It is known that the functions f x 2 are to the power of x 1 and 2 to the power of x

Solution: (1) Because f(x)=xf(x), f(-x)=-xf(-x), so f(x) is not equal to f(-x), so the function is not an even function; f(-x)=-xf(-x), and -f(-x)=xf(-x), so f(-x) is not equal to -f(-x), so the function is not odd either. In summary, this function is not odd or even.

2) No, when x=0, f(x) must be equal to 0, why is f(x) greater than 0?

f(x)=2^x+(1/2)^x-1，(1/2)^x=2^(-1)*x=2^(-x)

f(x)=x*f(x)=x(2 x+2 -x-1)f(-x) = -x(2 -x+2 x-1)f(x) = -x(2 x+2 -x-1) so f(-x) = -f(x), so the function is odd.

Since it is an odd function, f(0)=0, and this function is symmetric with respect to the origin.

If you let x>0, then f(x)>0 is true, but x<0, obviously f(x)<0 Okay.

In the end, I didn't let you be satisfied with it, and if you were satisfied, you would give it a dear!

Because x belongs to r and f(-x)=-f(x) (let x 0).

So it's an odd function.

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It is known that the function f(x)=(1 2)x to the power of the function f(x to the power of -1) Answer: f(-x)=a-1 2 (-x)+1=a-1 2 x+12 (-x)=2 x So x=0 odd function f(0)=0a-1+1=0 a=02 x is the increase function 1 2 x is the subtraction function -1 2 x is the increase function, so f(min)=f(-infinity)=-infinity f(max)=f(+infinity)=a+1 The range is (-infinity, a+1) Extended information: The domain of the function is the set of real numbers r, which is defined by the odd function, and when the odd function is meaningful at x=0, f(0)=0 is constant.

Therefore f(0) = a-1 2 = 0. Therefore a=1 2

It is known that the function f(x) = (1 2) x power is found to find the function f(x power -1) good afternoon <>

It is known that the function f(x)=(1 2)x to the power of the function f(x to the power of the shed-1) Answer: f(-x)=a-1 2 (-x)+1=a-1 2 x+12 (-x)=2 x So x=0 odd function f(0)=0a-1+1=0 a=02 x is the increase function 1 2 x is the subtraction function -1 2 x is the increase function, then f(min)=f(-no land fiber poor) =-infinity f(max)=f(+infinity)=a+1 The value range is (-infinity, a+1) Extended information: The domain of the function is the set of real numbers r, which is defined by the odd function, and when the odd function is meaningful at x=0, f(0)=0 is constant.

Therefore f(0) = a-1 2 = 0. Therefore a=1 2

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f(x+1)=x^2-1

Let t=x+1

Then x=t-1

So f(t)=(t-1) 2-1=t 2-2t, so chain base f(x)=x 2-2x

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(1)f(x)= [2^x-2^(-x)]/[2^x+2^(-x)]，62616964757a686964616fe59b9ee7ad9431333330323339

For any x r, 2 x>0 ,2 (-x)>0,2 x+2 (-x)>0, the domain of the function is r

Let f(x)=y, then y= [2 x-2 (-x)] [2 x+2 (-x)][2 x+2 (-x)]y=[2 x-2 (-x)][2 (2x)+1]y=[2 (2x)-1]2 (2x)=(1+y) (1-y).

When x r, 2 (2x) 0 , 1+y) (1-y)>0, the solution is -10

f(x1)-f(x2)<0, i.e. f(x1).< f(x2) f(x) is a monotonically increasing function over the defined domain.

f(-x)=(1-1 2 x) (1+1 2 x) to become (1 2 x-1) (1 2 x+1)=-f(x);

That is, f(-x) = -f(x), odd functions.

1-1 2 x is always less than 0, 1+1 2 x is always greater than 0, so the value of the hole is (- 0).

2^x-2^(-x) x≥0

f(x)=0 x<0

f（x）=2, 2^x-2^(-x)=2

So 2 x=( 3+1) 2 x=log2( 3+1) -12 t*[2 2t-2 (-2t)]+mt 0 so 2 t=u,u [2,4] t=log2 u then there is u*(u -1 u)+mlog2 u=u -1 u+mlog2 u 0

mlog2 u 1 u-u , m (1 u-u) log2 u easy to verify function g(u) = (1 u-u) log2 u is the subtraction function.

g(u)max=g(2)=1 2-8=-15 2, so m g(u)max=-15 2

The value range of m is [-15 2,+

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

Simplify the big card f(x) + f(-x) = -1 is not equal to 0, so when a=2, f(x) is not an odd function.

2) Suppose m>n, f(m)-f(n)=(2 m-a) (2 m+1)+[2 n-a] [2 n+1].

f(m)-f(n)=(a+1)(2 m-2 n) [2 m+1)(2 n+1)]

Because a>-1, a+1>0, m>n, (2 m-2 n) >0

And [(2 m+1)(2 n+1)]>0

So f(m)-f(n)>0, this function is a monotonic increasing function.

3) If this function is odd, then f(x)+f(-x)=0, i.e., (2 x-a) (2 x+1)+[2 (-x)-a] [2 (-x)+1]=0

Simplification yields 1-a=0, so a=1

f(x) x 2-4x+m is held at [-2,2] constantly, i.e., (2 x-1) (2 x+1) x 2-4x+m

i.e. (x 2-4x+m-1)(2 x+1) 0

Since (2 x+1) Evergrande is above 0, for the above inequality to hold, then (x 2-4x+m-1) 0

i.e. (x-2) 2+m-5 0, i.e. (5-m) (x-2) 2

Because x [-2,2], the maximum value of (x-2)2 is 4 (at x=0).

So 5-m 4m 1