It is known that the odd function f (x) = - X's square + 2x (x > 0) 0 (x = 0) x's Square = MX (x)

It is known that the odd function f (x) = - X's square + 2x (x > 0) 0 (x = 0) x's Square = MX (x)

If f (x) is an odd function, then f (- x) = - f (x)
Let - x > 0, then f (- x) = - x ^ 2-2x
Then f (x) = x ^ 2 + 2x (x < 0)
So m = 2
Second, from the function expression, we can know that the function increases monotonically between - 1 and 1, then the absolute value of - 1 < a - 2 < = 1
The range of a is [- 3, - 1) and (1,3]

Using the method of drawing points in a list, the image of function y = 1 / (absolute value of K) is drawn

When k is greater than 0, the image is in 1,3 quadrant

Draw an appropriate function image to find the solution of x2 = x 3
x+3

x^2＝x+3
x^2-x＝3
x^2-x+1/4＝13/4
(x+1/2)^2＝13/4
x+1/2＝±√13/2
∴x＝(-1+√13)/2
Or x = (- 1 - √ 13) / 2

F (x) = - x2 + 2 | x | + 3 how to draw this function graph?
ditto...

There are two situations!
1.x>=0
Draw an image of F (x) = - x2 + 2x + 3
2.x