1: The function f (x) = 1 / 2 x ^ 4 - 2x ^ 3 + 3M is known. If f (x) + 9 > = 0 is constant, the value range of real number m is? A.m >=3/2 B.m > 3/2 C.m

1: The function f (x) = 1 / 2 x ^ 4 - 2x ^ 3 + 3M is known. If f (x) + 9 > = 0 is constant, the value range of real number m is? A.m >=3/2 B.m > 3/2 C.m

(1) F (x) + 9 = 1 / 2x ^ 4-2x ^ 3 + 3M + 9 ≥ 0 is constant, that is, if 1 / 2x ^ 4-2x ^ 3 ≥ - 3m-9 is constant, G (x) = 1 / 2x ^ 4-2x ^ 3, then G (x) min is needed to find the derivative G '(x) = 2x ² (x-3) according to the derivative function, G (x) has a minimum value when x = 3. G (3) = (- 27 / 2) ≥ - 3m-9. The solution is m ≥ 3 / 2 (2) know the derivative function from the subject

Make a square bottomed uncovered water tank with a volume of 256 liters (the bottom is square), then its height is______   The material is the most economical

If the height of the water tank is x and the bottom edge length is a, a2x = 256,
Its surface area s = 4ax + A2 = 4A × two hundred and fifty-six
a2+a2=1024
a+a2=512
a+512
a+a2≥33 five hundred and twelve
a × five hundred and twelve
a × a2
=3 × 26=192．
If and only if a = 8, i.e. H = 256
When 82 = 4, s gets the minimum value
So the answer is 4

Solve a derivative problem The rotation angle of the straightened wheel is directly proportional to the square of time. If it takes 0.8 seconds for the first rotation after the wheel is started, calculate the instantaneous angular velocity at 3.2 seconds after the rotation starts

The rotation angle of the wheel is directly proportional to the square of time. Let the scale coefficient be K, then
Angle θ= kt^2
The angular velocity is obtained by the first derivative of angle with time ω= 2kt
Find the second derivative to obtain the angular acceleration a = 2K
Rotate once, i.e. 2 π = K × 0.8^2
After the coefficient K is calculated, it can be substituted into the angular velocity formula

Quadratic function f (x) = ax ^ 2-2x + C f '(- 1) = - 3, then f' (3)= My answer to 1 is 3. I don't know what's wrong

f(x)=ax^2-2x+c
Derivation
f'(x)=2ax-2=2(ax-1)
f'(-1)=2(-a-1)=-2a-2=-3
a=1/2
therefore
f'(x)=x-2
When x = 3, f '(3) = 3-2 = 1
Is the answer wrong

Two curves y = x are known ²- 1 and y = 1-x ³ If the tangent at point x0 is parallel, x0 is?

Answer:
y=x ²- 1. Derivation: y '(x) = 2x
y=1-x ³, Derivation: y '(x) = - 3x ²
When x = x0, if the tangents are parallel, the derivatives are equal:
y'(x0)=2x0=-3x0 ²
(3x0+2)x0=0
The solution is: x0 = 0 or x0 = - 2 / 3

Given that the tangent equation of the image with the function f (x) = ax-6 / x square + B at point m (- 1, (F-1)) is x + 2Y + 5 = 0, find the analytical formula of y = f (x) of the function. Just tell me the idea.

Derivative: a + 12 / X Cubic = k tangent
K cut = 1 / 2, bring in - 1
∴a=12.5
F（-1）=-a+6-b
Bring in a = 12.5, i.e. - 6.5 + B
So - 1 + 2 (- 6.5 + b) + 5 = 0
So B = 4.5
∴f(x)=12.5x-6/x ²+ four point five