Find the monotonic interval under the root number x 2x 3 Find the monotonic interval under the roo

You should finish the topic. If I'm not mistaken, you should be trying to express the monotonic interval of y=root (x +2x-3). If it's just your title, it doesn't mean anything, it's just an algebraic formula.

y=root(x +2x-3)=(x +2x-3) 1 2, so obviously this is a composite function.

Use mantras'Same increase and different subtraction'

y=x1 2 is monotonically increasing on x>=0.

Now it's just a monotonic interval of x +2x-3.

However, due to the root number, it should be greater than or equal to 0

So solve x<=-3 or x>=1

Then the axis of symmetry of x +2x-3 is -1

Therefore, the function decreases monotonically at x<-1 and increases conversely.

So in summary, when x<=-3 the original function decreases monotonically and x>=1 the original function increases monotonically.

The second function should be the same.

Try it yourself first and don't ask me again.

Under the root number (x + 2x-3), the domain is defined as x>=1, or x<=-3, and y=the root number u is monotonically increasing, as long as the monotonicity of u=x +2x-3 is considered.

u=x +2x-3=(x+1) 2-4, so x>=-1, monotonically increasing, x<-1 monotonically decreasing, therefore, y is monotonically increasing at x>=1 and x<=-3 is monotonically decreasing.

y=1 u is monotonically decreasing in the interval, so y is monotonically decreasing at 0>x>=-1, x>0 is monotonically decreasing, and x<=-1 is monotonically decreasing.

Under the root number (x +2x-3).

[(x+1) 2-4] under the root number

x+1）^2-4>=0

x<=-3,x>=1

x<=-3, and the root number (x +2x-3) is the subtraction function.

x>=1 (x +2x-3) is the increment function.

The lower (1 x +2x-3) and the lower (x +2x-3) are the reciprocal of each other.

x<=-3, and the root number (x +2x-3) is the increment function.

x>=1 (x +2x-3) is the subtraction function.

First, the function f(x)=x +2x-3=(x+3)(x-1)>=0, i.e., x>=1 or x<=-3, and the decreasing interval of the function f(x) is (negative infinity, -1) and the increasing interval is [1, positive infinity) Therefore, the monotonically decreasing interval under the root number (x +2x-3) is (negative infinity, -3] and the monotonically increasing interval is [1, positive infinity).

Solution: u=-x +2x

There is also u 0, which is -x +2x 0

i.e. x -2x 0

That is, 0 x 2, and there is a function early rise y = -x square + 2x under the root number

This is the function y= u

The axis of symmetry by u=-x +2x=-(x-1) +1 is x=1The opening is downward and you know that u is an increasing function on [0,1].

u is a subtractive function on [1,2].

And y= u is an increasing function.

That is, the increasing interval of -x square + 2x under the function y = root number is [0,1], and the decreasing interval is [1,2].,5,gaara610 report.

x Zhenru -2x 0 is not x 2 0 How did you get the solution x -2x 0 is not x 2, from x -2x 0 to get x(x-2) 0 i.e. x 0, x-2 0....1) or x 0, x-2 0....2) Solution (1) gives 0 x 2 (2) gives x does not exist, so x(x-2) 0 gives 0 x 2.

This is the solution for junior high school, and according to the solution for high school, you can get 0 x 2 directly from x -2x 0. ,

root, so -x 2+2x is greater than zero or equal to zero.

So 0 x 2

And y's most rocky and thick front value is 1

Axis of symmetry x = 1

Therefore, according to the coefficient of the quadratic term is less than zero, its image can be known.

The base is increased by [0,1] in the monotonic interval. (1, 2] minus.

f(x)=root number(x2+2x—3) Since x2+2x—3>=0, then there is an image opening of x greater than or equal to 1 or x less than or equal to -3 and y=x2+2x—3 When x is less than or equal to -3, the value of the function decreases with the increase of x, that is, f(x) decreases When x is greater than or equal to 1, the value of the function increases with the increase of x, that is, f(x) increases; So the decreasing interval is negative infinity to minus three ; The monotonically increasing range is 1 to positive infinity.

y=under the root number(x 2+2x-3) = under the root number[(x-1)(x+3)] The domain is defined as: x <= -3 or x >= 1 The monotonic interval is: when x <= -3, subtract the function, when x >= 1, increase the function.

First, the domain is defined as [-3,1], and the axis of symmetry is x=-1, then the decreasing interval is [-3.].-1], with an increasing interval of [-1,1].

From the monotonicity of the composite function, it can be seen that from negative infinity to -1 is monotonically decreasing, and from -1 to positive infinity is monotonically increasing.

First of all, it is obtained by the root number x 2-2x-3 greater than or equal to 0, x greater than or equal to 3 or x less than or equal to -1.

x 2-2x-3=(x-1) 2-4, the function increases when x is greater than or equal to 1 and decrements when it is less than or equal to -1.

So the function decrements when x is less than or equal to -1 and increases when x is greater than or equal to 3.

Because x 2 + 2x-3

x+3)(x-1)≥0

So the monotonic interval is (-infinity, -3】u[1,+infinity)

Monotonicity of composite functions.

Let t=x +2x-3, then y= t increase monotonically.

t 0 solution x -3 or x 1

The axis of symmetry is x=-1, t=x +2x-3 decreases monotonically at (- 3] and increases monotonically at [1,+.

So the function f(x) has a monotonically increasing interval of [1,+

y=sqrt(x^2+2x-3)

First, there is x 2+2x-3>=0 (greater than or equal to 0 under the root number) to find the defining domain (- 3] [1,+

The outer function is the increasing function (y=sqrt(m)).

Therefore, it is required that the decreasing interval m=(x+1) 2-4 of the inner function m=x 2+2x-3 is decreasing at (- 1) and then intersecting with the defined domain, so that the monotonic decreasing interval is (- 3).

y = root number x +2x-3

Root number [(x+1)(x-3)].

The monotonic reduction interval is x<-1

Solution: y = root number (-x 2+2x+3).

Define the domain -x 2+2x+3>=0

x^2-2x-3<=0

x+1)(x-3)<=0

x∈[-1,3]

Let g(x)=-x 2+2x+3

g'(x)=-2x+2

Ream'(x)=-2x+2<=0

x>=1

So the decreasing interval of g(x) is [1, positive infinity).

Combine to define domains.

Then the monotonically decreasing interval under the y-root (-x 2x+3) is [1,3].