Calculation Of Local Maximum And Minimum Points With Characteristic Polynomial And Prime-Like Numbers

Local Maximum and Minimum Points

2. is the derivative of  with respect to

power rule, subtraction and additional rule we obtain        

For

 is the derivative of  with respect to

, addition and the subtraction rule we obtain   

Since  then

 for

1. The stationery points are at point

Hence the stationery points will be at the roots of

Using the quadratic equation

We obtain the roots of the function at the points

The value of  at the stationery points will be        

1. Finding local minimum and local maximum

The stationery points are at

Now we use the sign test to determine if the points are maximum or minimum

	-2	-1	0
	0		-12
	+ve		-ve

Using -2 and 0 to do the sign test we can see that the sign changes from positive to negative. When the sign changes from positive to negative, then this indicates that we have a local maximum. Hence point (-1,7) is a local maximum point.

Now using the test sign, we test point (-2,6) using the values -2 and 0

	-3	-2	-1
	-2		0
	-ve		+ve

The sign is changing from negative to positive hence the point (-2,6) is a local minimum.

1. Graphing the function f gives

 From this graph we can observe two points which can be classified as global maximum and minimum. That is point; (-1,7) and (-2,6)

2. Eigenvalues of M

Using the characteristic polynomial       

 eigenvalues as    

And        

1. Finding eigen vector

For every  we find its own vectors      

, so, we have a homogeneous system of linear equations we solve it by Gaussian Elimination  

      (1)

Find the variables from the equation of the system    (1)

, we have a homogeneous system of linear equations, we solve it by Gaussian Elimination to obtain

      (1)          

1. Diagonalize M

The diagonal matrix (the diagonal entries are the eigenvalues

The matrix with the eigenvectors ( as its columns                           

1. Calculate probability 

1. The profit function is 

The optimal points are at the derivative of the profit function equals to zero.

The derivative with respect to b gives  

The derivative with respect to a      

Now

Now  

we obtain b

Since the value of b is minimum we take -3

This gives the value of a

1. Not divisible by 2 and not divisible by 5.
2. This means the last digit cannot be

Now we must pick 4 digits from a bucket of 10

This can be picked in the following way

The fist number can be picked in 9 ways

2^nd in 10 ways

3^rd in 10 ways

4^th in 4 ways

Hence the total prime like numbers between 1000 and 9999 is  

1. Has at least one even digit

Here we select 4 numbers from a bucket of 10 in the following way

1^st number 9 ways

2^nd number 10 ways

3^rd number 10ways

4th number 5 ways

Now let’s assume that the even digit is in the 1^st number then the numbers will be        

Leta now assume the even number is in the second digit        

Let’s assume the even number is in the 3^rd digit        

The even digit cannot be in the fourth number as this will make the number prime like anymore

In total the numbers are

1. Prime like numbers not divisible by 3

Here we must choose the four numbers such that the sum of the digit is not divisible by 3

1^st digit 9 ways

2^nd digit 10 ways

3^rd digit 10 ways

4^th digit 3 ways

The numbers are

1. Y uniformly distribute

Y	-1	0	1	2
	2	0	0	2

Since Y is uniformly distributed all the values occur with equal probability the expected value of
expected value of

For

Y	-1	0	1	2
	0	0.5	0	0.5

 Since all the values occur with uniform probability the expected value will be

1. Variance of          

The variance is 0

Variance of         

The variance is 0

1. Testing independence of and

No, the equations are dependent as they share similar variance

1. A fair 6-sided dice is rolled 3 times
2. Probability that sum is 8

Possible outcomes are

Each roll must have a value between 1 and 6       

The only way we can have a sum of 8 is

The roll is 3 times so there are

Thus, the probability is

1. Probability of 1 given sum of 8

1. Drawing graph

1. The adjacency matrix is given by the graph does not have a spanning tree.
2. No, there is no connection that can allow all eight students to seat in a way that any of them knows both neighbours.

References

McQuarrie, D., 2003. Mathematical Methods for Scientists and Engineers, s.l.: University Science Books.

Salas, S. L., Hille, E. & Etgen, G. J., 2007. Calculus: One and Several Variables. 10th ed. s.l.:Wiley.