Let f (x) = xsinx, then f '(2 / π)=

Let f (x) = xsinx, then f '(2 / π)=

F '(x) = SiNx + X * cosx, just bring in the value!
1. Let f (x) be a monotonically decreasing odd function defined on R. if X1 + x2 > 0, X2 + X3 > 0, X3 + X1 > 0, then f (x1) + F (x2) + F (x3) 0, X2 + X3 > 0, X3 + X1 > 0, then f (x1) + F (x2) + F (x3)
1. Because f (x) is an odd function defined on R, so f (x) = - f (- x), so f (x1) = - f (- x1), f (x2) = - f (- x2), f (x3) = - f (- x3), and f (x) is a monotonically decreasing odd function defined on R, X1 + x2 > 0, X2 + X3 > 0, X3 + X1 > 0, so f (x1) = - f (- x2), f (x3) = - f (- x3) is a monotonically decreasing odd function defined on R
Let f (x) = xsinx, f '' (2 / x)=
Let f (x) = xsinx, X ∈ [- π / 2, π / 2], if f (x1)
Let f (x) be a monotonically decreasing odd function defined on R, if X1 + x2 > 0, X2 + X3 > 0, X3 + X1 > 0, then ()
A. f（x1）+f（x2）+f（x3）＞0B. f（x1）+f（x2）+f（x3）＜0C. f（x1）+f（x2）+f（x3）=0D. f（x1）+f（x2）＞f（x3）
∵ X1 + x2 ＞ 0, X2 + x3 ＞ 0, X3 + x1 ＞ 0, ∵ x1 ＞ - X2, X2 ＞ - X3, x3 ＞ - x1, and f (x) is a monotonically decreasing odd function defined on R, ∵ f (x1) ＜ f (- x2) = - F (x2), f (x2) ＜ f (- x3) = - f (x3), f (x3) ＜ f (- x1) = - f (x1), ∵ f (x1) + F (x2) ＜ 0, f (x2) + F (x3) ＜ 0, f (x3) + F (x1) ＜ 0（ X1) + F (x2) + F (x3) < 0, so select B
Let f (x) = xsinx, then f (x) is ()
A. Odd function
B. Even function
C. Non odd non even
D. None of the above is true
F (x) = f (- x) function is even if f (- x) = - f (x) function is odd
Let f (- x) = (- x) * sin-x) = (- x) * (- SiNx) = x * SiNx = f (x)
The function is even
f(-x)=-x*(sin-x)=(-x)*(-sinx)=x*sinx=f(x)
The function is odd
Odd * odd or odd
F (- x) = (- x) sin (- x) = xsinx = f (x), select B
It is known that the function f (x) is an odd function defined on R. when x ≥ 0, f (x) = x (1 + x), the image of function f (x) is drawn and the analytic expression of function f (x) is obtained
∵ when x ≥ 0, f (x) = x (1 + x) = (x + 12) 2-14, f (x) is an odd function defined on R, ∵ when x < 0, - x > 0, f (- x) = - x (1-x) = (X-12) 2-14 = - f (x), ∵ f (x) = - (X-12) 2 + 14 ∵ f (x) = (x + 12) & nbsp; 2-14x ≥ 0 - (X-12) & nbsp; 2 + 14x < 0
The function f (x) is a function with a period of 10 π when - 5 π
f(-19.5π)=f(-9.5π)=f(0.5π)=0.5π*sin(0.5π)=0.5
Let f (x) be an odd function defined on R. when x ≤ 0, f (x) = 2x ^ 2-x, then the analytic expression of F (x) is?
A:
F (x) is an odd function
So: F (- x) = - f (x)
When x = 0, - x = 0, f (x) = - 2x & # 178; - X
So:
x=0,f(x)=-2x²-x
In the derivatives of higher numbers: the derivative of function y = x [x] at x = 0? [x] represents the derivative of absolute value xsinx, where SiNx is the superscript position
F (x) = x | x |, when x > = 0, f (x) = x & sup2;, when x
Let f (x) be an odd function defined on [- 1,0) ∪ (0,1]. When x ∈ [- 1,0), f (x) = 2x + x ^ 2. Find the analytic expression of F (x) when x ∈ (0,1)
Suppose x ∈ (0,1), then - x ∈ [- 1,0]
So f (- x) = 2 (- x) + (- x) ^ 2 = - 2x + x ^ 2 = - f (x)
So f (x) = 2x-x ^ 2
Simple enough, detailed enough