Let f (x) = ax ^ 7 + BX ^ 3 + cx-5, where ABC is a constant and f (- 7) = 7, then f (7) is equal to

Let f (x) = ax ^ 7 + BX ^ 3 + cx-5, where ABC is a constant and f (- 7) = 7, then f (7) is equal to

f(x)=ax^7+bx^3+cx-5
f(-7)=a(-7)^7+b(-7)^3+c(-7)-5=7
a(-7)^7+b(-7)^3+c(-7)=12
a×7^7+b×7^3+7c=-12
therefore
f(7)=a×7^7+b×7^3+7c-5
=-12-5
=-17

We know that f (x) = ax ^ 3 + BX ^ 2 + CX (a is not equal to 0) obtains the extreme value when x = plus or minus 1, f (1) = - 11. Find the value of constant a, B, C

f(1)=a+b+c=-11.1
f(x)'=3ax^2+2bx+c
The extreme value is obtained when x = 1
f(1)'=3a+2b+c=0.2
f(-1)'=3a-2b+c=0.3
From 1,2,3, a = 5.5, B = 0, C = - 16.5

The vertex coordinates of quadratic function image are (1, - 6) through (2, - 8), and the analytic expression of quadratic function is obtained

Let y = a (x-1) ^ 2-6, (2, - 8) be substituted to get a = - 2, so y = - 2 (x-1) ^ 2-6

Given that the vertex coordinates of quadratic function image are (2,4) and pass through points (1,8), the analytic expression of quadratic function is obtained

∵ the vertex coordinates of quadratic function are (2,4)
Let y = a (X-2) ^ 2 + 4
∵ image passing point (1,8)
∴a（1-2）^2+4=8
The solution is a = 4
∴y=4（x-2）^2+4
Standard problem solving format

According to the following conditions, the analytic expressions of quadratic functions are obtained: (1) the vertex coordinates of known images are (- 1, - 8) and pass through points (0, - 6)
(2) It is known that the image passes through points (3,0), (2. - 3) and takes the line x = 0 as the symmetry axis

1. Let the quadratic function relation be y = a (x + 1) & sup2; - 8
When x = 0, y = A-8 = - 6
So a = 2
So y = 2 (x + 1) & sup2; - 8
2. Take the line x = 0 as the axis of symmetry, that is, the axis of symmetry is the Y axis
So B = 0
So let y = ax & sup2; + C
Take points (3,0), (2. - 3) into
A = 3 / 5, C = - 27 / 5
So y = 3 / 5x & sup2; - 27 / 5

The image passes through the point (6,0), and the vertex coordinates are (4, - 8)

Vertex coordinates are (4, - 8)
y=a(x-4)²-8
Passing point (6,0)
0=a(6-4)²-8
a=2
So y = 2x & # 178; - 16x + 24

Square of 64x - 81 = 0, is x nine eighths or nine eighths

Plus or minus 9 / 8 is OK

It is known that the quadratic function y = the square of negative x + X + 2. [1] in what range is x taken, and the image of quadratic function is above the x-axis? [2] in what range is x taken, and the image of quadratic function is below the x-axis?

(1) - 1 < x < 2, image above
(2) When x ＜ - 1, X ＞ 2, the image is below

If the image of quadratic function y = (M + 5) x2 + 2 (M + 1) x + m is all above the x-axis, then the value range of M is___ ．

∵ the images of quadratic function y = (M + 5) x2 + 2 (M + 1) x + m are all above the x-axis, ∵ (M + 5) > 0, ∵ 0, ∵ m ＞ - 5, 4 (M + 1) 2-4 (M + 5) × m ＜ 0, so m ＞ 13

Finding the range and definition of function y = 1 / 3arcsin1 / X

|1/x|≤1
Domain x ≤ - 1 or X ≥ 1
range
-π/6≤y≤π/6 and y≠0
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