Find the maximum and minimum of function y = 2x3-3x2-12x + 5 on [0, 3]

Find the maximum and minimum of function y = 2x3-3x2-12x + 5 on [0, 3]

Let ∵ f ′ (x) = 6x2-6x-12, let ∵ f ′ (x) = 6x2-6x-12 = 0, obtain x = - 1 or x = 2, and the list is as follows: x0 (0,2) 2 (2,3) 3F ′ (x) - 0 + F (x) 5 decreasing minimum - 15 increasing - 4, so the decreasing interval of function y on [0,3] is [0,2], increasing interval is [2,3], so the
The minimum value on the closed interval [- 2,3] of the function y = 2x ^ 3-3x ^ 2-12x + 13
Y '= 6x ^ 2-6x-12, let y' = 0, the inflection point x = - 1, x = 2
f(-2)=9,f(-1)=10,f(2)=-7,f(3)=4.
So the minimum value of the function is - 7
The known function f (x) = 1 / 3x ^ 3-4x + 4
If the equation f (x) = k about X has three real roots, then the value range of real number k is?
The derivative of the function is obtained, and its maximum and minimum are obtained. The range of K is in it
f'(x)=x^2-4
Let f '(x) = 0
The solution is x = 2 or - 2
Bring in the original function to get f (2) = and f (- 2)=
The range of K is between
If the maximum value of the function y = - 3x3 + 6x2 + m is equal to 13, then the real number m is equal to?
On the first floor
On the second floor, y = - 3x ^ 2 (X-2) + m is Shenma
The first reciprocal of the maximum is zero and the second reciprocal is less than zero
Y '= - 9x & sup2; + 12x = 0, the solution is x = 0 or x = 4 / 3
y''(4/3)
Y = - 3x ^ 2 (X-2) + m, the extremum of X has x = 0, x = 2, then the interval from 0 to 2 is monotone recursive subtraction, then x = 0 has the maximum, then M = 13
The derivative y '= - 9x ^ 2 + 12x = 0, x = 0 or 4 / 3
And y '' = - 18x + 12,
y''(4/3)=-12
Finding the maximum of function f (x) = x3-6x2-15x + 2
f(x)=x³-6x²-15x+2
f ‘(x) = 3x²-12x-15 = 3(x+1)(x-5)
When x ＜ - 1, it increases monotonically, and when - 1 ＜ x ＜ 5, it decreases monotonically
Maximum f (- 1) = - 1-6 + 15 + 2 = 10
If the function f (x) = x (x-m) ^ 2 has a maximum at x = 2, then the value of the constant M
∵ f (x) = x (x-m) ^ 2 has a maximum at x = 2
∴==>f'(x)=(x-m)^2+2x(x-m)=0
And ∵ x = 2,
∴f'(2)=(2-m)^2+2*2(2-m)=0
==>m=2,or,m=6;
To talk about
If the function f (x) = x (x-m) 2 has a maximum at x = 2, then the value of the constant M is?
Why 6 is not 2?
If 2 is the extreme value, then the derivative of F (x) at 2 is 0, and M = 6
f(x)=x(x-m)^2
=x(x^2-2mx+m^2)
=x^3-2mx^2+xm^2
Seeking derivative
f‘（x）=3x^2-4mx+m^2
f'(2)=3*4-8m+m^2=0
(m-6)(m-2)=0
m1=6
f(2)=2*(2-6)^2=32
m2=2
f（2）=2*0=0
f(x)=x(x-m)^2
=x(x^2-2mx+m^2)
=x^3-2mx^2+xm^2
Seeking derivative
f‘（x）=3x^2-4mx+m^2
f'(2)=3*4-8m+m^2=0
(m-6)(m-2)=0
f(x)=x(x-6)^2
F (x) = x (X-2) ^ 2 where x = 2 and x = 2 have maxima and minima
I don't see your question very clearly. What is it?
If the function f (x) = x (x-C) square has a maximum at x = 2, then the value of constant C is?
Thank you! It's urgent!
F '(x) = (x-C) square + X * 2 (x-C) = 3x square - 4xc + C square = (x-C) (3x-c)
If x = 2, f '(x) = 0, then (2-C) (6-C) = 0, C = 2 or 6
If C = 2, the square of F (x) = x (x-C) is not the maximum at x = 2, so C = 6
F'(X)=2x-c=0.x=c/2=2.c=4
It is shown that f (x) = x2 + CX is a quadratic function. If there is a maximum, the opening is upward. The vertex coordinates are [- B / 2a, (4ac-b ^ 2;) / 4A], so - C / 2 * 1 = 2, so C = - 4
F(X)=X(X-C)
=x^2-cx
So - B / 2A = C / 2 = 2
C=4
C = 6 (this is a derivative problem, in front of the C = 4 children's shoes are omitted to see you write the square, the steps are the same as the number of people's practice)
If there is only one element in the set a = {x | AX2 + 2x + 1 = 0}, then the value of a is ()
A. 0b. 0 or 1C. 1D. Not sure
If the set a = {x | AX2 + 2x + 1 = 0, a ∈ r} has only one element, then the equation AX2 + 2x + 1 = 0 has only one solution. When a = 0, the equation can be reduced to 2x + 1 = 0, which satisfies the condition. When a ≠ 0, the quadratic equation AX2 + 2x + 1 = 0 has only one solution, then △ = 4-4a = 0, and the solution is a = 1, so the value of a satisfying the condition is 0 or 1, so B is selected
If the function f (x) = x (x-C) ^ 2 has a maximum at x = 2, then the constant C?
F '(x) = (x-C) ^ 2 + 2x (x-C) = (x-C) (3x-c) = 0, x = C, C / 3
f"(x)=6x-4c