Parity of functions "beginner and intermediate counseling"

First, the function of the parity may also be given in what form?

1, give directly;

The function \ (f (x) \) is an odd function on a range \ (d\) .

2, the definition of the given;

such as \ (\forall x \in d,f (×) =-f (x) \), then it is an odd function. such as functions \ (f (x) =x^3\),

3, the definition of the variable form given;

such as \ (\forall x \in d,f (x) + f (×) =0\),\ (\cfrac{f (-X)}{f (x)}=\pm 1 (f (x) \neq0) \).

"Apply ①"

such as functions \ (f (x) =ln (\sqrt{x^2+1}-x) \),

(f (-X) =ln (\sqrt{x^2+1}+x) \),

i.e. \ (f (x) +f (-X) =ln1=0\), i.e. function \ (f (x) \) is an odd function;

What about the function \ (f (x) =ln (\sqrt{x^2+1}-x) +1\) ?

The same can be done,\ (f (x) +f (×) =2\),

That is, the function \ (f (x) \) is symmetric about point \ ((0,1) \) .

"Apply ②"

such as functions \ (g (x) =lg (\sqrt{sin^2x+1}+sinx) \),

It is known that \ (g (-X) =lg (\sqrt{sin^2x+1}-sinx) \),

(g (x) +g (-X) =lg1=0\), that is, the function \ (g (x) \) is an odd function;

4, given in the form of images;

For example, a function image about the origin of symmetry, a function image about \ (y\) axisymmetric.

5, with the characteristics of the parity of the application of the form of the conclusion given;

In the public domain, the conclusion is established and can be proved.

\ ("Odd + odd" \) is odd,

such as \ (f (x) =x+sinx\);\ (g (x) =x^3+2sinx\);\ (h (x) =x+\cfrac{1}{x}\);\ (h (x) =2x+\cfrac{3} {x}\);

\ ("odd-odd" \) is odd,

such as \ (f (x) =x^3-sinx\);\ (h (x) =x-\cfrac{2}{x}\);

\ ("Odd \cdot qi" \) is even,

such as \ (f (x) =x\cdot sinx\);\ (f (x) =x^3sinx\);

\ ("Odd ÷ odd" \) is even;

such as \ (f (x) =\cfrac{sinx}{x}\);

\ ("even + even" \) is even,\ ("Even-even" \) is even;
\ ("Even \cdot even" \) is even,\ ("Even ÷ even" \) is even;
\ ("Odd \cdot even" \) is odd,\ ("Odd ÷ even" \) is odd;

If \ (f (x) \) is even function, the function \ (g (x) =xf (x) \) is the odd function.

• Special case, originally did not have the parity function, carries on the arithmetic, also has the parity sex.

such as \ (f (x) =e^x+\cfrac{1}{e^x}=e^x+e^{-x}\), even function;

such as \ (f (x) =e^x-\cfrac{1}{e^x}=e^x-e^{-x}\), odd function;

Add a simple proof model,
Known $f (x), G (x) $ are all odd functions, proving that $h (x) =f (x) +g (x) $ is an odd function;
Analysis: $f (x), G (x) $ are all odd functions, then the definition field of the function $h (x) $ must be symmetric about the origin.
and satisfies $f (-X) =-f (x) $, $g (-X) =-g (x) $,
Then $h (-X) =f (-X) +g (-X) =-[f (×) +g (x)]=-h (x) $,
So $h (x) $ is an odd function.

6, based on the image transformation to give,

If the symmetric axis of \ ( f (x-1) \) is \ (x=1\), then the symmetric axis of \ (f (x) \) is the \ (y\) axis, i.e. \ (f (x) \) is even function;

7. The combination of periodicity and symmetry to the surprisingly even;

The known function \ (f (x) \) has a period of 2 and satisfies \ (f (2+x) =f (x) \), which is inferred function \ (f (x) \) is even function.

The specific variants are as follows:

by \ (f (x+2) =f (x) \) and \ (f (2+x) =f (-X) \),

Get \ (f (-X) =f (x) \),

So the function \ (f (x) \) is even function.

8, with the combination of assignment method given;

Known function \ (f (x) \) satisfies \ (f (1) =\cfrac{1}{2}\), and \ (f (x+y) +f (x-y) =2f (×) f (y) \), it is inferred function \ (f (x) \) for even function.

The specific variants are as follows:

Order \ (x=y=0\), then there \ (2f (0) =2f^2 (0) \), get \ (f (0) =0\) or \ (f (0) =1\);

Re-order \ (x=1,y=0\), there is \ (2f (1) =2f (1) f (0) \), get \ (f (0) =1\);

Also the topic known \ (f (1) =\cfrac{1}{2}\), obtained \ (f (0) =1\)[\ (f (0) =0\) ;

Re-order \ (x=0\), you get \ (f (Y) +f (-y) =2f (0) F (y) =2f (y) \),

So \ (f (Y) =f (y) \), the function is known as even function.

9, in the form of a constructor to give or get the odd-even nature "difficult";

For specific examples, see: Example 2

Second, the common conclusion of the parity of the function:

1, two functions \ (y=ax^2+bx+c (a\neq0) \) are even functions of the necessary and sufficient condition is \ (b=0\) ,

Proof: The axis of symmetry is \ (x=-\cfrac{b}{2a}\),

Adequacy: by \ (b=0\), get the axis of symmetry ( x=0\), that is, the \ (y\) axes.

Necessity: By the function for even function, the symmetric axis is \ (x=-\cfrac{b}{2a}\), gets \ (b=0\).

The following conclusions are derived from this generalization:

2, the polynomial function \ (y=f (x) =ax^4+bx^3+cx^2+dx+e\) is an odd function of the necessary and sufficient conditions is \ (a=c=e=0\)

Description: \ (f (-X) +f (x) =0\) constant set up,

That is \ ([A (-X) ^4+b ( -X) ^3+c (-X) ^2+d (-X) +e]+ (ax^4+bx^3+cx^2+dx+e) \)
\ (=2ax^4+2cx^2+2e=0\),

That \ (ax^4+cx^2+e=0\) to \ (\forall x\in r\) are established, so \ (a=c=e=0\).

For example, the known function \ (f (x) =x^3+ (A-1) x^2+ax\) is an odd function, then \ (a=1\);

3, polynomial function \ (y=ax^4+bx^3+cx^2+dx+e\) is even function of the necessary and sufficient conditions is \ (b=d=0\)

The example above can be explained.

4, the derivative function of the odd function is even function;

Liberal Arts: For example, the function \ (f (x) =sinx\) is an odd function whose derivative function is \ (f ' (x) =cosx\) is even function;

Science: logical proof, set function \ (f (x) \) is an odd function, its derivative function \ (f ' (x) \),

Remember \ (f ' (x) =g (x) \), the function \ (f (x) \) is the odd function,

Then there is a \ (f (-X) +f (x) =0\), which is derivative of the two sides,

\ (-F ' (-X) +f ' (×) =0\), i.e. \ (-G (-X) +g (×) =0\),

That is \ (g (-X) =g (x) \),

That is , the function \ (g (x) \) is even function;

5, the derivative function of the even function is an odd function;

Liberal Arts: For example, the function \ (f (x) =cosx\) is even function, whose derivative function is \ (f ' (x) =-sinx\) is an odd function;

Science: logic proof, set function \ (f (x) \) is even function, its guide function is \ (f ' (x) \),

Memory \ (f ' (x) =g (x) \), the function \ (f (x) \) is even function,

Then there is \ (f (-X)-F (x) =0\), which is derivative of the two sides,

\ (-f ' (-X)- F ' (x) =0\), which is \ (-g (-)-g (x) =0\),

That is \ (g (-X) =-g (x) \),

That is , the function \ (g (x) \) is an odd function;

6. If the function is an odd function, the value of the Guide function at the two point at which the origin is symmetric is equal.

Liberal arts: such as \ (f (x) =x^3\), then \ (f ' (x) =3x^2\), so \ (f ' ( -1) =f ' (1) =3\);

Science: if function \ (f (x) \) satisfies \ (f (x) +f (x) =0\), the derivative of the two sides is obtained,

\ (-F ' (-X) +f ' (×) =0\), then there \ (f ' (X_0)-F ' (-X_0) =0\)

Extension: symmetry, if function \ (f (x) +f (2-x) =2\), then function \ (f (x) \) about point \ (( ) \) symmetry,

and \ (f ' (x_0) =f ' (2-x_0) \), such as \ (f ' (0) =f ' (2) \);

7, if the function is even function, then the value of the Guide function at the two point of symmetry of the origin is opposite to each other.

Liberal arts: such as \ (f (x) =x^2\), then \ (f ' (x) =2x\), so \ (f ' ( -1) =-2,f ' (1) =2\);

Science: if function \ (f (x) \) satisfies \ (f (-X)-F (x) =0\), the derivative of the two sides is obtained,

\ (-F ' (-X)-F ' (x) =0\), then there \ (f ' (x_0) +f ' (-x_0) =0\)

Extension: symmetry, if the function \ (f (x) =f (2-x) \), then the function \ (f (x) \) about the line \ (x=1\) symmetry,

and \ (f ' (x_0) =-f ' (2-x_0) \), such as \ (f ' (0) =-f ' (2) \);

Parity of functions "beginner and intermediate counseling"