The linear equation with slope of 3 and tangent to circle x2 + y2 = 10 is______ .

The linear equation with slope of 3 and tangent to circle x2 + y2 = 10 is______ .

Let the equation of the straight line be y = 3x + B, that is, 3x-y + k = 0, then the distance from the center of the circle (0, 0) to the straight line is equal to the radius, then | 0 − 0 + K | 9 + 1 = 10 can be obtained, k = 10 or K = - 10 can be obtained, so the equation of the straight line is 3x-y + 10 = 0 or & nbsp; 3x-y-10 = 0, so the answer is: 3x-y + 10 = 0 or & nbsp; 3x-y-10 = 0
Given circle C1: (x + 3) 2 + y2 = 1 and circle C2: (x-3) 2 + y2 = 9, the moving circle m is circumscribed with circle C1 and circle C2 at the same time
The center of moving circle m (x, y) is set M (x, y), and the tangent points of moving circle m and C1, C2 are respectively a and B, then | MC1 | - ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||c2 | = 6, which is defined by hyperbola: the locus of the moving point m is determined by
Let a be a real number, f (x) = 2x ^ 2 + (x-a) | x-a | let H (x) = f (x), X ∈ (a, + ∞), find the solution set of inequality H (x) ≥ 1
Because x > A, so h (x) = 2x ^ 2 + (x-a) ^ 2 = 3x ^ 2-2ax + A ^ 2, then the inequality H (x) ≥ 1 is reduced to (x-a / 3) ^ 2 ≥ 1 / 3-2 / 9A ^ 21, when 1 / 3-2 / 9A ^ 2 ≤ 0, that is, when a ≤ - √ 6 / 2 or a ≥ √ 6 / 2, (x-a / 3) ^ 2 ≥ 1 / 3-2 / 9A ^ 2 is constant, so the solution x ∈ (a, + ∞) 2, when 1 / 3-2 / 9A ^ 2 > 0
It is proved that f (a) = 0 (x + 2) is a monotone decreasing function
How can I prove that it is monotonic increasing
A = 2 can be obtained from F (1) = f (2), so f (x) = x + 2 / X,
The derivation is: F '(x) = 1-2 / x ^ 2 = (x ^ 2-2) / x ^ 2
When x belongs to (0, radical 2], f '(x)
Given the function f (x) = | X-1 | (x + 3), (1) find the monotone interval of function f (x), and prove the monotone decreasing interval;
(2) Finding the maximum value of function f (x) in the interval [- 3,0]
When x ≥ 1, f (x) = (x-1) (x + 3) = (x + 1) & # 178; - 4 is a decreasing function on (- ∞, - 1], and an increasing function on [- 1, + ∞). Because x ≥ 1, f (x) is an increasing function on X ≥ 1. When x ≤ 1, f (x) = - (x-1) (x + 3) = - (x + 1) & # 178; + 4 is an increasing function on (- ∞, - 1], and an increasing function on [- 1, + ∞]
Finding the period and monotonicity of the function y = SiNx + cosx
y=sinx+cosx=y=√2sin(x+π/4)
Period T = 2 π
Y = SiNx, X ∈ (- (Π / 2), Π / 2), find the monotonicity of this function. Please write the detailed steps
Draw!
Or the derivative is cosx, in this interval, cosx is always greater than 0, so it increases monotonically
Given the function f (x) = 2x + 1 / 2x-1, try to discuss the monotonicity of function f (x)
In that f (x) = 2x + 1 / 2x-1, X is the exponent and 2 is the base
Forget it. Don't answer. Goodbye
Monotonically increasing function: as x increases, f (x) tends to be 1
The monotone increasing interval of function y = 2x + SiNx is______ .
The definition domain of y = 2x + SiNx is r, ∵ y ′ = 2 + cosx, and the monotone increasing interval of cosx ∈ [- 1, 1] ∵ y ′ > 0 ∵ function y = 2x + SiNx is (- ∞, + ∞), so the answer is (- ∞, + ∞)
The monotone increasing interval of function y = 2x + SiNx is______ .
The definition domain of y = 2x + SiNx is r, ∵ y ′ = 2 + cosx, and the monotone increasing interval of cosx ∈ [- 1, 1] ∵ y ′ > 0 ∵ function y = 2x + SiNx is (- ∞, + ∞), so the answer is (- ∞, + ∞)