The 5th power of function ax + the 3rd power of BX + Cx. If f (- 7) = 7, what is f (7) = 5 F (- 7) = 7 indicates that f (- x) = f (x) should not be an even function

The 5th power of function ax + the 3rd power of BX + Cx. If f (- 7) = 7, what is f (7) = 5 F (- 7) = 7 indicates that f (- x) = f (x) should not be an even function

Minus 7
F (x) = ax ^ 5 + BX ^ 3 + CX, by
f(-x)=a(-x)^5+b(-x)^3+c(-x)
=-ax^5-bx^3-cx
=-(ax^5+bx^3+cx)
=-f(x)
Is an odd function
Odd negative even positive, odd function and opposite, is negative

Given the function f (x) = ax to the third power - BX + 2, if f (2) = 5, then f (- 2) is equal to______

f(2)=a*2^3-b*2+2=5
a*2^3-b*2=3
f(-2)=a*(-2)^3-b*(-2)+2
=-a*2^3+b*2+2
=-(a*2^3-b*2)+2
=-3+2
=-1

Given the function f (x) = the 5th power of X + the 3rd power of AX + bx-2, if f (- 2) = 0, find the value of F (2)

G (x) = f (x) + 2 is an odd function
So g (2) = - G (- 2)
That is, f (2) + 2 = - f (- 2) - 2
So f (2) = - f (- 2) - 4 = - 4

Let f (x) = the x power of E / 1 + ax * 2, where a is a positive real number. When a = 4 / 3, find the extreme point of F (x),

1) Find the derivative and get f '(x) = e ^ x {1 + (4 / 3) x ^ 2 - (8 / 3) x} / {1 + (4 / 3) x ^ 2} ^ 2
Because the extreme point is calculated, x = 0.5 or 1.5
0, x = 0.5 or 1.5
So the extreme point is x = 0.5 or 1.5
(2)f'(x)=e^x(ax^2-2ax+1)/(1+ax^2)^2
Because it is a monotone function, as long as ax ^ 2-2ax + 1 is constant greater than 0 or constant less than 0
When a = 0, the condition is satisfied
When a > 0, the minimum value 4ac-b ^ 2 / 4A > 0 is 0

Function f (x) = 1 3x3-x2 + AX-1 has extreme points, then the value range of a is () A. （-∞，0） B. （-∞，0] C. （-∞，1） D. （-∞，1]

∵ function f (x) = 1
3x3-x2 + AX-1 has extreme points,
The derivative f '(x) = x2-2x + a = 0 of F (x) has two real roots,
∴△=4-4a＞0，∴a＜1，
Therefore, C

If the function y = ax ex has an extreme point less than zero, the value range of real number a is () A. （0，+∞） B. （0，1） C. （-∞，1） D. （-1，1）

∵y=ax-ex，
∴y'=a-ex．
According to the meaning of the question, a-ex = 0 has a real root less than 0, let Y1 = ex, y2 = a, then the intersection of the two curves is in the second quadrant,
Combined with the image, it is easy to get 0 < a < 1,
Therefore, the value range of real number a is (0, 1),
Therefore: B