Proof of solving a calculus mean value theorem~ Let f (x) be continuous on [0, a], differentiable in [0, a], and f (a) = 0 prove the existence of B such that 3f(b)+bf'(b)=0 F '(b) means the derivative of the function in B

Proof of solving a calculus mean value theorem~ Let f (x) be continuous on [0, a], differentiable in [0, a], and f (a) = 0 prove the existence of B such that 3f(b)+bf'(b)=0 F '(b) means the derivative of the function in B

Let g (x) = x ^ 3 * f (x), then G (x) is continuous on [0, a], differentiable in [0, a], and G (0) = g (a) = 0. Therefore, it is known from the mean value theorem that there is 0 = 1 and there is always 0

Solve a calculus proof problem, mean value theorem F (x) is continuous on [0, a], (0, a) is internally differentiable, and f (a) = 0.. proves that there is a point ξ, Belong to (0, a) make f（ ξ)+ξ f’( ξ)= 0

In fact, it is very simple to let H (x) = XF (x), then H (0) = 0f (0) = 0h (a) = AF (a) = 0, then according to the Lagrange mean value theorem: if the function is continuous on the closed interval [a, B] and differentiable in the open interval (a, b), then there is at least one point C in (a, b), so that f (b) - f (a) = (B-A) f '(c) holds. In this problem, because H (a) = H (0), then

Calculus is proved by Lagrange's theorem If x → 0 + limf (x) = f (0) = 0, and f '(x) > 0 when x > 0, then f (x) > 0 when x > 0

When x > 0, there is f (x) - f (0) = f '(m) m, where m is on (0, x). Since f (0) = 0, there is f (x) > 0

Proving kinetic energy theorem

Content of kinetic energy theorem:
The work done by a force to a body in a process is equal to the change of kinetic energy in the process
The work done by the combined external force (the sum of the external forces on the object, the direction and magnitude of the final resultant force of the object can be calculated by the orthogonal method according to the direction and magnitude of the force) on the object is equal to the change of the kinetic energy of the object
Particle kinetic energy theorem
expression:
W1 + W2 + W3 + W4... = △ w = ek2-ek1 (K2) (K1) is the subscript
Where Ek2 represents the final kinetic energy of the object and EK1 represents the initial kinetic energy of the object. △ W is the change of kinetic energy, also known as the increment of kinetic energy, and also represents the total work done by the combined external force on the object
The expression of kinetic energy theorem is scalar. When the external force does positive work on the object, Ek2 > EK1, the kinetic energy of the object increases; Conversely, if EK1 > Ek2, the kinetic energy of the object decreases
The displacement, initial and final kinetic energy in the kinetic energy theorem should be relative to the same reference frame
The object formula of energy theorem is a single object, or the formula can be called a single object system
The calculation formula of kinetic energy theorem generally takes the ground as the reference system
The kinetic energy theorem is applicable to the linear motion of an object and also to the curve motion; It is suitable for constant force work and variable force work; The force can act in sections or at the same time, as long as the positive and negative algebraic sum of each force can be obtained, which is the advantage of the kinetic energy theorem
Group kinetic energy particle group kinetic energy theorem
The sum of work done by all external forces plus the sum of work done by all internal forces is equal to the change of the total kinetic energy of the particle system
Like the particle kinetic energy theorem, the particle system kinetic energy theorem is only applicable to the inertial system, because the work done by the external force on the particle system is related to the selection of the reference system, while the work done by the internal force is independent of the selected reference system, because the forces always appear in pairs. The algebraic sum of the work done by a pair of forces and reactions (internal forces) depends on the relative displacement, and the relative displacement is independent of the selected reference system
The content of kinetic energy theorem: the total work of all external forces on the object (also known as the work of combined external forces) is equal to the change of the kinetic energy of the object
Mathematical expression of kinetic energy theorem: W total = square of 1 / 2m (V2) - square of 1 / 2m (V1)
The kinetic energy theorem is only applicable to the macro low-speed situation, while the momentum theorem can be applied to any situation in the world (provided that the sum of external forces in the system is 0)
1) Kinetic energy definition: the energy of an object due to motion. It is expressed in EK
The expression EK = 1 / 2mV ^ 2 can be both scalar and procedural
Unit: Joule (J) 1kg * m ^ 2 / S ^ 2 = 1J
(2) Content of kinetic energy theorem: the work done by combined external force is equal to the change of kinetic energy of an object
Expression w combination= Δ Ek=1/2mv^2-1/2mv0^2
Scope of application: constant force work, variable force work, subsection work and whole process work

A proof of the mean value theorem in calculus ~ it's actually quite simple ~ please~ It is known that the function f (x) is continuous on [0,1], differentiable in (0,1), and f (1) = 0. It is proved that there is a point C in (0,1), so that f '(c) = - f (c) / C It shouldn't be difficult ~ but I proved incompetent... Please help me write the process ~ thank you in advance^^~

prove:
Let f (x) = XF (x)
Then the function f (x) is continuous on [0,1], differentiable in (0,1), and f (1) = f (0) = 0
Therefore, there must be a point C such that f '(c) = 0, that is, [XF (x)]' = 0, CF '(c) + F (c) = 0, f' (c) = - f (c) / C

What is the content of kinetic energy theorem? How to prove it?

Theoretical analysis and demonstration: the object with mass m starts to move along the horizontal plane with speed V1 under the action of a constant tension F, and the speed increases to V2 after displacement s. It is known that the friction between the object and the horizontal plane is constant F, and try to derive the relationship between the work done by the combined external force and the change of the kinetic energy of the object. (sorry, the picture can't be transmitted)
Niu Erde: f-f = ma
Kinematic formula: V1 ^ 2-v2 ^ 2 = 2As (displacement s can be calculated)
Therefore, work done by external force: W external = (f-f) s = 1 / 2 * MV2 ^ 2-1 / 2 * MV1 ^ 2
Namely: W outer = Ek2 － EK1 = △ EK
The work done by the external force on the object is equal to the change of the kinetic energy of the object