point
In summary, the critical point of a function is where its derivative equals 0 or does not exist. In the given question, the critical points are x=0.071, -14.071, 3, and -2. However, when asked to find the stationary point, only x=0.071 and -14.071 should be considered as x=3 and -2 are excluded. Additionally, the attached solution is incorrect as the function given can be rewritten and would result in division by zero at x=-2 and x=3.
  • #1
goldfish9776
310
1

Homework Statement

the critical point is the point which the f'(c) = 0 or f'(c) = doesn't exist . How if I'm asked to find the stationary point .

In this question , the critical point is x=0.071 , -14.071, 3 , and -2

So, if i am sked to find the stationary point , it should be only x=0.071 , -14.071 , am i right ? x=3 and -2 are excluded ?

Homework Equations

The Attempt at a Solution

 

Attachments

  • IMG_20151011_124811.jpg
    IMG_20151011_124811.jpg
    37.1 KB · Views: 474
Physics news on Phys.org
  • #2
Yes. What do you think about the reason?
 
  • #3
goldfish9776 said:

Homework Statement

the critical point is the point which the f'(c) = 0 or f'(c) = doesn't exist . How if I'm asked to find the stationary point .

In this question , the critical point is x=0.071 , -14.071, 3 , and -2

So, if i am sked to find the stationary point , it should be only x=0.071 , -14.071 , am i right ? x=3 and -2 are excluded ?

Homework Equations

The Attempt at a Solution


The attachment is incorrect: the function
[tex] f(x) = \frac{x^2+1}{x^2-x-6} [/tex]
can be re-written as
[tex] f(x) = \frac{x^2+1}{(x+2)(x-3)} [/tex]
Your attachment says that ##f(-2)## and ##f(3)## exist, but that is nonsense: you would be dividing by zero in both cases.

BTW: please refrain from using such attachments---just type things out. I find that when I open your attachment, I have no way of getting back to PF: I need to log off then log back on again.
 

FAQ: Finding Stationary Points of a Function

What is a critical point?

A critical point is a point on a graph or function where the slope (or derivative) is equal to zero. This means that the rate of change is neither increasing nor decreasing, and the function may have a local maximum, minimum, or saddle point at this point.

What is a stationary point?

A stationary point is a point on a graph or function where the slope (or derivative) is equal to zero. This is the same definition as a critical point. However, the term "stationary point" is often used in optimization problems, while "critical point" is used more generally in calculus.

What is the difference between a critical point and a stationary point?

There is no difference between a critical point and a stationary point. Both terms refer to a point on a graph or function where the slope (or derivative) is equal to zero. The difference lies in their usage, with "critical point" being a more general term and "stationary point" being used in optimization problems.

How do you determine if a critical/stationary point is a local maximum, minimum, or saddle point?

To determine the type of critical/stationary point, you can use the second derivative test or the first derivative test. The second derivative test involves evaluating the second derivative at the critical/stationary point. If the second derivative is positive, the point is a local minimum. If the second derivative is negative, the point is a local maximum. If the second derivative is zero, the test is inconclusive and you may need to use the first derivative test. The first derivative test involves evaluating the first derivative at points close to the critical/stationary point. If the first derivative changes from negative to positive, the point is a local minimum. If the first derivative changes from positive to negative, the point is a local maximum. If the first derivative does not change sign, the point is a saddle point.

How are critical/stationary points used in optimization problems?

Critical/stationary points are used in optimization problems to find the maximum or minimum value of a function. By finding the critical/stationary points and determining their type (using the second derivative test or first derivative test), we can determine the maximum or minimum value of the function. This is useful in various fields such as economics, engineering, and physics.

Similar threads

Back
Top