-- Take the first derivative of the function with respect to the variable.
-- Set the derivative equal to zero.
-- Solve the equation [ first derivative = 0 ] for the variable.
-- The solution for the variable, along with the value of the function for that value of
the variable, gives you a point that is either a minimum or a maximum of the function.
By looking at another point just before or just after that value of the variable, you
should be able to decide whether it's a max or a min. If you still can't tell which it
is, then there's a way to do it that involves the second derivative of the function,
but I don't think I want to go there just now.
* * * * *
The above is an excellent answer for basic calculus but there are some limitations. The method requires the function to be differentiable throughout its domain. As a result it cannot deal effectively with discontinuities.
Also, it cannot deal with the extrema being at either end of the domain. This can happen in two ways which are illustrated below:
Consider f(x) = 2x+3 for 0<x<1. Trivially, the minimum and maximum values, (3 and 5) are limiting values of the function over its domain and are never actually attained. It is easy enough to sketch the graph and conclude that the minimum value of 3 is attained when x = 0 and the maximum value, of 5 is attained at x = 1. Except that these values of x are in the domain for x. But the derivative of f(x) is 2, and f'(x) = 0 simply gives 2 = 0 and THAT has no solution!
But I saved the biggest problem to the last. Consider f(x) = x3 - x where the domain is the whole of the real line. Differentiating it gives f'(x) = 3x2 - 1. And solving f'(x) = 0 gives x = ±sqrt(1/3). Substituting these values of x in f(x) you can evaluate the "maximum" and "minimum". Why the quotation marks? Because what you have calculated are the local extrema, not the global extrema. The true minimum for the function is "minus infinity" and the maximum is "plus infinity". The cubic curve starts at "minus infinity", its gradient gradually deceases until, at x = -sqrt(1/3) it levels off and starts falling. It falls until it reaches +sqrt(1/3) and then starts rising again and keeps rising - forever. So x = -sqrt(1/3) was only a turning point: a local extremum, not a global one. Likewise the local minimum at x = +sqrt(1/3).
take the max plus the min and divide the anser by to
max=number min=number counter=1 start while(counter<=3) read number start if max<number max=number else if min>number min=number end while write max write min /*modify the counter switch according to the number of values being compared if required*/
#include<stdio.h> #include<conio.h> void main(void) { int max=0,min=0; int temp; int n,i; clrscr(); printf("what is the lenght of number set?\n"); scanf("%d",&n); printf("\n\nNow enter the numbers\n"); for(i=1;i<=n;i++) { scanf("%d",&temp); if(temp>max)max=temp; if(i==1)min=temp; if(temp<min)min=temp; } printf("\n\nThe range of set is %d",max-min); getch(); }
Add the min and max numbers and divide that sum by two. Here is an example: The data is {1, 1, 3, 50,60} The midrange is 30.5 since that is is (1 + 60) / 2 = 30.5.
You would calculate 2.3/7 . That number will be the rate in km per min.
The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)The MIN function returns the lowest value from a set of values. The MAX function returns the highest value from a set of values.=MIN(A2:A20)=MAX(A2:A20)
Use the function MAX to find the largest value. Use the function MIN to find the smallest value. If you want to find the value in a row, use the range of the cells in the row; for column, use the range of cells in the column. =MAX(A1:A12) will find the largest value in column A (from row 1 through 12). =MIN(A1:M1) will find the smallest value in row 1 (from column A through M).
The MIN function is the function that you would use to find the lowest value in a set. You could also use the SMALL function, but that is more when you are looking for a value other than the lowest, such as the second lowest or third lowest. That is specified by a number at the end. Both of the following will find the lowest value in the range of A2 to A16: =MIN(A2:A16) =SMALL(A2:A16,1)
The MAX function gets the highest value in a range and the MIN function gets the lowest. If there were values in all the cells from A2 to A20, you could use the two functions to get the highest and lowest values in the range:=MAX(A2:A20)=MIN(A2:A20)
If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value
The range is the difference between the highest and the lowest values. You can find the highest using the MAX function less the value found by the MIN function. So if you wanted to get the mathematical range for the cell range A2 to A30, you would do this: =MAX(A2:A30) - MIN(A2:A30)
Average Sum Count Max value Min Value
Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. sonika aggarwal GNIIT
MIN is used to find the lowest number in a range. You would use MAX or LARGE to find the highest number in a range.
In Calculus, to find the maximum and minimum value, you first take the derivative of the function then find the zeroes or the roots of it. Once you have the roots, you can just simply plug in the x value to the original function where y is the maximum or minimum value. To know if its a maximum or minimum value, simply do your number line to check. the x and y are now your max/min points/ coordinates.
MIN is a function that gets the lowest value from a set of values or a range. It is short for minimum. So if you wanted to find the lowest value in the range from A2 to A25, you would type the function in like this: =MIN(A2:A25)
You take the integral of the sin function, -cos, and plug in the highest and lowest values. Then subtract the latter from the former. so if "min" is the low end of the series, and "max" is the high end of the series, the answer is -cos(max) - (-cos(min)), or cos(min) - cos(max).