Let X1 and X2 (x1 ≠ x2) be two extremal problems of function f (x) = ax ^ 3; + BX ^ 2-A ^ 2x (a > 0) If X1 + x2 = 2 times the root 2, find the maximum value of B

Let X1 and X2 (x1 ≠ x2) be two extremal problems of function f (x) = ax ^ 3; + BX ^ 2-A ^ 2x (a > 0) If X1 + x2 = 2 times the root 2, find the maximum value of B

f'(x)=3ax^2+2bx-a^2
x1+x2=-2b/3a x1*x2=-a/3
From the above formula
2b=-3a(x1+x2)=9x1x2(x1+x2)
∵|x1|+|x2|≥x1+x2
That is, max (x1 + x2) = 2 √ 2
|x1|+|x2|≥2√(|x1||x2|)
=>2≥|x1||x2|≥x1x2
That is, max (x1x2) = 2
max(b)=9[max(x1x2)*max(x1+x2)]/2=9*2*2√2/2=18√2

It is known that x1, X2 (x1 ≠ x2) are two extreme points of the function f (x) = ax ^ 3 + BX ^ 2 + (a ^ 2) x (a > 0)
Question: if | x1 | + | x2 | = 2 √ 2, find the maximum value of B
Part of the answer goes like this
∵f'(x)=3aX^2+2bX-a^2(a>0)
According to the meaning of the title, X1 and X2 are two parts of the equation f '(x) = 0 and | x1 | + | x2 | = 2 √ 2
∴(X1+X2)^2-2*X1*X2+2|X1*X2|=8
I don't understand this part. What's the formula?

X1 and X2 are two parts of the equation f '(x) = 0 and | x1 | + | x2 | = 2 √ 2
∴(X1+X2)^2-2*X1*X2+2|X1*X2|=8
This part is the formula of the sum of squares. It's just for the sake of WIDA's theorem
4b^2/(9a^2)+4a/3=8
Multiply both sides by 9A ^ 2 to get 4B ^ 2 + 12a ^ 3 = 72A ^ 2
b^2=-3a^3+18a^2
Let y = - 3A ^ 3 + 18a ^ 2
Derivation y '= - 9A ^ 2 + 36a
When 0

Given that the function FX = ax ∧ 3 + BX ∧ 2-2x obtains the extremum at X1 = - 2 and X2 = 1, the analytic expression of the function FX is obtained

f'(x) = 3*a*x^2 + 2*b*x -2 =0
x1=－2,x2=1
Substituting into the two equations of a and B

What does PI mean

Second, the phosphate group in biology
Third, PI is pi
Fourthly, pi = proforma invoice
Fifth, product identity
……
Many of the meanings are English abbreviations
I wish you a happy study!

1 / 1 + 2 + 1 / 1 + 2 + 3 + 1 / 1 + 2 + 3 + 4 +... + 1 / 1 + 2 +... + 99 =? Simple calculation

(1+3+5+.+97+99)-(2+4+6+.+96+98)
=(1-2)+(3-4)+(5-6)+...+(97-98)+99
=-1-1-1-...-1+99
=-49+99
=50

limx→0 (tanx-sinx)/[(2+x^2)^(1/2)]*{[e^(x^3）]-1}=?

Your {[e ^ (x ^ 3)] - 1} should be in the denominator, or there is no answer LIM (x → 0) (TaNx SiNx) / {[(2 + x ^ 2) ^ (1 / 2)] * [e ^ (x ^ 3) - 1]} = LIM (x → 0) 1 / √ 2 * (TaNx SiNx) / x ^ 3 (0 / 0, using the law of lobita) = LIM (x → 0) 1 / √ 2 * (SEC ^ 2x cosx) / (3x ^ 2) = LIM (x →

How to calculate 1 / 5 * 1 / 10 * 75 * 7 / 45 with a simple method?

After the start of the approximate points, in order to calculate

RT
Let a and B be two points on the hyperbola x ^ 2-y ^ 2 / 2 = 1, and the point m (1,2) be the midpoint of the line ab

The fastest way to use the slope formula of the midpoint chord is as follows:
∵x^2-y^2／2=1,∴2x^2-y^2-2=0
∴k=1
Y-2 = X-1, that is, X-Y + 1 = 0

X-8 / 3x-2 / 2x = 100 solutions

（1-3/8-1/2）x=100
1/8x=100
x=100÷1/8
x=800

On the system of equations x, y, where the square of X + the square of y = 2, x + y = B has a set of real number solutions and inverse scale function
On the system of equations x2 + y2 = 2 of X, y, ① x + y = B, ② has a set of real number solutions, and the image of inverse scale function y = 2B / x increases with the increase of X in the quadrant where the image is located, then does the image of function y = - X-2 intersect with the image of function y = 2B / x? Please explain the reason

Simultaneous: X & # 178; + Y & # 178; = 2 (1) x + y = B (2): y = b-X is obtained from (2) and brought into (1): X & # 178; + (b-X) &# 178; = 2, that is: 2x & # 178; - 2bx + B & # 178; - 2 = 0, because the system of equations has a set of real number solutions, so: △ = 4B & # 178; - 8 (B & # 178; - 2) = 0, then: B & # 178; = 4B = 2 or - 2, because of the inverse ratio function