Discuss the monotonicity of the function f x 2 x 1 x on x 0

f'(x)=2-1 x 2=(2x 2-1) x 2, let f'(x)=0: x= 2 2 x (0, 2 2 ) f'(x)<0,x ( 2 2, + f'(x) >0, so f(x) decreases on (0, 2 2) and increases on (2 2, +).

First, the derivative of this function is given, and the result is f'(x)=2—(1 x) squared, and then let f'(x)=0, the solution is x=1 2

Therefore, when x is greater than 0 and less than 1 2, f'(x) is less than 0, then f(x) is monotonically decreasing;

When x is greater than 1 2 and f'(x) is greater than 0, then f(x) increases monotonically at this time.

f, the derivative of f(x).'(x) = 2-1 x2, let f'(x) = 0 to get: x = 2 2 under the root number

The list shows that f(x) is a subtraction function on (0, 2 2 under the root number).

is an increasing function on (2 2 under the root number, positive infinity).

To do it with the definition of a monotonic function, let x1 x2, f(x1)-f(x2)=2x1 2x2-2x1x2 2+x2-x1 tidy up: the first term and the last term are combined, and the other two items are combined.

Solution: f(x)positive elimination 2x (x+1).

Let u 2x, v x 1, then u 2, v 1f (x) (u v-uv) v 2

2（x＋1）-2x]／（x＋1）^2

2 (x Lift Burn 1) 2

x∈（0，＋∞x＋1）^2＞0

f＇（x）＞0

When x (0, , f(x) 2x (x 1) is a monotonic increment.

To judge the monotonicity of the function f(x) =x +2x + 1, we need to find its derivative.

Finding the derivative gives f'(x) =2x + 2.

When the derivative f'(x) jujube is earlier than zero, the function f(x) increases monotonically; When the nano number of the leading rocks is less than zero, the function f(x) decreases monotonically.

Now let's look at the sign of the derivative f'(x):

When 2x + 2 > 0, i.e., x > 1, the derivative is greater than zero, and the function f(x) increases monotonically in the interval of x > 1.

When 2x + 2 < 0, i.e., x < 1, the derivative is less than zero, and the function f(x) decreases monotonically over the interval of x < 1.

To sum up, the number f(x) =x +2x + 1 increases monotonically in the interval of x > 1 and decreases monotonically in the interval of x < 1.

Here's how to do it, please make the first to check the ginseng test:

If it helps, the reeds are destroyed.

The quadratic function chain ante f(x)=x +2x+1=(x+1) 2, the quiet sail opens upward, and the symmetry axis of the shed is buried x=-1

f(x) decreases monotonically at (-1) and increases monotonically at (-1,+).

f(x)=x²-4x+3

x²-4x+4-1

x-2)²-1

The axis of symmetry of the function is x=2, and the opening of the image is on the large code of the light, and the number of functions is monotonically decreasing in (- 2).

First, the domain of the balance solution f(x) is defined as x is not equal to 0

f‘(x)=1-a/x^2

Let f'(x)=0 give x= a or - a

When x<-a, f'(x) >0 and f(x) increases monotonically.

When -a a is done blindly, f'(x)>0, f(x) is monotonically increasing.

Let's talk about the situation of x>0 first:

f(x)=x+a/x

Order 00 so when 00

Therefore, (x1x2-a)(x1-x2)<0

So when x a, f(x) increases monotonically.

When x < 0, because f(-x).

x-a x=-f(x), the function is an odd function, and the image is symmetrical with respect to the origin.

So when- a

At x<0, f(x) is monotonically decreasing, and when x - a, f(x) is monotonically increasing.

This is equivalent to the derivative of f(x)*g(x), which is f"(x)=lnx+x*(1/x)=lnx+1

When f"(x) > 0, its function is monotonically increasing, i.e., lnx>-1, so x increases monotonically on (1 e, + infinity).

When f"When (x) < 0, its function is monotonically decreasing, i.e., lnx>-1, so x is monotonically decreasing at (0,1 e).

In the problem, x is greater than or equal to one, so it is a monotonically increasing function.

First, the derivative of this function is lnx+1, when x>=1 is evergrande to 0, so the function increases monotonically.

Derivation! f(x)=xlnx

x>=1

f'(x)=lnx+x(1/x)

lnx+1 and then draw the image of lnx+1 shows that when x>=1, the function is greater than 0, so the monotonicity of the original function in the interval is monotonically increasing! Thank you!

For x1, x2 0Consider making x1 less than x2

f（x1）-f（x2）=4/x1

4/x24（x2-x1）/xix2

Because the yard branch hall x1x2 has.

f（x1）＞f（x2）

Defined by lazy monotonicity.

Know. f(x) burns monotonically decreasing at x>0.

Solution: f(x)=ax (x-1).

The domain is defined as.

Negative infinity, 1).

1, positive infinity).

f(x)=ax/(x-1)=a(x-1+1)/(x-1)=a+a/(x-1)

f(x) is in the interval.

Negative infinity, 1).

Monotonic subtraction, f(x).

In the intervals. 1, positive infinity).

Monotonous minus.

Personally, I think it's a monotonically decreasing function.

f(x)=ax (x-1)=a+a(x-1)1 (x-1) is a subtractive function on r.

And a>0

a (x-1) is also minus.

So f(x) is a subtraction function.