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Derivative relationship between The displacement of a particle, the velocity is the derivative of the displacement, The above words appear in the content derivative of mathematics elective 2-1 taught by senior two But I don't quite understand the relationship between them
R represents displacement, V represents velocity, a represents acceleration, T represents time... The meaning of derivative, for example, Dr / dt... There is a functional relationship between R and t Δ r/ Δ T is what high school students can understand... It means that the change of displacement in a period of time is greater than that in this period of time... That is, the average velocity... Derivative and its meaning
Evaluate a domain Example: y = x ^ 2-2x + 3, value range of - 1 ≤ x ≤ 2 I always think we should substitute - 1 and 2 directly... But it's not right. What increase interval and decrease interval are required? Why? Just tell me how to calculate it. I don't need to calculate examples. I want a general solution,
It is known that the value range of the definition field must be increased or decreased, because like the quadratic function in the example (the image of the quadratic function should be taught), you can draw a curve of the quadratic function with the opening upward. It can be seen that there is a lowest point, that is, the minimum value. The left is decreased, the right is increased, and the definition field given to you may include the X when taking the minimum value, At this time, the minimum value of the value range of the function is not the value of the function when x is equal to 2 endpoints
Let me give you a simple example. Y = x ^ 2, the minimum value on [- 1,1] is when x = 0 is the 0, not at both ends
How to find the definition domain and value domain of a given function in Higher Mathematics? Senior one compulsory (1) Chapter 2 function content
Starting from the concept of definition domain and according to the conditions of function analytical formula, the definition domain is given
The value range is obtained according to the value of the definition field
y=3X y=8/x y=-4x+5 Square of y = x - 6x = 7 The above definition fields and value fields
R R
R R\{0}
R R
R [- 2, + infinity)
The definition field of function y = 2arcsin (1 / x) is --, and the value field is——————
The domain of 1 / X is not equal to 0
The domain of arcsinx is
Definition domain and value domain of function y = 2arcsin (TaNx)? Function y = arccos (SiNx) (- π / 3)
(1) Y is meaningful, then - 1 ≤ TaNx ≤ 1, so the definition field is [- π / 4 + K π, π / 4 + K π], and K is an integer
The range is [- π, π]
(2) Because - π / 3 and SiNx increase on (- π / 3, π / 2) and decrease on [π / 2,2 π / 3],
Therefore - √ 3 / 2 makes t = SiNx, then - √ 3 / 2 because y = arccost is a subtractive function on (- √ 3 / 2,1),
Thus 0 ≤ y < 5 π / 6
RELATED INFORMATIONS
- 1. Monotone interval of function y = root sign (square of x-2x-3)
- 2. F (x) is an odd function defined on R, and f (x + 2) = - f (x). When 0 ≤ x ≤ 1, f (x) = x ^ 2 + X ① find the period ② find the expression at - 1 ≤ x ≤ 0 ③ Find f (6.5)
- 3. It is known that the definition field of function f (x) is all real numbers of X ≠ O. for any X1 and X2 in the definition field, f (x1 · x2) = f (x1) + F (x2), and when x > 1, f (x) > 0, f (2) = 1 (1) Verify that f (x) = f (- x) (2) F (x) is an increasing function on (0, + infinity) (3) Solving inequality f (|x| + 1) < 2
- 4. If the function f (x) = log defined in (- 1,0) takes 2A as the base and X + 1 as the true number If f (x) is greater than 0, the value range of a is
- 5. If the domain of the function f (x) = x2 + 2x + A / 1 is r, the value range of the real number a is (to the process)
- 6. Y = f (x) is an even function defined on R, and f (x + 1) = f (x-1) holds for any real number. When x ∈ [1.2], f (x) = log a X If the maximum value of the function y = f (x) is 1 / 2, the inequality f (x) > 1 / 4 about X is solved on the interval [- 1,3] Can you tell me in detail with pictures? I didn't see what you said
- 7. The even function f (x) always defined on R is a subtractive function at [0, positive infinity). If f (m-1) - f (2m-1) > 0, the solution set of real number m is
- 8. The value range of 2cosx-1 under the function y = root is
- 9. Find the definition field and value field of the following functions (1) y = SiNx + 3 / SiNx + 2 (2) y = under the root sign (1 + SiNx)
- 10. Find the extreme value of function f (x, y) = (6x-x2) (4y-y2)
- 11. If the equations of X and y, x = 3-2t, Y-5 = 3T, then use the algebraic formula containing x to express y, which is correct
- 12. Area enclosed by star line x = ACOS ^ 3T, y = asin ^ 3T
- 13. （x+2y)/2+(3x-y)/3=5 ① 0.2(x+2y)+0.8(3x-y)=4 ② Calculate this binary linear equation, and add money when friends answer it in a minute
- 14. The general solution of the second derivative of Y + the first derivative of 2Y - 3Y = 6x + 1 The corresponding homogeneous equation is y '' + 2Y '- 3Y = 0, and the characteristic equation is γ^ 2+2 γ- 3 = 0, the general solution is y = C1E ^ x + c2e ^ (- 3x). Since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1. Therefore, a = - 2, B = - 5 / 3, y * = - 2x-5 / 3, then the general solution of the original equation is y = C1E ^ x + c2e ^ (- 3x) + 6 --- 2x 5 / 3 What I don't understand is "since 0 is not the root of the characteristic equation, let the general solution of the non-homogeneous equation y '' + 2Y '- 3Y = 6x + 1 be y * = ax + B, substitute it into the equation, and get 2a-3ax-3b = 6x + 1, so what does it mean by a = - 2, B = - 5 / 3, y * = - 2x-5 / 3" What 0 is not the root of the characteristic equation? What is 0? How did ax + B come from? How did you bring the general solution y * = ax + B into the equation? What should a and B do? Please give me your advice. I can't understand it
- 15. Y = 3x ^ 4 * e ^ 10 find the 10th derivative of Y
- 16. Solution of univariate cubic equation I want to ask about the root formula ~ don't copy it from the encyclopedia ~ I can't understand ~ I've read it! I wonder if the cubic equation can only be solved in some special form? What is that root expressed by those corresponding coefficients?
- 17. A problem about the derivative of univariate function The question is still unclear Taking y as an independent variable and X as a dependent variable, the transformation equation is proved {(dy/dx) * [(dy)^3/d(x^3)]} - 3 {[(dy)^2/d(x^2)] ^2} = x One question: The answer is: [(dy)^3/d(x^3)] = - d/dy{(dx/dy)^(-3)*[(dx)^2/d(y^2)]}*(dy/dx) ={3(dx/dy)^(-4)*[(dx)^2/d(y^2)]^2-[(dx/dy)^(-3)]*(dx)^3/d(y^3)}*（dx/dy）^(-1) =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3) But how do I figure it out =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3)*[(dx)^2/d(y^2)] In other words, it is multiplied by one more [(DX) ^ 2 / D (y ^ 2)] I do this: let DX / dy = t So (Dy) ^ 3 / D (x ^ 3) =-d/dy[t^(-3)*t']*t^(-1) =[3t^(-4)*t'-t^(-3)*t'']*t'*t^(-1) =[3t^(-5)*t'^2-t^(-4)*t''*t']*t' I've done it several times. Why is there more t 'in the last step Please kindly answer it. Don't worry. The steps are a little troublesome. Take your time Thanks a lot^^~~
- 18. Find the value range of the following function: y = x + √ (4-x) ²）
- 19. Example of absolute limit 1. Take an example to show that | f | continuity does not mean f continuity 2. For example, function f is discontinuous everywhere, but | f | is continuous everywhere
- 20. The absolute value of X-2 divided by X-2. What is the limit when x approaches 2