It is known that f (x) is an odd function defined on R, and when x > 0, f (x) = X3 + 2x2-1, the analytic expression of F (x) can be obtained

It is known that f (x) is an odd function defined on R, and when x > 0, f (x) = X3 + 2x2-1, the analytic expression of F (x) can be obtained

X=0
f(-0)=f(0)=-f(0)=0
f(0)=0
When x > 0, f (x) = X3 + 2x2-1
When X0
f(-x)=-x^3+2x^2-1=-f(x)
f(x)=x^3-2x^2+1
Piecewise function
f(x)={
f1=x3+2x2-1(x>0)
f2=0(x=0)
f3=x^3-2x^2+1(x
Let a = {x | x square minus 3x plus 2 = 0}, B = {x | x square minus ax plus (a minus 1) = 0}, C = {x | x square minus BX plus 2 = 0, X belongs to R}
Given the set a = {x | x square minus 3x plus 2 = 0}, B = {x | x square minus ax plus (a minus 1) = 0}, C = {x | x square minus BX plus 2 = 0, X belongs to R}, if B (U upside down plus a horizontal) a, C (U upside down plus a horizontal) a, find the conditions that real numbers a and B should satisfy
∵A={x|x^2-3x+2=0}∴A={1,2}
B={x|x^2-ax+(a-1)=0}∴ B={a-1,1}
B is contained in a, then A-1 = 1 or A-1 = 2, so a = 2 or 3
C = {x | x ^ 2-bx + 2 = 0} C is contained in a
If 1 ∈ C, then 1-B + 2 = 0, B = 3, then C = {1,2}
To sum up
A = 2 or 3 B = 3
Is it included in?
If acab = 5 − 12 ≈ 0.618, then BCAC=______ ≈______ .
According to the concept of golden section, acab = BCAC = 5 − 12 = 0.618
Given that the function FX is an odd function in the domain R, and when x > 0, f (x) = x3-x + 1, the analytic expression of FX is obtained
For odd functions, then f (0) = 0
X0
So f (- x) = - X & # 179; + X + 1
So f (x0 = - f (- x) = x & # 179; - X-1
So f (x)=
x³-x-1,x0
X-square-a [3x-2a + b] - b-square
X squared - a [3x-2a + b] - B squared = 0, wrong number! There are many solutions. Why does 2A square become 2.25A square!
x^2-a(3x-2a+b)-b^2 =0
x^2-3ax+2a^2-ab-b^2=0
x^2-3ax+2.25a^2-0.25a^2-ab-b^2=0
(x-1.5a)^2-(0.5a+b)^2 =0
(x-2a-b)(x-a+b)=0
x-2a-b=0,x-a+b=0
x1=2a+b,x2=a-b
If point C is a point on line AB, and AC ratio AB = CB ratio AC, then point C is called the golden section point of AB, ab = 2, AC =?
AC = 3-radical 5, but I don't know the process
Let AC = X
cb=2-x
X / 2 = (2-x) / x, X & # 178; = 4-2x, the solution is x = radical 5-1 or - radical 5-1 (rounding off) BC = 3-radical 5
If the definition of X (x 3 + 0) or F (x 3 + 1) is known, is it odd?
Because the function is odd,
∴f（-x）=-f（x）
∴f（-0）=-f（0）
- f (0) = - 1 or 1
‖ f (0) = 1 or - 1
But remember the teacher said that as long as the function is odd and defined in F (0), then f (0) = 0
So how to do this problem
If x = 0 is an odd function, then f (0) = 0
Here, when x > 0, f (x) = x ^ 3 + 2x ^ 2-1
X
Let a = {x | x2-3x + 2 = 0}, B = {x | x2 + 2 (a + 1) x + (A2-5) = 0}, if u = R, a ∩ (∁ UB) = a, find the value range of real number a
∩ a ∩ (∁ UB) = a, ∩ a ∩ B = ∞, if B = ∞, then △ 0 {a ＜ - 3 is suitable; if B ≠ B, then a = - 3, B = {2}, a ∩ B = {2}, is not suitable; if a ＞ - 3, then 1 ∉ B and 2 ∉ B are needed, then 2 is substituted into B's equation to get a = - 1 or a = - 3; 1 is substituted into B's equation to get A2 + 2a-2 = 0 {a}
Given that point C is the golden section point of line AB, and AC > BC, then the ratio of line AC to BC is
Let AB = 1, AC = x, BC = 1-x,
AB/AC=AC/BC
1/x=x/（1-x）
x²=1-x
x²+x-1=0
x=（-1＋√5）/2
BC=1-（-1+√5）/2
=（3-√5）/2,
∴AC/BC
=（√5+1）/2
If f (x) is defined as an odd function, f (x) x 3 (x 2) = 0
F (x) is an odd function, so f (- x) = - f (x), f (0) = 0
When x > 0, the analytic formula is known,
The analytic expression of x < 0 can be obtained by F (- x) = - f (x)
When x 0, substitute it into the analytic formula F (x) = X3 (1-x2) where x > 0
f(-x) = (-x)^3 [1-(-x)²] = -x^3 (1-x²) = -f(x)