Knowing the function f x ln 2 1 x x 2 1 x, find the monotonic interval of the function f x

The domain of the function f(x) is defined as (-1,+.)

Let g(x)=2(1+x)ln(1+x)-x2-2x, then g'（x）=2ln（1+x）-2x．

Let h(x)=2ln(1+x)-2x, then.

When -1 x 0, h'(x) 0, h(x) is an increasing function on (-1,0), and when x 0, h'(x) 0, h(x) is a subtraction function on (0,+).

So h(x) achieves a maximum at x=0, and h(0)=0, so g'(x) 0(x≠0), the function g(x) is a subtraction function on (-1,+).

So when -1 x 0, g(x) g(0)=0, and when x 0, g(x) g(0)=0

So, when -1 x 0, f'(x) 0, f(x) is an increment function on (-1, 0).

When x 0, f'(x) 0, f(x) is a subtraction function on (0,+

Therefore, the monotonically increasing interval of the function f(x) is (-1,0) and the monotonically decreasing interval is (0,+).

First, find the derivative function, so that the derivative function is equal to zero to obtain 1+x=e x, and this formula has no solution.

Is there a problem with the way you write this question? What does ln 2(1+x) mean? There's no such way to write it.

Correction in contact.

First find the derivative, f'(x)=a x-x=(a-x 2) x, discuss the positive and negative values of the number of the sensitive pin pilot bridge and the denominator is the denominator, because it has always been greater than 0, and then classify and discuss a When a 0, then the derivative has been less than 0, then f(x) is monotonically decreasing at (0, positive infinity) When a > 0, then the derivative is a standby at the root number a and - the root number a, and a-x....

Because f(x)=1 2x 2+lnx

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

So, x=0 is a non-derivative of f(x).

There is no point where f(x)=0.

And because, on the interval (- 0), f'(x)

f'(x)=(2x-2x(1+2lnx)) x =-4lnx x let f'(x)=0 gives x=1

When x-book Huai disturbs (0,1), f'(x)>0

When x (1,+, f'State(x)<0

Therefore, the monotonically increasing permeability interval of f(x) is (0,1) and the monotonically decreasing interval is [1,+

Increment then f'(x) > perturbation 0

So ln(x+1)> slow closure 0

ln(x+1)>ln1

So x+1>1

The same is true for x>0, and the decreasing ln(x+1) >0

x0x>-1

In Zongling Town, the increasing range is (0,+

The decreasing interval is (-1,0).

f ' (x)=1-[aln(x+1)+a]=1+a-aln(x+1)＞0

Get aln(x+1) 1+a

If a 0, then ln(x+1) 1+1 a gives x e (1+1 a)-1 so (negative infinity, e (1+1 a)-1 ) is a monotonically decreasing interval.

Then (e (1 + 1 a)-1, positive infinity) is a monotonic increasing interval, if a=0, f(x)=x, then r is a monotonic increasing interval, if a 0, then ln(x+1) 1+1 a gets x e (1+1 a)-1, so (e (1+1 a)-1, positive infinity) is a monotonically decreasing interval.

then (negative infinity, e (1+1 a)-1) is a monotonic increasing interval.

Correct solution! If you have not learned to find a derivative, you can consider setting x1> x2, and use f(x1)-f(x2) to write the formula to determine the positive and negative.

The first time I received a request for help... But the answer is not timely and not powerful, forgive me.

Hello answer, glad to answer your question, 0, again 2>0, so 2 (1-x)>0 f'The (x)>0 function increases monotonically over (-1,1) with a monotonically increasing interval of (-1,1).

Solution: f(x)=[1+ln(x+1)] x

Find the definition domain first:

From ln(x+1) we get x+1 and 0 get x -1

x is the denominator, so it is not equal to 0

Define the domain as x -1 and x ≠0

Derivation, got. f'(x)=-1 x +[x (x+1)-ln(x+1)] x =[-1+x (x+1)-ln(x+1)] x =-[1 (x+1)+ln(x+1)] x (1) when x 0:

1 (x+1) 0, ln(x+1) 0, x 0, so that -[1 (x+1)+ln(x+1)] x 0 i.e., f'(x) 0 at x 0

So f(x) decreases monotonically on (0,+).

2) When -1-ln(x+1).

Thus -[1 (x+1)+ln(x+1)] x 0 i.e., -1 so that f(x) decreases monotonically on (-1,0).

In summary, f(x) is monotonically decreasing on (-1,0)u(0,+).

Solution: From the problem, we can know that the domain x>-1 is defined

f'(x)=1/(1+x)-1+kx

1+x)f'(x)=1-1-x+kx+kx 2 1+x>0 to f'(x) Positive and negative have no effect.

1+x)f'(x)=(kx+k-1)*x

If k=0, then when x (-1,0) increases x [0, dross + infinity) decreases.

If k=1, then x (-1, +infinite) will increase.

x1=（k-1）/k x2=0

If k (0, trembling Liang Kai1) then f(x) increases at (-1,0), (k-1) k], +infinity).

at (0,[-k-1) k]).

If k(1,+infinity) 'When k tends to infinity, -(k-1) k tends to -1, then f(x) increases at (-1,[-k-1) k]],0,+infinite).

At ([-k-1) k],0) minus.

It's been a long time since I've done the topic, and the draft is all typed on it, it's a bit messy, you can take a look.