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If the domain of functions f (x) = 3 ^ x + 3 ^ - X and 9 (x) = 3x-3 ^ - x is r, then a, f (x) and 9 (x) are even functions, B, f (x) are even functions, and 9 (x) If the domain of functions f (x) = 3 ^ x + 3 ^ - X and 9 (x) = 3x-3 ^ - x is r, then A, f (x) and 9 (x) are even functions B F (x) is an even function and 9 (x) is an odd function C, f (x) and 9 (x) are odd functions D f (x) is an odd function and 9 (x) is an even function
F(-x)=3^-x+3^x=F(x)
Even function
g(-x)=3^-x-3^x=-g(x)
Odd function
Choose B
If the function f (x) is an even function defined on R, a subtractive function on (- ∞, 0], and f (2) = 0, the value range of x where f (x) ＜ 0 is () A. （-∞，2） B. （2，+∞） C. （-∞，-2）∪（2，+∞） D. （-2，2）
When x ∈ (- ∞, 0], f (x) ＜ 0, then x ∈ (- 2, 0]
And ∵ even functions are symmetric about the Y axis
The solution set of F (x) ＜ 0 is (- 2,2),
Therefore, D
Given that the function f (x) is an even function defined on R, and is a subtractive function on (negative infinity to zero), and f (2) = 0, find f (x)
Because function f (x) is an even function defined on R and a subtractive function on (negative infinity to zero)
So f (x) is an increasing function on x > 0, f (2) = f (- 2) = 0, so f (x)
If the function f (x) is an even function defined on R, an increasing function on (- ∞, 0), and f (2) = 0, the value range of x where f (x) ＜ 0 is () A. -2＜x＜2 B. x＞2 C. x＜-2 D. X ＜ - 2 or X ＞ 2
∵ (x) is an even function defined on R and an increasing function on (- ∞, 0),
‡ f (x) is a subtractive function on (0, + ∞),
And f (2) = 0, ‡ f (- 2) = f (2) = 0,
Make a sketch of F (x), as shown in the figure:
From the image, f (x) < 0 ⇔ x < - 2 or x > 2,
Therefore, D
An even function defined on R satisfies that f (x + 2) = f (x) and f (x) is a subtractive function on [- 3, - 2], if α，β Is the two internal angles of an acute triangle, then () A. f（sin α）＞ f（cos β） B. f（sin α）＜ f（cos β） C. f（sin α）＞ f（sin β） D. f（cos α）＞ f（cos β）
Since f (x + 2) = f (x), the period of the function is 2. Because f (x) is a subtractive function on [- 3, - 2], f (x) is a subtractive function on [- 1,0]. Because f (x) is an even function, f (x) is a monotonic increasing function on [0,1]. Because in an acute triangle, π- α-β＜ PI 2, so α+β...
F (x) is an odd function with period T, and defines the domain [- t, t]. If f (T) = 0, f (x) has several zeros on [- t, t] Sit and wait for 30 minutes
(1) Because f (x) is an odd function and is defined at x = 0, f (0) = 0
(2) Because t is a period, for any x on [- t, t], f (x + T) = f (x), take x = - t / 2, then f (- t / 2 + T) = f (- t / 2),
That is, f (T / 2) = f (- / 2t),
(3) F (x) is an odd function, so there is f (- t / 2) + F (T / 2) = 0. Combined with the above f (T / 2) = f (- / 2t), f (T / 2) = f (- / 2t) = 0
(4) It is known that f (T) = 0, so f (- t) = - f (T) = 0 is obtained from the odd function
To sum up, f (x) has five zeros on [- t, t], which are x = - t, - t / 2,0, t / 2, t
Do you understand?
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