y x 1 x 3 function parity

Does x3 mean the cubic of x? If the odd function is f(x)=y=x+1 x3, then f(-x)=-x+1 (-x)3=-x-1 x3=-(x+1 x3) and the domain of the function is (negative infinity, 0)u(0, positive infinity) so, the odd function.

Let x=-x, substitute the formula y=-x-1 x3=-(x+1 x3)=-y,, so the function is odd.

Idea: The components of a function include a defined domain.

Analytic and ranges. Parity of the pair function.

Let f( x) = x +1

then f(-x)=(x) +1=-x +1

Because f(-x) ≠ f( x) and f(-x) ≠ f(x) y=x +1 is a non-odd and non-even function.

Odd and not even. It is arranged in an odd and even way, which is called parity. In general, 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.

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

You say y=x to the third power plus one of its parity. Singularity.

Solution: Let f(x)=y=x +1

x takes any real number, and the function expression is always meaningful.

The function defines the domain as r, symmetrically with respect to the origin.

f(-x)=(x)³+1=-x³+1

f(x)+f(-x)=x +1-x +1=2≠0, the function is not an odd function.

f(x)-f(-x)=x +1-(-x +1)=2x, which is not constant zero, and the function is not even.

Functions are non-odd and non-even.

Summary: Determining the parity of a function is a two-step process.

1. First, determine whether the definition domain is symmetrical with respect to the origin. If the defined domain is not symmetrical with respect to the origin, it is directly judged as a non-odd and non-even function.

2. On the premise of defining the symmetry of the domain with respect to the origin, we will examine f(x)+f(-x) and f(x)-f(-x) to determine whether the function is odd or even.

y=x²(3-x)

Domain. Yes (- 0], [0,2],[2, ) is not symmetrical about the origin.

So it's a non-odd and non-even function.

f（x）=3x²-x³

Although the domain is defined as r

But f(1)=3-1=2

f(-1)=3+1=4

f(1)≠f(-1）

It does not satisfy f(-x)=f(x) or f(-x)=f(x) for any x, so it is not an odd function or an even function.

Non-odd and non-even functions, defined as r, symmetric with respect to the origin, but f(-x)≠-f(x), and f(-x)≠f(x), then y=3x x is not an odd and non-even function.

Let y=f(x)=3x -x

The domain of f(x) is r

f(-x)=3· (-x) x) =3x +x ≠f(x)f(x)=-3x -x )=3x +x ≠f(-x)f(x) is not odd or even.