It is known that the function f (x) is an odd function defined on R. when x is greater than or equal to 0, f (x) = x (1 + x). The image of function f (x) is drawn and the analytic formula of the function is obtained

It is known that the function f (x) is an odd function defined on R. when x is greater than or equal to 0, f (x) = x (1 + x). The image of function f (x) is drawn and the analytic formula of the function is obtained

When x is less than 0, - x > 0, f (- x) = - x (1-x)
Because the function f (x) is an odd function over R, f (- x) = - f (x) = - x (1-x)
So f (x) = x (1-x)
So: X

As a function of y = 2x-1, what value of X is, when y > 5x, y is less than or equal to - 3, why x is, y = - 7

Solve three inequalities, one, two X-1 > 5x has X

When x is equal to, the value of function y = 5x + 3 is greater than 0, and when x is equal to, the value of function y = 5x + 4 is less than 0

When x > - 3 / 5
The value of y = 5x + 3 is greater than 0
When x < - 4 / 5
The value of y = 5x + 4 is less than 0

It is known that y = f (x) is a function defined on R, and its image is symmetric with respect to x = 2. When x is greater than or equal to zero, f (x) = 5x + 3

Is the title wrong
When x is greater than or equal to 2, f (x) = 5x + 3,
X2
f(x)=f(4-x)=5(4-x)+3=-5x+23

Make the image of function y = 0.5x-4, and answer the question according to the image: 1. When x takes what value, y is greater than 0? When - 1 is less than or equal to X and less than or equal to 2, find the range of Y

The function image is a straight line, which intersects with y axis at (0, - 4) and X axis at (8,0)
When x > 8, Y > 0
When - 1 is less than or equal to X and less than or equal to 2, - 11 / 2

Draw the graph of the function y = | X-1 | + | 2x + 3 |, and write its monotone interval and maximum value according to the image

You just need to change this function into a piecewise function. When x is greater than 1, when x is less than or equal to 1 and greater than or equal to - 2 / 3, and when x is less than - 2 / 3, we will discuss it in three paragraphs

Write the monotone interval of the function y = x? - 2x and the symmetry axis of the image. Observe: what are the characteristics on both sides of the function image

y = x² - 2x = (x - 1)² - 1
The monotone decreasing interval of function y is (- ∞, 1), and the monotone increasing interval is [1, + ∞)
The symmetry axis is x = 1. It is observed that the function values of the function image are equal on both sides of the symmetry axis

The graph of the function y = x? - 4|x | - 5 is made, and the monotone interval of the function is written

X>0
y=x²-4x-5=(x-2)²-9
X<0
y=x²+4x-5=(x+2)²-9
X=0
y=-5
 
X < - 2 minus
-2 < x < 0
0 < x < 2 decreased
x> 2

Let f (x) and G (x) = Log1 If the graph of 2x is symmetric with respect to the straight line y = x, then the monotone increasing interval of the function f (x2 + 2x) is______ .

∵ function f (x) and function g (x) = Log1
The image of 2x is symmetric about the line y = X,
∴f(x)=(1
2) x
The function f (x) decreases monotonically on R
∵t=x2+2x=(x+1)2-1,
ν t = x2 + 2x decreases monotonically on (- ∞, - 1]
The monotonic increasing interval of function f (x2 + 2x) is (- ∞, - 1]
So the answer is: (- ∞, - 1]

It is proved that the function f (x) = x 2 is an increasing function on the interval [0, + ∞) Ask for the great God sect

It is proved that: F (x) = x 2, let X1 > x2 > = 0f (x1) - f (x2) = X1? - x2 = (x1-x2) (x1 + x2) because: X1 > x2 > = 0, so: x1-x2 > 0, X1 + x2 > 0, so: F (x1) - f (x2) = (x1-x2) (x1 + x2) > 0, so: F (x1) > F (x2) so: F (x) = x ~ 2 is increasing in the interval [0, + ∞)