Derivative of high school mathematics known function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e is even function The tangent equation passing through point a (0. - 1) and at x = 1 is 2x + Y-2 = 0. Find the analytic formula of y = f (x)

Derivative of high school mathematics known function f (x) = ax ^ 4 + BX ^ 3 + CX ^ 2 + DX + e is even function The tangent equation passing through point a (0. - 1) and at x = 1 is 2x + Y-2 = 0. Find the analytic formula of y = f (x)

Substituting point a (0. - 1) gives e = - 1
From the tangent equation at x = 1 is 2x + Y-2 = 0, we know that f (x) passes through the point (1,0)
If f (x) is an even function, then f (x) passes through the point (- 1,0) and the tangent equation at this point is - 2x + Y-2 = 0
Substituting into the equation, a + B + C + D + e = 0; A-B + C-D + e = 0
Then a + C = 1, B + D = 0
F '(x) = 4x ^ 3 + 3x ^ 2 + 2x ^ + D
Then f '(1) = 4A + 3B + 2C + D = - 2; f' (- 1) = - 4A + 3b-2c + D = 2
A + B = - 2, C + D = 3; - A + B = 2
To sum up: a = - 2, B = 0, C = 3, d = - 3, e = - 1
Substituting it into the equation, it is obtained that:
f(x)= -2x^4+3x^2-3x-1
Note: I haven't touched these things for a long time, so you'd better just take a close look

Given the quadratic function f (x) = ax ^ 2 + BX (a ≠ 0), and f (x + 1) is an even function, the definition is: the real number x satisfying f (x) = x becomes the "fixed point" of function f (x), if the function f (x) has and only one fixed point Is there an interval [M, n] (M

F (x + 1) = ax ^ 2 + (2a + b) x + A + 1 is even function, that is 2A + B = 0
Because f (x) = x has only one solution, that is, the equation AX ^ 2 + BX = x has only one solution, △ = (B-1) ^ 2 = 0, so B = 1, a = - 1 / 2
f(x)=-x^2/2 + x
When the symmetry axis X = 1 ∈ [M, n], 3N in the range is 1 / 2 of the maximum value, i.e. n = 1 / 6. Obviously, it is impossible to make x = 1 ∈ [M, n]
When m > 1, f (m) = 3N, f (n) = 3M. Obviously, m and N are greater than 0. From F (x) > 0, we know that m, n ∈ [1,2], namely 3M, 3N ∈ [3,6], but the maximum value of F (x) does not exceed 1 / 2. Obviously, m and n do not exist at this time
When n < 1, f (m) = 3M, f (n) = 3N, the solution is m = - 4, n = 0

If f (x) = ax ^ 2 + BX + C (- 2a-3 ≤ x ≤ 1) is even, then a = B=

Let f (x) = ax ^ 2 + BX + C be even functions
So (x) = F
B=0
Even function definition domain symmetry
-2a-3 = - 1, so a = - 1

Given that f (x) = AX2 + BX is an even function defined on [A-1, 2A], then the value of a + B is () A. −1 Three B. 1 Three C. −1 Two D. 1 Two

According to the meaning of the title: F (- x) = f (x),  B = 0, and A-1 = - 2A, ﹥ a = 1
3，
∴a+b=1
3．
Therefore, B

Is f (x-1) = x ^ 2 an even function? Is the image symmetric about x = 1 or Y?

f(x-1)=x^2=（x-1)^2+2(x-1)+1
==>f(x)=x^2+2x+1=(x+1)^2
So the axis of symmetry is x = - 1

Why y = f (x + 8) is even function, then y = (x) image is symmetric about x = 8

Because if f (x + 8) is an even function, it shows that the symmetry axis of F (x) image after being shifted to the left by 8 units is the Y axis
Therefore, the original symmetry axis of F (x) image is x = 8

If the graph of even function y = f (x) is symmetric with respect to the straight line x = 2 and f (3) = 3, then f (- 1)=______ .

Because the image of the even function y = f (x) is symmetric with respect to the line x = 2,
So f (2 + x) = f (2-x) = f (X-2),
That is, f (x + 4) = f (x),
Then f (- 1) = f (- 1 + 4) = f (3) = 3,
So the answer is: 3

If the function y = f (x + 1) is even, then the image of F (x) is symmetric

Y = f (x + 1) is symmetric about x = 0
f(x)=f(x+1-1)
F (x) is symmetric about x = 1

The function y = x ^ 2 + (a + 1) x + B holds Y > = x for any real number x, and when x = 3, y = 3, find the value of a and B The function y = x ^ 2 + (a + 1) x + B holds Y > = x for any real number x, and when x = 3, y = 3, find the value of a and B

Y = x ^ 2 + (a + 1) x + b > = x x x = R is equivalent to x ^ 2 + ax + b > = 0 x is R. the opening of this function can be transformed into (x + A / 2) ^ 2 + (4b-a ^ 2) / 4 > = 0, that is, the minimum value (4b-a ^ 2) / 4 > = 0. If x = 3 and y = 3 are substituted into the original equation to get 3A + B + 9 = 0, then B = - 3a-9 is substituted into (4b-a ^ 2) / 4 > = 0 to obtain (a + 6)

If the function y = f (x) is even, then the image of y = f (x + 2) is symmetric about ()

The axis of symmetry of F (x) is x = 0
Move him two units to the left. It's f (x + 2)
Then the axis of symmetry is shifted to the left by 2 units
So x = - 2