Given the function f (x) = the third power of x-12x + 8, find the maximum and minimum of the function

Given the function f (x) = the third power of x-12x + 8, find the maximum and minimum of the function

F (x) = 3x ^ 2-12 for function f (x)
In addition, f (x) = 0
X = plus or minus 2
Then judge according to the image
First, find the derivative f (x) '= xsquare - 12 = 3 (x + 2) (X-2), when x > 2, f (x) > 0, f (x) monotonically increases, when - 2
Finding the maximum and minimum of function f (x) = 6-12x + x ^ 3
f'(x)=3x^2-12=3(x-2)(x+2)
When x 2, f '(x) > 0, f (x) increases monotonically
-2
Y = 1 + 3x-x cubic power has maximum value and minimum value
y'=3-3x²=0
x=±1
X 2, y '> 0, y is an increasing function
-1
The maximum of y = x cubic - 3x square + 7
The maximum value of function y = - 2x-1 / 2x ^ 2-2x + 3 is equal to
y=(-2x-1)/(2x^2-2x+3)
y*(2x^2-2x+3)=-2x-1
2yx^2+(2-2y)x+3y+1=0
It is regarded as an equation of X, and this equation has a solution on R
deta>=0
That is: (2-2y) ^ 2-4 * 2Y * (3Y + 1) > = 0
-1
When x=______ When y = 3x + 1, the function value of y = 2x - 4 is equal
From the meaning of the question: 3x + 1 = 2X-4, the solution: x = - 5, so the answer is: - 5
When x = -, the function y = 2x + 4 and y = 3x-3 have the same function value? This function value is——
Let 2x + 4 = 3x-3 get x = 7
Take x = 7 into any equation and get y = 18
When x = 7, the function value is 18
Write programs to achieve the following functions, function y = x (x)
#include
void main()
{
int x,y;
scanf("%f",&x);
if(x=10)
{
y=3*x-11;
printf("%f",y);
}
else
{
y=2*x-1;
printf("%f",y);
}
}
#include
#include
main()
{
float x, y;
Printf ("please enter the value of X and y)";
scanf("%f%f", &x, &y);
if(x
Find the maximum and minimum of function f (x) = 2x ^ 3 + 3x ^ 2-12x-1 on (negative infinity, 2]
Using derivative method, find f '(x) = 6x ^ 2 + 6x-12, when f "(x) > 0, x1, F" (x)
F '(x) = 6x ^ 2 + 6x-12
Finding the extreme value 6x ^ 2 + 6x-12 = 0
Find X
Substituting two X's and endpoint 2 into the original equation
Find the maximum and minimum
The maximum and minimum values of the function y = 2x ^ 3-3x ^ 2-12x + 5 on [2,3] are
*Cubic curves are flattened "s" shaped
First give y derivative and then = 0, x = - 1 or 2
-If 1 is not in the required range, it can be directly substituted into x = 2 and x = 3
Ymin = Y(x=2)= -15
Ymax = Y(x=3)= -4
y’=6x^2-6x-12=0
x^2-x-2=0
x1=-1
x2=2
In [2,3], when x = 2, there is an extreme value: - 15
When x = 3, y = - 4
The maximum and minimum values of the function y = 2x ^ 3-3x ^ 2-12x + 5 on [2,3] are - 4 and - 15, respectively