Ask for f x detailed file, thank you

Solution: The piecewise function calculates f(-x) and replaces all x's in it with -x. That's the trick f(-x)=(-x) 2=x 2 -x《=0 x>=0f(-x)=(-x) 2+(-x)=x 2-x -x>0 x<0 so x 2 x>=0

f(-x)= x^2-x x<0

First of all, this is a composite function derivative problem.

The function in the problem can be seen as the composite of the following two functions: f(u)=u 2, u=f(x).

Derivation using composite functions The derivative is 2u*u'That's 2f(x)f'(x)

f](sinx)=cos²x

Let t=sinx t [-1,1].

cos²x=1-sin²x=1-t²

i.e. f'(t)=1-t²

f'(x)=1-x² x∈[-1,1]

f(x)=∫f'(x) dx=∫ 1-x² dx=x-x³/3+c x∈[-1,1]

I'm glad to answer your questions and wish you progress in your studies! The Math Fun team will answer the questions for you.

If you don't understand, you can ask! If you approve of mine.

Please click the [Select as satisfied] button below, thank you!

f(x) 1 (x+1), so.

Integral ln(x+1) of f(x).

The value of 0-1 is ln2-ln1 ln2

It's not enough to find a derivative by interval. Will it be a derivative?

Let f(x) = f(x)-f(-x).

f(-x)=f(-x)-f(x)

[f(x)-f(-x)]

f(x) under the condition that the domain is symmetrical with respect to the origin.

y=f(x) f(-x) is an odd function.

If f(-x)=-f(x), then the function.

baif(x) is a strange letter.

Number. du let f(x) = f(x)-f(-x).

then f(-x)=f(-x)-f(x)=-[f(x)-f(-x)]=-f(x).

zhif(x)=f(x) f(-x) is an odd dao function.

Solution: Let g(x)=f(x)-f(-x).

Then there is: g(-x)=f(-x)-f(x).

g(x)+g(-x)=f(x)-f(-x)+f(-x)-f(x)=0, i.e.: g(x)=-g(-x).

So we get: y=f(x).

Specify f(-x) as an odd function! Genus.

Now it's about finding the parity of the function y=f(x)-f(-x), which is not easy to understand if we use y.

So let's f(x)=f(x)-f(-x).

then f(-x)=f(-x)-f[-(x)]=f(-x)-f(x).

f(x) Therefore, f(x) is a singular function.

Discussion of function parity knowledge answer notes:

1.It only makes sense to discuss parity if the interval of the independent variable of the function is symmetrical with respect to the y-axis.

2.Discussing parity is validation: f(x)=f(-x) even symmetry f(x)=-f(-x) oddsymmetry.

3.For odd-symmetric functions: if x=0 is in the defined domain, there is always f(0)=0.

So if some problem tells us that f(x) is an odd function and x=0 is in the definition domain, we should get the implicit message: f(0)=0. It may be used when solving for some parameters.

Using the relationship between infinitesimal and infinitesimal quantities, when x-->2, x-->2, (2-x 2) 4-->0 is infinitesimal and when x-->2, x-->2, the function limit f(x)=4 (2-x 2) is infinite.

The set of input values x is called the defined domain of f; The set of possible output values y is called the range of f. The domain of a function is the set of actual output values obtained by mapping f to all elements in the defined domain. Note that it is incorrect to call the corresponding domain a value range, and the value range of a function is a subset of the corresponding domain of the function.

In computer science, the data types of parameters and return values determine the defined and corresponding domains of the subroutine, respectively. Therefore, defining the domain and the corresponding domain is a mandatory constraint that is determined at the beginning of the function. On the other hand, the range is related to the actual implementation.

Functions are related to inequalities and equations (elementary functions). Let the value of the function be equal to zero, and geometrically, the value of the corresponding independent variable is the abscissa of the intersection of the image and the x-axis.

From an algebraic point of view, the corresponding independent variable is the solution of the equation. In addition, if you replace the "=" in the expression of the function (except for functions without expression) with "<" or ">" and replace the "y" with another algebraic formula, the function becomes an inequality and the range of independent variables can be found.

There are horizontal asymptottoes, vertical asymptotes and oblique asymptostotes, and it is easy to know that there can only be one type of horizontal asymptote and oblique asymptote on the same side.

1. Look at the horizontal asymptote first.

lim(x + f(x)=0, so there is a horizontal asymptote y=0 on the right

lim(x - f(x)=0, so there is a horizontal asymptote y=0 on the left

That is, f(x) has a horizontal asymptote x=0

2. Looking at the vertical asymptote (the vertical asymptote is generally the end of the break, that is, lim(x x0)f(x)= when x=x0 is the vertical asymptote).

The question f(x) has discontinuities x=-2 and x=2

lim(x -2-)f(x)=-, lim(x -2+)f(x)=+, we can see that x=-2 is the vertical asymptote of f(x).

lim(x 2-)f(x)=+ , lim(x 2+)f(x)=-, it can be seen that x=2 is also a perpendicular asymptote of f(x).

3. From 1, it can be seen that f(x) has horizontal asymptotic lines on the left and right sides, so there are no oblique asymptotic lines on the left and right sides.

In summary, f(x) has a horizontal asymptote line y=0 and a vertical asymptote line x=2 and x=-2