Let f (2x + 1) = 3x-2, then f (x) = what

Let f (2x + 1) = 3x-2, then f (x) = what

Let 2x + 1 = u, then x = (U-1) / 2
In the original equation, f (U) = [3 (U-1) / 2] - 2
Algebraic operation, f (U) = (3u-7) / 2
Let x = u, f (x) = (3x-7) / 2 = 1.5x-3.5
Given that the maximum value of quadratic function f (x) = ax & # 178; + BX + C is 8, the function g (x) = f (x) + 1, and G (x) has two zeros, which are 2 and - 1 respectively, try to determine the analytic expression of quadratic function f (x)
g(x)=f(x)+1
Therefore, the image of G (x) is obtained by translating the image of F (x) upward one unit
The axis of symmetry is invariant in the process of upward translation
If the zeros of G (x) are 2 and - 1, then its axis of symmetry is the midpoint of 2 and - 1, x = 1 / 2
So the axis of symmetry of F (x) is also x = 1 / 2
Because the maximum value of F (x) is 8
Then: the vertex of F (x) is (1 / 2,8)
We can write f (x) as the vertex formula: F (x) = a (x-1 / 2) &# 178; + 8
Then G (x) = a (x-1 / 2) &# 178; + 9
G (2) = 0, that is: 9A / 4 + 9 = 0
A = - 4
So, f (x) = - 4 (x-1 / 2) &# 178; + 8
That is: F (x) = - 4x & # 178; + 4x + 7
Let f (x) = 2x ^ 2-3x + 7, find f (x + 1)
Just replace X in the original function with x + 1: F (x + 1) = 2 (x + 1) & # 178; - 3 (x + 1) + 7 = 2x & # 178; + X + 6
It is proved that the function f (x) = ax ^ 2 + BX + C (a > 0), and f (1) = - A / 2. Has two zeros
From F (1) = - A / 2, a + B + C = - A / 2 is obtained
So B = - 3A / 2-c
So B ^ 2-4ac = 9A ^ 2 / 4 + 3aC + C ^ 2-4ac
=9a^2/4-ac+c^2
=2a^2+(a/2-c)^2
Because a > 0, the above formula is always greater than 0, so the original function and X axis have two intersections, that is, there are two zeros
Let f (x) = (1 / 3) ∧ X & # 178; - 3x + 1, G (x) = 3 ∧ 5-2x, find the range of x such that f (x) > G (x)
f(x)=3^(-x^2+3x-1),
So f (x) > G (x) is: - x ^ 2 + 3x-1 > 5-2x,
The solution is x ∈ (2,3)
Mathematics problem of senior one: we know that the quadratic function f (x) = ax & # 178; + BX (a, B are constants, and a ≠ 0) satisfies the condition f (x-1) = f (3-x) and the equation f (x) = 2x has equal roots. This is how I do ∵ f (x-1) = f (3-x) ∵ B = - 2A and ∵ the equation f (x) = 2x has equal roots
Given the function f (2x + 1) = 3x-2, find the analytic expression of function f (x)
Let 2x + 1 = t, x = (t-1) / 2
f(t)=3(t-1)/2-2=3t/2-7/2
f(x)=3x/2-7/2
Given the quadratic function f (x) = ax & # 178; + BX (a, B are constants and a ≠ 0), if f (2) = 0 and f (x) x = x has equal real roots, find f (x)
Given the quadratic function f (x) = ax & # 178; + BX (a, B are constants and a ≠ 0), satisfying the condition f (2) = 0 and f (x) x = x has equal real roots, the analytic expression of F (x) is obtained
The answer is that the ∵ equation AX & # 178; + (B-1) x = 0 has equal real roots, ∵ B = 1
I want to ask why B equals 1
Let the equation AX & # 178; + (B-1) x = 0 have equal real roots,
The discriminant △ = 0, that is, (B-1) ^ 2 - 4A · 0 = 0,
(B-1) ^ 2 = 0, that is, B-1 = 0, B = 1
If a quadratic equation of one variable has equal real roots, then the discriminant of the roots is B ^ 2-4ac = 0, where the discriminant of the roots is (B-1) ^ 2-4 · a · 0 = 0, s, so B = 1
x1=0,x2=-(b-1)/a
x1=x2,
So B = 1
Given that f (x) is an odd function on R, and when x > 0, f (x) = 2x & sup2; - 3x + 5, the analytic expression of F (x) is obtained
If f (x) is an odd function, then f (x) = - f (- x)
When x > 0, f (x) = 2x & sup2; - 3x + 5
When x = 0, f (0) = 0 (the definition of odd function)
When x
Did you copy it wrong
Function y = ax + BX + C (a ≠ 0) and AC
B squared minus 4ac > 0, so there are two zeros
Two. He's greater than zero