Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) given function y = cosx / X finding tangent equation of function at x = π Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) Given the function y = cosx / x, find the tangent equation at x = π

Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) given function y = cosx / X finding tangent equation of function at x = π Finding function derivative f (x) = (x + 1) (x-1) (x ^ 2 + 2) Given the function y = cosx / x, find the tangent equation at x = π

1,f(x)=(x^2-1)(x^2+2)=x^4+x^2-2
f'(x)=4x^3+2x
2,y=cosx/x x=π y=-1/π
y'=(-xsinx+cosx)/x^2 x=π 1/π^2
The tangent equation is y + 1 / π = 1 / π ^ 2 (x - π) x - π ^ 2y-2 π = 0

For the function f (x) = cosx + SiNx, the following four propositions are given,
1. There exists α belonging to (0, Pai / 2) such that f (α) = 4 / 3
2. If α belongs to (0, Pai / 2), then f (x + α) = f (x + 3 α) is constant
3. There exists θ∈ r such that the image of the function f (x + θ) is symmetric about the y-axis
4. The image of function f (x) is symmetric with respect to point (3pai / 4,0)

F (x) = cosx + SiNx = √ 2Sin (x + π / 4), when x ∈ (0, π / 2), (x + π / 4) ∈ (π / 4,3 π / 4), so sin (x + π / 4) ∈ (√ 2 / 2,1), f (x) ∈ (1, √ 2), and 4 / 3 ∈ (1, √ 2), so the first proposition holds. 2. F (x + α) = f (x + 3 α), which indicates that the period of function is 2 α

Given the function f (x) = cosx ^ 2 + SiNx, then the false proposition in the following proposition is
A (a) is neither an odd function nor an even function, and (b) has exactly a zero point on (0, π)
(C) Is a periodic function (d) is an increasing function in

First, f (x) = cosx ^ 2 + SiNx can be reduced to (1-sinx ^ 2) + SiNx = - (sinx-1 / 2) ^ 2 + 5 / 4
The new function f (T) = - (t-1 / 2) ^ 2 + 5 / 4 is a periodic function (c) pair; "SiNx" is an odd function, and - (t-1 / 2) ^ 2 + 5 / 4 form is not odd or even, so the composite function is ") neither odd nor even", (a) is correct. Let f (T) = 0, t = (1-radical 5) / 2, about - 0.6, The range of SiNx is (0,1), - 0.6 is not in the range. So there is no zero point on (0, π). (b) is a false proposition. About (d) is an increasing function on? I don't know what it means. Anyway, the answer is (b)

For the function f (x) = cosx + SiNx, the following four propositions are given: 1) there exists α∈ (0, π 2) so that f (α) = 43; 2) there exists α∈ (0, π 2) so that f (x + α) = f (x + 3 α) is constant; 3) there exists ϕ∈ r so that the image of function f (x + ϕ) is symmetric about y axis; 4) the image of function f (x) is symmetric about (3, π 4, 0). The correct proposition number is______ ．

The function y = SiNx + cosx = 2Sin (x + π 4), where α ∈ (0, π 2), y ∈ (1, 2], because (2) f (x + α) = f (x + 3 α) indicates that 2 α is the period of function, and the period of function f (x) is 2 π, which is obviously a false proposition; (3) there exists θ∈ r such that the image of function f (x + θ) is symmetric about the y-axis, and function f (x) is a periodic function and has a symmetry axis, which can be satisfied by proper translation (x) When x = 3 π 4, f (3 π 4) = 0, it satisfies the meaning of the question. If this option is true proposition, then the number of the correct proposition is (1), so the answer is: 1