If point C is the golden section of line AB, then AC of AB is equal to

If point C is the golden section of line AB, then AC of AB is equal to

AC of AB is equal to BC of AC, that is: the square of AC = AB times BC
That is: Half Root 5 minus one
Root 5-1
——
Two
Approximately equal to 0.618
If f (x) is an odd function defined on R, when x > 0, f (x) = - 1 / 3x ^ 2 + 2x, find the analytic expression of F (x)
f(0)=f(-0)=-f(0),2f(0)=0,f(0)=0
When x0 - f (x) = f (- x) = - x ^ 2 / 3-2x, so f (x) = (x ^ 2) / 3 + 2x, when x = 0, it also satisfies this formula
If it's written as a piecewise function
f(x) -1/3x^2+2x ,x>0
1/3x^2+2x ,x
An odd function has f (- x) = - f (x)
When x > 0, - x0, f (x) = - 1 / 3x ^ 2 + 2x
When x0, f (x) = - 1 / 3x ^ 2 + 2x
Then when X0
That is, f (- x) = - 1 / 3 (- x) ^ 2-2x
=-f(x)
So f (x) = 1 / 3x ^ 2 + 2x
So the analytic expression of function f (x) is f (x) = 1 / 3x ^ 2 + 2x
Finding f (x) with known f (x + 1) = 2x-3x + 1
Why can I change t into x at the last step? I don't understand that
Let x = x + 1, then x = X-1
f(X)=2(X-1)^2-3(X-1)+1
=2(X^2-2X+1)-3X+3+1
=2X^2-4X+2-3X+4
=2X^2-7X+6
That is, f (x) = 2x ^ 2-7x + 6
f(x+1)=2x^2-3x+1=2[(x+1)-2]^2+(x+1)-2
f(x)=2(x-2)^2+x-2=2x^2-7x+6
Let x + 1 = t, then x = T-1
The X of substituting T-1 into the formula is:
f(t)=2(t-1)²-3(t-1)+1
=2(t²-2t+1)-3t+4
=2t²-7t+6
F (T) is equivalent to f (x), that is, t is equivalent to X
So f (x) = 2x & sup2; - 7x + 6
Let x + 1 = t, x = T-1
f（x+1）=2x^2-3x+1
f(t)=2(t-1)^2-3(t-1)+1
=2(t^2-2t+1)-3t+3+1
=2t^2 -4t+2-3t+4
=2t^2-7t+6
So f (x) = 2x ^ 2-7x + 6
Given that C is a point on the line AB, ab = a, AC = B, and 1 / A + 1 / B-1 / A-B = 0, try to explain that point C is a golden section point of the line ab
Let a / b = k, then a = BK 1 / A + 1 / B-1 / (a-b) = 0 1 / BK + 1 / B-1 / b (k-1) = 0 k-1 + K (k-1) - k = 0 k-1 + K ^ 2-k-1 = 0 K ^ 2-k-1 = 0 K ^ 2-k + 1 / 4-1 / 4-1 = 0 (k-1 / 2) ^ 2 = 5 / 4 k-1 / 2 = +, - 1 / 2 * radical 5 because a and B are positive real numbers, a / b = k > 0, so K is AB / AC = A / b = 1 / 2 * (1 + radical
Let f (x) be an odd function defined on R. when x ″ = 0, f (x) = 3x_ 2X + a (a ∈ R), then f (- 2)=
Odd function f (0) = 0, so as long as f (x = 0) = 0 is calculated as a, and then x = - 2 is substituted to calculate
It is known that f (x) is an odd function. When x > 0, f (x) = 3x & # 178; - x + 1, we can find if X
A:
If f (x) is an odd function, then f (- x) = - f (x)
x> 0, f (x) = 3x ^ 2-x + 1
x0,f(-x)=3(-x)^2-(-x)+1=3x^2+x+1=-f(x)
So: X
Let x0, so f (- x) = 3x ^ 2 + X + 1 = - f (x)
F (x) = - 3x ^ 2-x-1
Given that C is the golden section point of line AB, find AC ratio AB and CB ratio AC!
CB / AC = AC / AB = radical 5-1 / 2
If x (x) is defined as an odd square function, then f (x) = 1
solution
Let - x = x
-f(x)=f(x)
therefore
When x
When x0 ∫ f (- x) = 3 (- x) ^ 2 + 2 (- x) - 1 = 3x ^ 2-2x-1 ∵ f (x) is an odd function ∫ f (x) = - f (- x) = - (3x ^ 2-2x-1) = - 3x ^ 2 + 2x + 1, X
F (x + 1 / x) = x & # 178; + 1 / X & # 178;. Find ① f (x) ② f (3x-2)
f(x+1/x)=x²+1/x²+2-2=(x+1/x)^2-2
Then f (x) = x ^ 2-2
f(3x-2)=(3x-2)^2-2
This is the integral substitution method of composite function
f(x+1/x)=x²+1/x²=(x+1/x)^2-2
f(x)=x^2-2
f(3x-2)=(3x-2)^2-2=9x^2-12x+2
It is known that l is the golden section of line AB, and AC of Ba is about 0.618?
When point C is on line AB, CB of AC = Ba of AC is about 0.618
When point C is on the extension line of segment Ba, CB of AC is 1.618/0.618