Critical Numbers

Critical numbers also know as critical points are numbers that can determine what is happening in the function using derivatives. Now that was poorly explained but i will demonstrate the uses.

Lets  find the critical points of the function f(x) = 3x^5 - 20x^3.

Step 1:

First find the first deriviative of f(x)

d(f(x))/dx = 15x^4 - 60x^2

Step 2:

Now set the derivative to zero

f(x)' = 15x ^4 - 60x^2 = 0

15x^2(x^2-4) = 0

15x^2(x+2)(x-2) = 0

so upon inspection we see that x =  -2 , 0 , 2

These x's are ciritical numbers.

Plugging in these found critical points into f(x) our original function we get

f(-2) = 3(-2)^5 - 20(-2)^3 = 64
f(0) = 3(0)^5 - 20(0)^3 = 0
f(2) = 3(2)^5 - 20(2)^3 = -64

These numbers determine where the function has a local max or local min.

Definition:

Local Maximum: The point where the function has the high mountain or peak in the local area of the function. Being explained the point where y is equal to  a high point of the function or possibly the highest point.


Local Minimum: The point where the function has a low valley or canyon. where y is equal to a lower point on the graph in the local area.

Both definitions are terrible, but if you graph the function using graphing technology you will know what i am talking about.

  A local maximum and local minimum can be determined if the critical points found from the first derivative are plugged into the function. Such that a

local max can be found if f(x) > 0 . By plugging in the critical points of the first derivative.
local min can be found if f(x) < 0 . By plugging in the critical points of the first derivative.

Concavity

concavity is just a fancy way of saying does the function at a point start to turn into a mountain or into a  valley. Also known as concave up or down.

To find concavity can be just as easy as finding the local min and max. Finding concavity is also another way of finding the local min and max. So you can really kill two birds with one stone by finding the second derivative. First find the second derivative of f(x).

d^2(f(x))/dx^2 = d(15x^4 - 60x^2)/dx = 60x^3 - 120x

So now all you have to do is plug in the critical points from the first derivative.

f(x)'' = 60x^3 - 120x

f(-2)'' = 60(-2)^3 - 120(-2) = -240
f(0)'' = 0 (which tells us nothing, you'll see why)
f(2)'' = 240

Just like the same way of determining the local min and max. This determines both that and concavity.


local max can be found if f(x)'' < 0 . By plugging in the critical points of the first derivative.
local min can be found if f(x)'' > 0 . By plugging in the critical points of the first derivative.

concave down can be found if f(x)'' > 0 . By plugging in the critical points of the first derivative.
concave up can be found if f(x)'' < 0 . By plugging in the critical points of the first derivative.

We know that the function is concave up at x = 2 and has a local min.
We know that the function is concave down at x = -2 and has a local max.

As far as x  = 0, this does not work for our rule/theorem. 

Absolute extreme values

Not so extreme. In a function there has to be a maximum value and a minimum value of the function. If the function was horizontal on a graph, there would be a tie for y = constant value.

so far we know that

We know that the function is concave up at x = 2 and has a local min.
We know that the function is concave down at x = -2 and has a local max.

 Now to find the absolute max and min. We just need to know where the function is at it's highest and where it is at it's lowest point.

So by comparison of all our values of critical points. We can see that our local min and local max are also are absolute min and absolute max.

Easy right!


Endpoint values:

 This is pretty much the simplest of all simple. Lets say our function f(x) was continuous between the closed interval of  [ -5 , 5]. For the endpoint values all you do is plug in these numbers into your equation.

If  f(5) > f(-5)  then 5 is an endpoint max else 5 is an endpoint min and -5 is an endpoint max.

simple!

Point of Inflection

A point of inflection is a point such that the function mirror's itself almost. Such a function is x^2 where a point of inflection is (0,0). You can determine a point of inflection easily. To do this you need to find an x such that the second derivative f(x)'' = 0.

So for our second derivative. A point of inflection can be determined by x = 0.

For f(0)'' =60(0)^3 - 120(0) = 0.

A point of inflection is a point such that ( f(x)'' = 0 , f(x)) this is equivalent to ( 0 , f(0))

Where f(0) = 0.

So our point of inflection is (0, 0). Which from observation of graphing technology makes sense. For the function mirrors itself at that point.

Conclusion : 
Knowing these things can help you sketch a graph of your function or understand what your function is doing at certain points. Making them of importance.

Well hopefully this blog helped you understand Critical points. Also i hope this helped me study for my calculus final exam.