It seems to be called Labrador's theorem, which is FX tangent and secant
Lagrange mean value theorem: if the function f (x) satisfies: 1) continuous on the closed interval [a, b]; 2) Differentiable in the open interval (a, b). Then: there is at least one point in (a, b) ξ (a
Can anyone use the derivative limit theorem? If a function is differentiable everywhere on interval I, is the derivative continuous? According to the derivative limit theorem, if the limit of the derivative function exists at a certain point, the derivative at that point must exist. On the contrary, if the derivative exists at a certain point, must the derivative be continuous at that point?
Two upstairs didn't figure out the problem
The relationship between derivative and limit is continuous for univariate functions. The derivation of bivariate or multivariate functions is independent of continuity
This question asks whether the derivative function is continuous, and the answer is not necessarily
If f (x) = kx-x + 1 is a subtractive function on R, find the value range of K?
If f (x) is a monotonically decreasing function, the coefficient of the first-order term is less than 0
If the function f (x) = kx3-x is a subtractive function in R, then the value range of K is ___
∵ if the function f (x) = kx3-x is a subtractive function in R, then f '(x) = 3kx2-1
‡ f ′ (x) ＜ 0, i.e. 3kx2-1 ＜ 0
Simplified: K < 1
So the answer is; k≤0
Given that y = KX is a subtractive function in the definition domain, the value range of K is _
Y = KX is a subtractive function in the definition field, then the value range of K is: K < 0
So the answer is: K < 0
Given the function f (x) = 3x ^ 2-2 (k ^ 2-k + 1) x + 5, G (x) = 2K ^ 2x + K Q: if the function f (x) is a monotone function on the interval (0,3), find the value range of F
f(x)=3x ²- 2(k ²- K + 1) x + 5 axis of symmetry x = (k) ²- K + 1) / 3 function f (x) is a monotone function on (0,3), that is, the axis of symmetry is not on the interval. (k ²- K + 1) / 3 ≤ 0 or (k) ²- k+1)/3≥3(k ²- k+1)/3=[(k-1/2) ²+ 3 / 4] / 3 constant > 0, inequality (k) ²- k+1)...