Urgent! Judge the parity of the function, judge the parity of the function y sinx x?

Even function: in the defined domain f(x)=f(-x).

Odd function: in the defined domain f(x)=-f(-x)Subtract function: in the defined domain a>0 f(x+a)Periodic function: In the defined domain f(x)=f(x+a) The minimum value of a is called the period of the function.

Now it's y=|x|Obviously any of the |x|=|-x|i.e. even function image is.

a Even function f(x) = f(-x), |x|=|-x|A correct.

b When x < 0, y=-x is a subtraction function, i.e., the larger x is and the smaller y is within and outside this range.

c is the same as b, x > 0, y=x is the increasing function.

The d periodic function satisfies f(x)=f(x+t), while y=|x|Monotonically decreasing at x<0 and increasing monotonically at x>0 is obviously not a periodic function.

Generally, the sine and cosine function is a periodic function, when x1>x2 f(x1)> f(x2) is an increasing function, if f(x1) where we let y=f(x).

I scored 112 points in the Chinese language entrance examination (haha, full score of 120) to judge the parity of the letter, I need to judge f(x) f(-x)x > 0, f(x) = x, f(-x) = x = f(x)x<0, f(x) = -x, f(-x) = (-x) = f(x).

So y=|x|(x≠0) is a even function.

If not, it is not a parity function, and there is no need to calculate it further.

2. If satisfied, then find f(-x), equal to f(x) is even, and equal to -f(x) is odd.

Title: Defining the domain: Symmetry with respect to the origin, then find f(-x)=|-x|=|x|=f(x), so it is an even function.

Hello. Because y=|x|

So f(-x)=|-x|=|x|=f(x)

So the function is even.

How to judge the parity of a function.

If you want to judge the parity of the function y=sin(x)+x3, you can do this:

First, the function y=sin(x)+x3 is a polynomial function, and its parity depends on whether the coefficient of the highest power is odd or not. In this example, the coefficient of the highest power is 1, so the function y=sin(x)+x3 is the odd function.

You can use the following ** to determine the parity of the function y=sin(x)+x3:

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

The function y sinx x 3 is an odd function.

Solution: The domain of the function y sinx x 3 is (- is noisy) then in x r any x, then -x r, gets.

f（-x）＝sin（-x）＋（x）^3

sinx-x^3

f(-x) f(x) sinx x 3 (-sinx-x 3) 0, i.e. f(-x) bumper -f(lift scrambled x).

y sinx x 3 is an odd function in x r.

f( x) = sin ( -x) + (-x) travel training.

sinx-x³

f(x)=-f(-x), i.e. f(x) is an odd function.

f(x)=sinx＋x³

f(-x)=sin(-x) leak Sakura Taka(-x) =sinx-x =-sinx x )

f(-x)=-f(x)

f(x)=sinx x is the number of strange letters returned.

In general, for the function f(x) (1) If there is f(-x)=-f(x) for any x in the function definition domain, then the function f(x) is called an odd function.

2) If there is f(-x)=f(x) for any x in the function definition field, then the function f(x) is called an even function. (3) If f(-x)=-f(x) and f(-x)=f(x) are true at the same time as any x in the function definition domain, then the function f(x) is both an odd and even function, and is called both an odd and even function.

4) If f(-x)=-f(x) and f(-x)=f(x) cannot be true for any x in the function definition domain, then the function f(x) is neither odd nor even, and is called a non-odd and non-even function.

Note: Odd and even functions are integral properties, and the definition domains of odd and even functions must be symmetrical with respect to the origin for the entire defined domain.

If the domain of a function is not symmetric with respect to the origin, then the function must not be an odd (or even) function. (Analysis: To judge the parity of a function, first test whether the definition domain is symmetrical with respect to the origin, and then simplify and sort it out in strict accordance with the definition of parity and evenness, and then compare it with f(x) to draw conclusions) The basis for judging or proving whether a function has parity is the definition of the function.

Parity function image.

Characteristics: The image of the theorem odd function is a centrally symmetrical graph with respect to the origin.

The image of the even function is axisymmetric with respect to the y-axis. Let f(x) be an odd function equivalent to f(x) in an image symmetrical with respect to the origin, then the points (x,y) (x,-y) are monotonically decreasing in its symmetrical interval because the even function increases monotonically in a certain interval. A singular function that increases monotonically over an interval is also monotonically increasing on its symmetrical interval.

P.S. It should be noted that the domain of the parity function must be symmetrical, for example, the interval is (-2,2). But functions are not necessarily symmetrical.

1) The answer given is incorrect! should be discussed.

When a=0, even; When a is non-zero, it is not odd or even. Methods such as the second floor.

2) Segmented Discussion:

When x>=a, f(x)=x +x-a+1=(x+, because -1 2 a, f(x) is incremented.

So, the minimum value of f(x) = f(a) = a +1;

When x<=a, f(x)=x -x+a+1=(, because a 1 2, f(x) decreases, so the minimum value of f(x) = f(a) = a +1;

So, the minimum value of f(x) = a +1

Solution: Substitute -x into f(x) to see if it is equal to f(x), or if it is the opposite of each other, if it is equal it is an even function, if it is the opposite of each other, it is an odd function. If it is neither, it is a non-odd and non-even function. From this, this function is a non-odd and non-even function.

Problem 1: Let x=1 and substitute f(x)=x +|x-a|The +1 result is: Because f(x) is not equal to f(-x) nor is it equal to -f(-x). So non-odd non-even functions.

Question 2: Drawing comes out as soon as you draw it, and drawing is the easiest.

f(x)≠f(-x)≠—f(x) are non-odd and non-even functions with quadratic function images.

Let x>0, then -x<0

f(-x)=(-x)^2-x=x^2-x=-(-x^2+x)=-f(x)

So f(x) is an odd function.

f(-x) = [1-(-x) 2]= (1-x 2)=f(x) so f(x) is an even function.

1. f(x)=5-2x f(-x)=5+2x, f(-x) is neither equal to f(x) nor equal to -f(x), so it is not odd or even. This question can also be decided, and it will be judged soon!

2、.f(x)=x2 + 5f(-x)=x+5 squared, f(-x) equals f(x), so even function!

The root number x is not odd or even because the domain is defined as x 0 and is not symmetric on the y axis.

f(-x)=1 x and f(-x) equals -f(x), so odd function!

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f(-x)=5+2x≠f(x);≠ f(x) so it is a non-odd and non-even function.

+5f(-x)=x +5=f(x);

So it's an even function;

root number xf(-x) = -x;

When domains are defined differently; f(x)=f(-x);

So it's an even function;

f(-x)=-1/(-x)=1/x=-f(x);

So it's an odd function.

Hello, I'm glad to answer for you, skyhunter002 for you to answer your questions, if you don't understand anything about this question, you can ask, if you are satisfied, remember to adopt if there are other questions, click on me to ask for help after this question, it's not easy to answer the question, please understand, thank you.

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This is not difficult, first use the characteristics of the curious even function, and then you can write it.