Find Local Extrema of f(x): f'(x) = x^2(x-1)^2(x-3)^2

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In summary, the function f(x) has four critical points at x = 0, x = 1, x = 3, and x = -1. These points exist on the domain of the function since f(x) is a polynomial. Additionally, the derivative of f(x) is always positive, so there are no changes in the sign of the derivative and no relative extrema. However, there are two saddle points at x = 0 and x = -1, which are stationary points that are neither maximum nor minimum values.
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Qube
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Homework Statement



If f'(x) = x^2(x-1)^2(x-3)^2, how many local extrema does f(x) have?

Homework Equations



Extrema occur at critical points.

Critical points are values of x such that f'(x) = 0 or = ±∞

The Attempt at a Solution



Not all values that zero the derivative are critical points. For example, if the problem had been f'(x) = -1/x^2, 1/x would have been the original function. 0 would seem like a critical point since it causes the derivative to approach negative infinity. However, it cannot be a critical point since 0 is not on the domain of the original function.

Is there any case to worry with this particular problem, however? I don't think so because the derivative appears to be a polynomial and the original function appears to also be a polynomial and polynomials are continuous through all real numbers.

Am I correct? Should I just say the critical points are x = 0, x = 1, and x = 3 and rest assured they exist on the domain of f(x) since f(x) appears to be a polynomial given its polynomial derivative?

----

Also, regardless of the critical numbers, the derivative appears to be all positive throughout its domain, making the discussion of relative extrema moot since there are no changes in the sign of the first derivative. All the terms are squared and we know that any term squared is a positive number.
 
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  • #2
Qube said:
Is there any case to worry with this particular problem, however? I don't think so because the derivative appears to be a polynomial and the original function appears to also be a polynomial and polynomials are continuous through all real numbers.

Am I correct?
Yes.

Should I just say the critical points are x = 0, x = 1, and x = 3 and rest assured they exist on the domain of f(x) since f(x) appears to be a polynomial given its polynomial derivative?
The domain is not an issue, right.

Also, regardless of the critical numbers, the derivative appears to be all positive throughout its domain, making the discussion of relative extrema moot since there are no changes in the sign of the first derivative. All the terms are squared and we know that any term squared is a positive number.
Good, as this settles the question about saddle points.
 
  • #3
What's a saddle point? I'm getting something about a stationary point ... what does stationary mean in this context?
 
  • #4
Qube said:
What's a saddle point? I'm getting something about a stationary point ... what does stationary mean in this context?

A saddle point is a stationary point that is neither a maximum nor a minimum; for example, the point x = 0 is a saddle point of f(x) = x^3.

Anyway, you have missed two other critical points.
 
  • #5
Ray Vickson said:
Anyway, you have missed two other critical points.
Be careful, the function given in the first post is f', not f itself.
 

1. What is the function being used to find local extrema?

The function being used to find local extrema is f(x) = x^2(x-1)^2(x-3)^2, with a derivative of f'(x) = x^2(x-1)^2(x-3)^2.

2. How do you determine the local extrema of a function?

The local extrema of a function can be determined by finding the critical points (where the derivative is equal to 0 or undefined) and using the first or second derivative test to classify them as either a maximum or minimum point.

3. What is a critical point?

A critical point is a point on a function where the derivative is either equal to 0 or undefined. This is where the slope of the function changes and can indicate a local extrema.

4. How do you use the first derivative test to classify a critical point?

The first derivative test involves plugging in values on either side of a critical point into the first derivative. If the derivative changes from positive to negative, the critical point is a local maximum. If it changes from negative to positive, the critical point is a local minimum.

5. How do you use the second derivative test to classify a critical point?

The second derivative test involves plugging in the critical point into the second derivative. 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 the second derivative is 0, the test is inconclusive.

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