Design functions, find the factorial of integer n, and call the function in the main function (through cyclic structure) to compute the following polynomial: y=1! +3! +5! +7!

Design functions, find the factorial of integer n, and call the function in the main function (through cyclic structure) to compute the following polynomial: y=1! +3! +5! +7!

#include"stdio.h"
long int cal(long int a)
{
for( long int i=1,sum=1;i
Given the factorial of n! What is the factorial of 0! 0?
It's so simple!
The factorial of 0 is 1, which is defined
It is known that the set a = {x | - 2 is less than or equal to x, less than or equal to 5}, B = {x | - M + 1 is less than or equal to x, less than or equal to 2m + 1} satisfies that B belongs to the range of a and real number M
m+1》-2
2m+1《5
The solution is: - 3 "m" 2
Set can not use belong, belong is element, if it is true subset
Then - 2
When judging the monotonicity of function, how to judge whether the interval is open or closed
Example: how to judge the monotone interval of the function y = x-178; + 2x + 3
Monotonically decreasing on [- 1,1) and [3, positive infinity], monotonically increasing on (negative infinity, - 1) and [1,3]
Why not some [- 1,1] (3, positive infinity) (negative infinity, - 1) (1,3]
The end point of monotone interval can be ignored, because the end point is often the dividing point of increasing and decreasing interval (for continuous function), which side the end point belongs to does not matter
If the set a = {X / - 2 ≤ x ≤ 5}, B = {X / M + 1 ≤ x ≤ 2m-1}, and B is included in a, then the value range of real number m is?
Why is it - 2 ≤ m + 1 and 5 ≥ 2m-1 instead of - 2 < m + 1 and 5 > 2m-1? If it is equal, then a = B?
Incomplete solution
First consider the case where B is an empty set
① B is an empty set
m+1＞2m-1
m＜2
② B is not an empty set
M ≥ 2, - 2 ≤ m + 1 and 5 ≥ 2m-1
SO 2 ≤ m ≤ 3
[it can be taken as equal, because even if a = B, it is also a subset]
So m ≤ 3
If you don't understand, please hi me, I wish you a happy study!
Yes, a = B. we can say that B contains a, but we can't say that B really contains a
m－3＋﹙－1﹚=0 m=4
B is contained in a, that is, B is a subset of a, and a can be equal to B
Judge the monotonicity of function f (x) = 1 - (1 / x), and prove your conclusion
When x0, it increases
Proof: Let f (x) increase singly over 00
The same is true for X
Function f (x) = 2x ^ 3-3x ^ 2-12x on interval [- 1,2]
A: There is a maximum of 7 and a minimum of - 20
D: There is neither a maximum nor a minimum
Question: does extreme value not exist in monotone interval?
f'(x)=6x²-6x-12=0
x²-x-2=0
x1=-1,x2=2
There is no extremum
You are right. There is no extremum in monotone interval
There must be extremum in a closed interval, but it's not monotonic
The existence of extremum depends on whether all the function values near the stable point are less than or greater than him. Note that the extremum is different from the maximum value only near the stable point
D. is related to the definition of extremum, so it can be traced back to neighborhood. In the definition of neighborhood, delta lowercase (no mathematical formula editor is not convenient to input) > 0. Therefore, there is neither maximum nor minimum in the interval [- 1,2]
Given the function f (x) = 3 / X-2, it is proved by definition that f (x) is a decreasing function on (2, positive infinity)
Any 20
f(x1)>f(x2)
So f (x) is a decreasing function on (2, positive infinity)
You mean f (x) = 3 / (X-2). It's very simple
Let X '> - x > 2 and prove that f (x) - f (x') > 0
It's too easy
Let the image of quadratic function f (x) = x2 + 2x + B (x ∈ R) have three intersections with two coordinate axes, and the circle passing through these three intersections is marked as C
Suppose that the image of quadratic function f (x) = x2 + 2x + B (x ∈ R) has three intersections with two coordinate axes, and the circle passing through these three intersections is denoted as C. find the value range of real number B and the equation of circle C, ask whether circle C passes through a certain point (its coordinate is independent of B), and prove it
Δ = 4-4b > 0, B range B
It is proved that the function f (x) = x & # 179; - 2x & # 178; + 2x-7 increases on R
Don't use the derivative method. I haven't learned it yet. Use the difference method. Thank you
prove:
Let a < B. (a, B ∈ R)
f(a)-f(b)
=(a³-2a²+2a-7)-(b³-2b²+2b-7)
=(a³-b³)-2(a²-b²)+2(a-b)
=(a-b)[(a²+ab+b²)-2(a+b)+2]
=(a-b)[a²+(b-2)a+b²-2b+2]
=(a-b){[a+(b-2)/2]²+(b²/2)+[(b-2)²/4]}
Obviously, f (a) < f (b)
On R, the function f (x) is incremented