Derivatives+-+Critical+points,+points+of+inflection+and+maximum+and+minimum+values,+with+optimization

=**Critical points, points of inflection and maximum and minimum values, with optimization**=

Critical points, points of inflection and maximum and minimum values (by Nate Kresh):
Critical Point Definition- A critical point is any point on a function, f(x), where the derivative of that function is zero or is undefined. In addition, for a point x=c, the function value for that point needs to be defined. The term "critical point" can also refer to the point on the x-axis as opposed to the actual point on the graph, so the particular point one is trying to find relies on the context of the problem. In this function, Y=(x-1)2+1, the derivative becomes Y'=2(x-1). Therefore, in order to find the critical point, one must substitute zero into the equation. 0=2(x-1) 0=(x-1) x=1 y=1 Since the derivative is zero at x=1, that means that the critical point is (1,1). That is the algebraic method of finding the critical point. If you have the graph of the function, you can see which point has a horizontal tangent line, and that is the critical point. That is the graphical method of finding the critical point.
 * Example**

However, exactly what type of critical point (1,1) is is unknown. This is where maximum and minimum values enter the equation.


 * Maximum Values-** Maximum values come in two varieties. There is the global maximum, which is the largest value in the entire function, and then the local maximum, which means that f(c) is the largest value when x is kept in a neighborhood of c, so essentially the highest value in an interval as opposed to the entire function.


 * Minimum Values-** Minimum values are also both of the local and global variety, substituting the word smallest for largest. That should serve as enough explanation.

In order to find out whether or not a critical point is a minimum or a maximum, the graphical method would simply involve looking at the graph. If we take the example from before, you can see in the graph that the point (1,1) is a minimum. The algebraic method is not terribly difficult either. In general, a critical point is a maximum if the points to the left of it have a positive slope and the points to the right of it have a negative slope, which would mean that the graph is going up, reaching the critical point, and going down. The critical point is the summit of the imaginary hill. A critical point is a minimum if the points to the left of it have a negative slope and those to the right have a positive slope. The graph is going down, reaches the critical point, and then starts going up again, so the point is the bottom of an imaginary valley.

Let's look at our example from before, perhaps pretending that we did not have the graph and inherently know that the critical point is a minimum.

Y'=2(x-1); x=0.9 Y'=2(x-1); x=1.1 Y'=2(.9-1) Y'=2(1.1-1) Y'=-0.2 Y'=0.2 Since the derivative of a point to the left of the critical point, with an x value of 0.9, has a negative value, that means that the slope is negative. Since the derivative of a point to the right of the critical point, with an x value of 1.1, has a positive value, that means that the slope is positive. Therefore, this critical point fits the requirement of being a minimum value.

Additionally, in order to determine whether or not a point is a maximum or a minimum, you can use something called the **second derivative test**. Essentially, once you have established that there is a critical point, you can take that x-value and plug it into the second derivative. If the value of the second derivative is positive at this point, then the critical point is a minimum value. If the second derivative is negative, then the critical point is a maximum value. In the case of our example, Y''=2, and then by plugging in the x-value for the critical point, the second derivative is shockingly STILL 2, meaning that it is a positive value, so that means that the critical point is a minimum value.


 * Point of Inflection Definition**- Also known as the inflection point, it is essentially the point at which the concavity of a graph changes signs. Points of inflection work in a very similar way to critical points, in that the points of inflection are the critical points not of the function itself, but of the derivative of the function. Therefore, potential points of inflection are any places that the second derivative is zero. Additionally, in order for a point to be a point of inflection, the value for the second derivative needs to switch signs going from the left to the right of the critical point, or vice versa.

**Optimization (****by Aasim Rawoot):**
In optimization problems the goal is to find the largest or the smallest value a function, in which the function is subject to some kind of constraint, by using derivatives. Optimization problems are usually associated with real life situations, one of the most used examples is an area enclosed by a set amount of fencing material.

A field needs to be enclosed by a fence. We have 500 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the **LARGEST** area.
 * Example Problem:**


 * STEP 1 - Make a sketch of the situation if possible**


 * [[image:http://tutorial.math.lamar.edu/Classes/CalcI/Optimization_files/image001.gif]]

STEP 2 - Determine the functions needed**



We want to maximize the area of a field and we know that will use 500 ft of fencing material. So, the area will be the function we are trying to optimize and the amount of fencing our constraint.


 * STEP 3 - Solve for on of the two variables for the Constraint equation**

In this case it is easier to solve for x,




 * STEP 4 - Substitute the function you've just found into the function you want to optimize**

In this case we are going to substitute, into the Area function giving you the Area of the field as a function of y...




 * STEP 5 - Determine your interval**

To determine the interval find the zero's of //A(y)//. In this case //A// = 0 when //y// = 0 and //y// = 250. If you notice, these are the two extreme cases of y, where when y = 0 there are no sides to the fence and when y = 250 there are only two sides and no width to the field.


 * STEP 6 - Derive and find the critical points**

The derivative of the area of the field as a function of y turn out to be... Setting this equal to zero gives you the critical point, y = 125 which also happens to be within the interval you determined in the last step. To make sure that this is actually a max (as opposed to a min) we can use the second derivative test... //A''(y)// = -4 The second derivative is concave down for all real values of y making y = 125 an absolute max.


 * STEP 7 - Solve for x (or the other missing variable)**

We can obtain the //x// value by plugging in our //y// into the constraint.




 * STEP 8 - Conclusion**

The dimensions of the field that will give the largest area, with consideration to the fact that 500 ft of fencing material is available, are 250 x 125.

For another great step-by-step example on optimization check out this video...

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Works Cited: mathwarehouse.com The textbook

Topic: Critical Points Group Members: Nate and Aasim

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I like the graphs, the organization and the thoroughness of the explanations. Could use a little more graphical evidence. Also, in the optimization, this could have a numerical or graphical representation of the Area in terms of x etc. Decent work.

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