Help determine the parity of the function!! Fast!

Haha, the function f(x) is an odd function, let a 0, and a≠1, if g(x) = (a-1)f(x)[1 (a's x power) minus 1)+1 2], find g(x) parity.

It's yours, right? Already.

Let h(x)=1 (a x-1)+1 2, then g(x)=(a-1)f(x)h(x).

h(x)=1/(a^x-1)+1/2=(a^x+1)/[2(a^x-1)]

h(-x)=1/[a^(-x)-1]+1/2

1/[(1/a^x)-1] +1/2

1/[(1-a^x)/a^x]+1/2

a^x/(1-a^x) +1/2

a^x+1)/[2(1-a^x)]

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

g(-x)=(a-1)f(-x)h(-x)=(a-1)[-f(x)][h(x)]=(a-1)f(x)h(x)=g(x)

g(x) is an even function.

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.

The original function f(x) = y=e to the power of -x, to the power of -x, to the power of -1 of e+1=1 (to the power of e x)-e x+1

f(-x)=e^x+1/(e^x)+1

f(x)≠f(-x)≠-f(-x)

So it's not odd or even (not so sure).

Even functions. When x!=0（!= is not equal to), f(0+x) = f(0-x), symmetrical with respect to the line x=0.

When x=0, f(x)=0, on a straight line x=0.

So it's an even function.

x takes any real number, the function expression is always meaningful, and the function definition domain is (- Symmetry with respect to the origin.

f(-x)=a⁻ˣ+

f(x)+f(-x)=a + a +a>0, >0,a, a, uniformity"0,f(x)+f(-x)≠0, the function is not an odd function.

f(x)-f(-x)=a + a + f(x)-f(-x)=0 if and only if x=0, i.e. f(x)-f(-x) is not constant zero and the function is not even.

In summary, the resulting function f(x) is a non-odd and non-even function.

u=x-tu can be deformed into.

u=x/〈1+t)=f(x)

Obviously, u=f x) is an odd function—not a non-odd and non-even function as you say, f(u)=u 10 t).

uf(u)=u2 (1+t) u2 denotes the square of u, and it is clear that uf(u) is an even function about u.

and since uf(u)=x 2 1+t) 3

Obviously, uf(u) is also an even function about x.

When analyzing the parity of functions, it is necessary to clarify the independent and dependent variables, the correspondence, and the function of who is who.