What is the geometric meaning of differentiability? Is that understandable z=f(x,y) (x,y)-->(x+Δx,y+Δy) It can be expressed as: (DZ =) Δ Z ≈ a * Δ x + b * Δ y Where a and B are the partial derivatives of Z to X and Y respectively, that is, (x, y) -- > (x + Δ x, y + Δ y) the distance between two points can be expressed as the result of summation of a * Δ X in X direction and B * Δ y in Y direction

The geometric meaning of differentiability is that there is a tangent at this point
The distance between two points can be expressed as: first in the X direction, that is, a * Δ x, and then in the Y direction, b * Δ y
Then it is not the result of summation, and Δ Z cannot be expressed by a and B being the partial derivatives of Z to X and Y respectively
It is three-dimensional and has derivatives for X and y, not necessarily in other directions

What is the algebraic meaning of the absolute value of a number
It's algebraic

Distance from the origin
That's to remove the symbol

Given that the two imaginary roots of equation 3x2-6 (m-1) x + M2 + 1 = 0 are α, β, and | α | + | β | = 2, find the value of real number M

α, β are conjugate virtual roots each other (2 points) then | α | = | β |, α β = | α | 2 = M2 + 13 (6 points) | α | = | β | = M2 + 13 From | α | + | β | = 2, M2 + 13 = 1, M = ± 2 Because when m = - 2, △ 0, it is not suitable for the problem, so m = 2 (12 points)

Given that the maximum value of y = a-bcos2x (x ∈ [0, π / 3]) is 1 and the minimum value is - 1 / 2, find the values of real numbers a and B

From X ∈ [0, π / 3], we get 2x ∈ [0,2 π / 3], cos2x ∈ [- 1 / 2,1],
If b > 0, then A-B = - 1 / 2, a + 1 / 2B = 1, then a = 1 / 2, B = 1
If B

If the absolute value of X is π / 4 and f (x) = cos Λ x-acosx, the range of F (x) is obtained when a = 4
(2) If the minimum value of F (x) is - 1 / 4, find the value of A

There's something wrong with your title. I can't understand it

The radius of circle O is r, and point P is a certain point. A straight line passing through point P intersects circle O at two points ab. it is proved that PA multiplied by Pb equals the absolute value of the square of OP minus R (in the case of points)

(1) P outside the circle, make Pt tangent to t, PA * Pb = Pt ^ 2, Pt ^ 2 = Po ^ 2-ot ^ 2, PA * Pb = | Po ^ 2-r ^ 2 | (2). P inside the circle, make chord EF perpendicular to op through P, then PE = PF, PA * Pb = PE * pf = PE ^ 2 = OE ^ 2-po ^ 2, PA * Pb = | Po ^ 2-r ^ 2 | (3). P on the circle, PA * Pb = 0, Po ^ 2-r ^ 2 = 0, PA * Pb = | Po ^ 2-r ^ 2 |

Point P is on the straight line L: y = X-1, if there is a straight line intersection parabola y = x ^ 2 of P at two points a and B, and the absolute value of PA is equal to the absolute value of Pb, then point P is called a good point
All the points on the line L are "good points". Why? Please give the idea and process of solving the problem

First, find the position relationship between the straight line and the parabola, and let C be the coordinate of its intersection point. According to the meaning of the problem, C satisfies the equations (1) y = X-1 and (2) y = x ^ 2, that is, x ^ 2 = X-1
According to the root formula: x = [- B ± √ (b ^ 2-4ac)] / (2a), x = 1 / 2 ± √ (1-4) / 2. There is no real root, so there is no intersection point between straight line and parabola, that is, there is no point C. on the other hand, it can be proved that parabola is monotone convex curve
The line L1 passing through any point P on the line l can be expressed as: L1: y = ax + B, then there is a point P on L1 satisfying the equation of L, that is, ax + B = X-1 has unique solution, so we have, (A-1) x = - (B + 1). According to the requirements of the problem, L1 intersects two parabolic points, that is, ax + B = x ^ 2 has two different solutions, that is, a-4b > 0 in the equation x ^ 2-ax + B = 0, and the X coordinates of the two points are respectively: A / 2 ±. 5 * √ (a-4b)
According to the above results, we can get the coordinates of two intersections, (x1, Y1), (X2, Y2), and the coordinates of P (x, y), which are functions of a and B, and a-4b > 0
It can be proved that there can be no straight line with the same absolute value of PA and Pb (the proof and discussion are omitted), unless a-4b = 0, that is, the straight line passing through P is tangent to the parabola - in fact, it is easy to distinguish by drawing method, because the straight line does not intersect the parabola
If we don't consider that a and B must be different, then only the tangent point can meet the requirements of the problem. So the problem is transformed into whether any point on the straight line can make a straight line tangent to the parabola?
We can have two ideas. One is to prove that there is at least one group (a, b) satisfying the above requirements by using the above method, let a-4b = 0
Another idea is to find the tangent equation of any point on the parabola. Obviously, the equation is a function of the point (x0, x0 ^ 2) on the parabola, and then prove that the tangent equation has a solution to the line L
The conclusion is that the answer is right, and two such lines can be made through all points on L, which satisfy the definition that P is a good point

As shown in the figure, a and B are on both sides of the line L, and find a point P on the line L to make the value of | pa-pb | maximum

Make a symmetric point a 'of point a with respect to line L, connect a' B and extend the intersection line L to P

1 | 4 | a + 5 | + 1 / 3 | A-1 | + 1 / 6 | A-4 |, find the minimum value of A. (| is the absolute value) \ \ is the semicolon
Please, I'm going to school tomorrow, my life is in your hands
1 / 3,,, is one third.

No technical content, classification discussion, a less than or equal to - 5, - 5 less than or equal to a less than 1, 1 less than or equal to a less than 4, a greater than or equal to 4, to discuss

Absolute value of X-1 + absolute value of x-3, Q: is there a minimum value? If so, please calculate. If not, please explain the reason

When x ≤ 1
|x-1|+|x-3|=-x+1-（x-3）=-2x+4|≥2
When x ∈ (1,3)
|x-1|+|x-3|=x-1-x+3=2
When x ≥ 3
|x-1|+|x-3|=x-1+x-3=2x-4≥2
So there's a minimum of two

What is the algebraic meaning of the absolute value of a number It's algebraic