The minimum and maximum of function f (x) = x2 ax (a) in the interval [0.1]

The minimum and maximum of function f (x) = x2 ax (a) in the interval [0.1]

This paper discusses the relative position relationship between the axis of symmetry and the interval of a function by using the image of quadratic function

In the interval [- 1,2], the minimum value of the function y = x2-ax + 30 is 5, and the value of a is obtained

y=f(x)=x²-ax+30
=(x-a/2)²+30-a²/4
The axis of symmetry is x = A / 2
(1) When a / 24, the function decreases monotonically on [- 1,2], and the minimum value is f (2) = 2 & # 178; - 2A + 30 = 34-2A
Let 34-2A = 5 get a = 29 / 2
In conclusion, the value of a is - 26 or 29 / 2

Known 0

y=2^x+3-4^x=-2^(2x)+2^x+3
Let t = 2 ^ x because 0

Given the function f (x) = 4 ^ X / 1-2 ^ X / 1 + 1 = x ∈ [- 3,2], find the minimum and maximum of F (x)

If x ∈ [- 3,2], then t ∈ (0,1) ∪ (1, + ∞) then G (T) = T & sup2; - t + 1 = (t-1 / 2) & sup2; + 3 / 4G (T) derivative function is g '(T) = 2t-1, if G' (T) = 0, then t = 1 / 2. Therefore, when t = 1 / 2, that is, x = - 1, the function has the minimum value

Finding the function f (x) = x + 4 / X (1)

x>0
Then x = √ (4 / 1) = 2 is the minimum
f(2)=4
The maximum is at the boundary
f(1)=1+4=5
f(3)=3+4/3=13/3
5>13/3
But x = 1 cannot be taken
therefore
The minimum is 4, there is no maximum

(1 / 2) it is known that the maximum value of a-bcos2x (b > 0) is 3 / 2 and the minimum value is - 1 / 2
(1 / 2) it is known that the maximum value of a-bcos2x (b > 0) is 3 / 2, and the minimum value is - 1 / 2. The period, maximum value and maximum value of y = - 4asin (3bx + 3 / C) are obtained

Then: y = - 2Sin (3x + π / 3): period 2 π / 3 = 2 π / 3, maximum 2, when 3x + π / 3 = 2K π - π / 2, that is, x = (2 / 3) k π - 5 π / 18, K ∈ Z}

A cubic function has a maximum when x = 1, a minimum when x = 3, and the function passes through the origin?
After solving three equations, it will become two of the same!

Let f (x) = ax ^ 3 + BX ^ 2 + CX + D. the function passes through the origin, so d = 0. Then the derivative function f '(x) = 3ax ^ 2 + 2bx + C. ∵ x = 1, x = 3 has extremum, ∵ 3A + 2B + C = 0,27a + 6B + C = 0. When x = 3, there is a minimum value 0, indicating that the point (3,0) is on f (x), that is 27a + 9b + 3C = 0

C language, write a program, input the value of X with scanf function, calculate and output the value of Y
#include
void main()
{
int x,y;
scanf("%d",&x);
if x=100);
y=3*x+1；
printf("%d",y);
}
What's wrong with me?
#include
void main()
{
int x,y;
scanf("%d",&x);
if(x

else y=3*x+1；
The title on the back is in Chinese

How to calculate the maximum and minimum value of the function y = x & # 178; - 2x + 3?

Function y = x & # 178; - 2x + 3 = (x-1) ^ 2 + 2 > = 2, so the minimum value is 2,

Proposition p: the function f (x) = X3 + AX2 + ax-a has both maxima and minima. Proposition q: the line 3x + 4Y-2 = 0 has a common point with the curve x2-2ax + Y2 + A2-1 = 0. If the proposition "P or Q" is true and "P and Q" is false, try to find the value range of A

From the meaning of the title, we can see that P and Q must be true and false
If P is true and Q is false, then
Proposition p: F (x) = X3 = AX2 = ax-a is differentiable, because there are maxima and minima, let the derivative function f '(x) = 3x2 + 2aX + a = 0, then △ > 0, and the solution is: A3
Proposition q: sorting out the curve, we can see that the curve is a circle (x-a) 2 + y2 = 1, the center of the circle is (a, 0), and the radius is 1
The distance from the center of the circle to the straight line is | 3a-2 | / 5, because there is no common point, then | 3a-2 | / 5 > 1, and the solution is A7 / 3
The intersection of them is A3
If q is true and P is false, then p is true and Q is false
Then proposition p: 0 ≤ a ≤ 3 proposition q: - 1 ≤ a ≤ 7 / 3
The intersection is 0 ≤ a ≤ 7 / 3
In conclusion, the value range of a is A3