It is proved that the function f (x) = 2 / 2 of X is an increasing function on (- infinity, 0)

It is proved that the function f (x) = 2 / 2 of X is an increasing function on (- infinity, 0)

It is proved that & # 402; (x) is an increasing function, that is, on the premise of x1 ＜ X2, it is proved that & # 402; (x1) - & # 402; (x2) ＜ 0, i.e., & # 402; (x1) ＜ & # 402; (x2) it is proved that if X1 and X2 are any two real numbers on (- infinity, 0) and have x1 ＜ X2, then there is & # 402; (x1) - & # 402; (x2) = (2 / X1 & # 178;) - (2 /

It is proved that the function f (x) = 3 ^ x-x & sup2; has and has only one zero point in the interval [- 1,0]

If this function has zeros, then f (x) = 3 ^ X and f (x) = x ^ 2 have and only have one intersection on [- 1,0]
The range of F (x) = 3 ^ x on [- 1,0] is [one third, 1], and the function increases monotonically; the range of F (x) = x ^ 2 on [- 1,0] is [0,1], and the function decreases monotonically
So this function has only one zero point in the interval [- 1,0]

If the only zero point of function f (x) is in the interval (0, 16), (0, 8), (0, 4), (0, 2), then the correct one in the following proposition is ()
A. The function f (x) has no zero in the interval (0,1). B. the function f (x) has zero in the interval (0,1) or (1,2). C. the function f (x) has zero in the interval (1,16). D. the function f (x) has no zero in the interval (2,16)

It can be determined that the only zero point of F (x) is in the interval (0,2), so there is no zero point in the interval [2,16]. D is correct, a cannot be determined, the zero point in B may be 1, and C cannot be determined. So D is selected

Let f (x) = 3x & # 178; - 1, then the value of F (a) - f (- a) is

f（-a）=3(-X)^2-1=3X^2-1
f（a）-f（-a）=3x²-1-(3X^2-1)
=3X^2-1-3X^2+1
=0
If you don't understand, you are welcome to ask,