A question about the function f x xsinx

Derivatives of functions.

for f'(x) = sinx + xcosx let f'(x) = 0 gives x = -tanx, so x has only one solution on [- 2, 2], x = 0, so (1) is wrong.

2) Let x=(n+1 2) n be a positive integer, then f(x) = (n+1 2) is obviously infinity.

So (2) wrong.

3) f(x) is obviously f(x) >0 on (0, ), but because it is an open range.

So there is no minimum.

And on (0, ) f'(x) = 0 has a unique solution x where f(x) takes the maximum value.

4) The center of symmetry of f(x) is (0,0), but f(x) is not periodic, so (,0) is not a center of symmetry of the function f(x) image.

f(x) is an image of .

1 False, this function is an even function, and it cannot be monotonic in this symmetry interval; 2 False, when x tends to positive infinity, f(x) has no upper limit, and it is impossible for m to satisfy; 3 pairs, f(0)=f(pie)=0, and x on (0, pie) f(x)>0, think about its image, f(x) has no minimum value, there is a maximum value; 4 is false, if (pie, 0) is the center of symmetry, then f(x) = -f(2 faction-x), xsinx is unequal - (2 faction -x) sin(2 faction -x), so it is false.

1. After finding the derivative, we can know that the derivative of f(x) in this region is not as high as 02, which is obviously wrong because f(x) does not converge.

3. Derivative, f(x) derivative is greater than 0 at (0, 2) and less than 0 at x= There must be zero in the middle, so there must be a maximum value and no minimum value.

4. This is an even function.

So pick 3

After studying the function f(x)=xsinx, a student came to the following conclusions:

The function f(x) is monotonically incremented on [- 2, 2].

There is a constant m 0 so that f(x) m is true for a real number x.

The function f(x) has no minimum value on (0, ), but there must be a maximum value.

The point ( ,0) is a symmetry center of the function f(x) image.

The ordinal number of the correct proposition is ......

Inverse functions. Let arctanx=y, then x=tany and other stoves such as Na shout both sides of x derive, implicit function derivative, implicit opening then 1 y'(tany)'＝y'sec^2y

So y'=1/sec^2y

Since tan 2+1 = sec 2

So y'=1/(1+tan^2y)

It says x=tany

So y'=1/1+x^2

Since the function f(x)=x-sinx, e.g. and f(x)=1-cosx

Therefore, the answer is: 1-cosx Qi withered.

Summary. Yes.

The known function f(x)=(xsin(1-x)) x|(x 2-1 has **.) There are 2 breaks to go to this problem and 1 jump break.

It is obtained by drawing.

I'm guessing that's the answer you want.

Wrong. I'll take another look.

Can you take a look at these two?

Received ha. Yes.

16 questions B, 18 B

Why is this wrong?

This option D

Summary. Hello, dear, the known function f(x)=x sinx, then f(x) sinx+xcosx, then f(0) 0 oh pro [eat whales].

If you know the function f(x) = x sinx, find f(x) and f(0).

Hello, dear, the known function f(x)=x sinx, then f(x) sinx+xcosx, then f(0) 0 oh pro [eat whales].

The title is the derivative of the product function Oh, f=ym, then f=y'm+ym'oh pro [eat whales].

There are a few more. Dear, we can only ask one question at a time, if you still need to ask questions, you need to upgrade the service, oh pro [eat whales].

1c, 2c, 3c, 4b, 5b, 6b, 7b, oh pro [eat whales].

Thank you (*°3.)

Summary. f(0)=0f(1)=1+sin(1) set the function f(x)=x+sinx, then f()=f(1 Hello dear, can you send me the complete question.

f(0)=0f(1)=1+sin(1)

f'(x)=1+cosx

∫f(x)f'(x)dx is equal to 1 2*(cosx) 2 (1+sinx) 4+c.

Solution: Since one of the original functions of f(x) is sinx (sinx+1), then f(x) = (sinx (sinx+1)).'=cosx/(1+sinx)^2。

and f(x)f'(x)dx

f(x)df(x)

1/2*(f(x))^2+c

1/2*(cosx/(1+sinx)^2)^2+c=1/2*(cosx)^2/(1+sinx)^4+c

It should be 1 2*[f(x)] 2+c, the original function of f(x) is known, and the derivative of it can be used to know f(x).