If f (3x) = the square of 2x - 1, then the analytic expression of F (x) is

If f (3x) = the square of 2x - 1, then the analytic expression of F (x) is

analysis
f(3x)=2x^2-1
Let 3x = t
x=t/3
f(t)=2t^2/9-1
X T interchange
f(x)=2x^2/9-1
Let x = x / 3 be substituted
f(x)=2(x/3)^2-1=2x^2/9-1
Hello,
Let u = 3x, then x = u / 3
f(3x)=2x²-1
f(U)=2（U/3）²-1=2U²/9-1
U→x
f(x)=2x²/9-1
Let 3x = X
Then x = x / 3;
f（X）= （2* X/3）^2 - 1 ;
f(x) = 4x^2/9 - 1 ;
It is known that the quadratic function y = AX2 + BX + C of X (the image with a not equal to 0 passes through the point (- 1,3) (1,1)
If 0 is less than C and less than 1, find the value range of A
3=a-b+c
1=a+b+c
Add
2a+2c=4
a=2-c
Zero
F (2x-1) = x ^ 2-3x + 2, find f (x)
Let t = 2x-1
Then x = (T + 1) / 2
f(t)=1/4*(t+1)^2-3(t+1)/2+2=1/4*t^2-t+3/4
f(x)=1/4*x^2-x+3/4
The quadratic function y = AXX + BX + C (a is not equal to 0), if 2A + B = 0, and when x = - 1, y = 4, then when x = 3, y = ()
Because 2A + B = 0
So 2A = - B - 2A / b = 1
So the axis of symmetry is x = 1
Because 1 - (- 1) = 3-1 means that the distance between the two points and the axis of symmetry is equal
And because it's a parabola,
So when x = 3, y = 4
Because when x = - 1, y = 4,
So substituting it into the analytical formula, we get A-B + C = 4.
Then substituting x = 3 into the analytic formula, we get y = 9A + 3B + C.
And because 2A + B = 0,
So 8A + 4B = 0.
So y = 9A + 3B + C - (8a + 4b) = A-B + C = 4.
2a+b=0
==>b=-2a
x=-1,y=4
==>A * (- 1) * (- 1) + b * (- 1) + C = 4, that is, A-B + C = 4,
Substituting B = - 2A gives 3A + C = 4
When x = 3,
y=9a+3b+c=9a+3*（-2a）+c=3a+c=4
f(3x+1)=x^2-2x,f(4)=?
3x + 1 = 4, x = 1. X-2x = 1-2 * 1 = - 1, so f (4) = - 1
The quadratic function y = ax ^ 2 + BX + C (a is not equal to 0) if 2A + B = 0 and x = - 1, y = 6, then when x = 3, y = 6=_____
When x = 3, y = 9A + 3B + C
From the question: 2A + B = 0 (1), A-B + C = 6 (2)
Formula 1 * 4 + 2: 9A + 3B + C = 6
Let f (x) = (2x + 5) ^ 3 (3x-10) ^ 4, find f '(x)
lgf=lg((2x+5)^3*(3x-10)^4)lgf=lg(2x+5)^3+lg(3x-10)^4lgf=3lg(2x+5)+4lg(3x-10)f'/f=3*1/(2x+5)*2+4*1/(3x-10)*3f'/f=6/(2x+5)+12/(3x-10)f'=(2x+5)^3*(3x-10)^4 *(6/(2x+5)+12/(3x-10))
6(2x+5)^2(3x-10)4+12(2x+5)^3(3x-10)^3
The quadratic function f (x) = ax ^ 2 + BX + C satisfies 2A + C / 2 > b and CB and C
When x = - 2,
F (- 2) = 4a-2b + C, and because 2A + C / 2 > b, so 4A + C > 2B, f (- 2) = 4a-2b + C > 0
When x = - 0,
f（0）=c
A
Because f (0) = C0, f (0) x f (- 2)
Let f (x) = (2x + 5) ^ 2 * (3x-1) ^ 4, find f '(x)
f(x)=(2x+5)^2*(3x-1)^4
f(x)=[(2x+5)^2]'*(3x-1)^4+(2x+5)^2*[(3x-1)^4]'
=2(2x+5)*(3x-1)^4*(2x+5)'+(2x+5)^2*4(3x-1)^3*(3x-1)'
=4(2x+5)*(3x-1)^4+12(2x+5)^2*(3x-1)^3
4（2x+5）（3x-1）^4+12(2X+5)^2(3x-1)^3
f'(x)=[(2x＋5)²]'×[(3x－1)^4]＋[(2x＋5)²]×[(3x－1)^4]'
=[4(2x＋5)](3x－1)^4＋(2x＋5)²×[12(3x－1)³]
=4(2x＋5)(3x－1)³×[(3x－1)＋3(2x＋5)]
=8(2x＋5)(3x－1)³(5x＋7)
General idea: remember y = f (x), take ln for both sides at the same time, and then take the left and right derivatives respectively
ln y=ln （2x+5）^2*(3*x-1)^4
After derivation, y '/ y = Write it yourself. There is a derivation formula in the book. It's very troublesome to type it out
(given the quadratic function f (x) = AX2 + BX + C.)
Given the quadratic function f (x) = AX2 + BX + C & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (1) if a & gt; B & gt; C and f (1) = 0, it is proved that f (x) has two zeros; (2) if x1, X2 ∈ R, X1 < X2, f (x1) ≠ f (x2), it is proved that the equation f (x) & # 8722; 1 / 2 [f (x1) + F (x2)] = 0 has a real root in the interval (x1, x2)
 
 
Let g (x) = f (x) &; 1 / 2 [f (x1) + F (x2)], then G (x 1) = f (x1) &; 1 / 2 [f (x1) + F (x2)] = 1 / 2 [f (x1) &; f (x2)] & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; this step is not understood
The first question I will, the second one
Let g (x) = f (x) & _; 1 / 2 [f (x1) + F (x2)], then G (x 1) = f (x1) - 1 / 2 [f (x1) + F (x2)] = f (x1) - 1 / 2F (x1) - 1 / 2F (x2) = 1 / 2 [f (x1) & _; f (x2)] g (x2) = f (x2) - 1 / 2 [f (x1) + F (x2)] = f (x2) - 1