If the function f (x) = log (x2 ax + 1 / 2) has a minimum value, then the value range of real number a is 2. It is known that the domain of function f (x) is the set of real numbers R, m and N are all real numbers. If the inequality f (m) - f (n) > F (- M) - f (- n) holds, then the following inequality holds Am-n0 C m+n

If the function f (x) = log (x2 ax + 1 / 2) has a minimum value, then the value range of real number a is 2. It is known that the domain of function f (x) is the set of real numbers R, m and N are all real numbers. If the inequality f (m) - f (n) > F (- M) - f (- n) holds, then the following inequality holds Am-n0 C m+n

The compound function "same increase and different decrease!", when 0 < a < 1, the logarithmic function is a subtractive function, but the quadratic function is infinite. At this time, there is no minimum value, or - ∞. So you can only a > 1. I don't know that your problem conditions are completely missing. If any x exists, you must require the quadratic function (4ac-b ^ 2) / 4A to be less than 0, then you get a [root 2

The differential of the - x power of e multiplied by cosx

D (the - x power of e multiplied by cosx)
=cosx*d(e^(-x))+e^(-x)dcosx
=[-e^(-x)cosx-sinx*e^(-x)]dx
=-e^(-x)(cosx+sinx)dx

On DX / DT It is said that the coffee grinding pot is composed of oil. On the top is an inverted cone with a bottom diameter and height of 6cm, and on the bottom is a cylinder with a bottom diameter of 6cm. The speed of coffee dropping from the top is 10cm3 / min, Q: at what rate is the coffee in the cylinder rising? At what rate is the coffee in the cone decreasing? Wrong, sorry, look here At what rate does the coffee in the cone decrease when it is 5cm deep from the coffee? At this time, at what speed is the coffee in the cylinder rising?

Your question is still unclear
That's all I can understand
Hope to understand is right
When the coffee in the cone is 5cm deep
According to the similarity principle, the bottom diameter of the cone liquid level is also 5cm
The area of the bottom surface of the liquid level is 25pai / 4 cm2 (PAI represents PI)
The volume of coffee in the cone is 125 Pai / 12 cm2
The dropping speed of coffee is v = 10cm3 / min
Set the dropping time as t
Then the liquid level height of the cone is H1 = (125pai / 12-10t) / 25pai / 4
By deriving from the above formula, the falling speed of the liquid level in the cone is V1 = - 8 / 5pai
That is, the falling speed of liquid level is 8 / 5pai cm / min
The height of cylinder liquid level rise is h2 = VT / s
Where s is the bottom area of the cylinder 3pai
Then h2 = 10t / 3pai
Derivative V2 = 10 / 3pai cm / min

How to prove ∫ (upper limit x, lower limit a) f (T) DT = f (x). Why is the definition of calculus so defined? ∫ (upper limit x, lower limit a) f '(T) DT = f (x) should be f' (T) instead of F (T)

Let g (x) = ∫ (upper limit x, lower limit a) f (T) DT because it can be written as ∫ (upper limit x, lower limit a) f (T) DT, f can be integrated. Let f (x) '= f (x) g's derivative at any point m be g (m)' = {Lim Q - > 0} [g (M + Q) - G (m)] / [(M + Q) - M] = {Lim Q - > 0} ∫ (upper limit m + Q, lower limit m) f (T) DT] / [(M + Q) - M] = {

A proof of calculus F (x) is defined in (- ∞, + ∞). For any x1, X2, f (x1 + x2) = f (x1) + F (x2), and f (x) = 1 + XG (x), Lim g (x) = 1. It is proved that f (x) is everywhere differentiable in (- ∞, + ∞) Lim g (x) = 1 x is the limit when it approaches zero. I'm sorry to play less x->0

impossible
Hair f (0) = 1 + 0 * g (0) = 1
Let X1 = 0, f (x2) = f (0) + F (x2) = 1 + F (x2), contradiction

A proof of University Calculus The intermediate value theorem is used to prove: F ∈ C [a, + ∞), f (a) 0. It is proved that there is m ∈ (a, + ∞), so that f (m) = 0

When x → + ∞, f (x) → a > 0
For E = A / 2, there is x > 0. When x > x, there is | f (x) - a | f (x) > A / 2 > 0
-->f(X+1)>0
F (a) has m ∈ (a, + ∞), so that f (m) = 0