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It is known that the function f (x) is an odd function defining the domain on R. when x is less than or equal to 0, f (x) = - 4x / x + 4 ① Find the analytical formula of F (x) ② If f (2m + 1) + F (m ^ 2-2m-4) > 0, find the range of M It is known that the function f (x) is an odd function on R. when x is less than or equal to 0, f (x) = - 4x / x + 4. ① find the analytical formula of F (x) ② if f (2m + 1) + F (m ^ 2-2m-4) > 0. Find M
(1) When x ≤ 0, f (x) = - 4x / (x + 4), - x ≥ 0, f (x) is an odd function, then f (- x) = - f (x) = 4x / (x + 4), f (x) = 4x / (x-4). Therefore, f (x) = {- 4x / (x + 4), X ≤ 0; 4x/(x-4),x>0.
（2）f(2m+1)+f(m ²- 2m-4)>0,f(2m+1)>-f(m ²- 2m-4)=f(-m ²+ 2m + 4). F (x) monotonically decreases at R, only 2m + 1 < - M ²+ 2m + 4, the solution is - √ 3
The function f (x) defined on R is an odd function, and when x is greater than or equal to 0, f (x) = x ^ 2 + 4x 1. Find the analytical formula of F (x) 2. If inequality f (T ^ 2-2) + F (T)
When x is less than 0, - x > 0, f (- x) = (- x) ^ 2 + 4 (- x) = x ^ 2-4x = - f (x), so f (x) = - x ^ 2 + 4x2. When x is greater than or equal to 0, f (x) = x ^ 2 + 4x is an increasing function (because f '(x) = 2x + 2 > 0); when x is less than 0, f' (x) = - 2x + 4 > 0 is also an increasing function, f (0) = 0. The whole function is continuous, so f (
The known function f (x) is an odd function defined on R. when x is greater than or equal to 0, f (x) = x (1 + x). Find the analytical formula of the function
The function f (x) is an odd function f (- x) = - f (x) defined on R
When x is greater than or equal to 0, f (x) = x (1 + x)
When x is less than or equal to 0, - x < 0, f (- x) = - x [1 + (- x)] = - f (x)
So when x is less than or equal to 0, f (x) = x [1 + (- x)] = x (1-x)
When x is greater than or equal to 0, f (x) = x (1 + x)
It is proved that the function f (x) = - 2 + 1 is a subtractive function on R Senior one compulsory 1 function exercise
It should be f (x) = - 2x + 1
Order x10
I.e. x1f (x2)
So it's a subtraction function
X = 0 is the removable breakpoint of the first type of discontinuity of sh8i function f (x) = xsin1 / x? Isn't it both left and right limits = 0? Isn't that continuous? Is it undefined when x = 0?
It's very smart. When f (x) = 0, it's continuous. Brother f (x) = 1, I neglected to be embarrassed
Why is x = 0 the second type of breakpoint of function f (x) = 1 / x? For example, when question ~ x = 0, don't both left and right limits exist?
When x approaches + 0, the function f (x) = 1 / x approaches + ∞
When x approaches - 0, the function f (x) = 1 / x approaches - ∞
The left and right limits of the function do not exist, so it is the second type of discontinuity
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