Let f (x) be differentiable at point x = a, and find [f (a + 5H) - f (a-3h)] / 2H

Let f (x) be differentiable at point x = a, and find [f (a + 5H) - f (a-3h)] / 2H

f(a)‘=[f(a+5h)-f(a-3h)]/8h
[f(a+5h)-f(a-3h)]/2h=4f(a)‘

Why can derivative be written as f '(a) = LIM (x → a) f (x) - f (a) / x-a
Shouldn't it be written as f '(x) = LIM (△ x → 0) (x + Δ x) - f (x) / Δ x?

Different symbols mean different things

Find LIM (1 + x) ^ (1 / x) - e when x tends to 0,
Wrong number. When x tends to 0, find Lim [(1 + x) ^ (1 / x) - e] / X,

The original formula = Lim {e ^ [ln (1 + x)] / x-e} / x = e * Lim {e ^ [ln (1 + x) / X-1] - 1} / x = e * Lim [ln (1 + x) / X-1] / x, Infinitesimal Substitution = e * Lim [ln (1 + x) - x] / X & # 178;, lobita's rule = e * Lim [1 / (1 + x) - 1] / (2x) = (E / 2) * LIM (1-1-x) / x = (E / 2) * - 1 = - E / 2

Let f (x) be differentiable on the interval (0.1) and f (1) = 2 ∫ (0.51) XF (x) DX. It is proved that there exists ξ∈
It is proved that there exists ξ ∈ (0,1) such that f (ξ) + ξ f '(ξ) = 0

Let f (x) be continuous on [0,1]
It is proved that the function g (x) = XF (x) is continuous on [0,1] and differentiable in (0,1)
The condition f (1) = 2 ∫ XF (x) DX is transformed into G (1) = ∫ g (x) DX / (1-0.5)
According to the first mean value theorem for integrals in the open interval version, there exists C ∈ (0.5,1) such that G (c) = ∫ g (x) DX / (1-0.5) = g (1)
According to Rolle's mean value theorem, there exists ξ ∈ (C, 1) such that G '(ξ) = 0, that is, f (ξ) + ξ f' (ξ) = 0
The reason why we use the open interval version of the first integral mean value theorem is to ensure C

In F (1) = 2 ∫ XF (x) DX
The upper limit is 0.5
The lower limit of the integral is 0
Let f (x) be differentiable on [0,1] and satisfy the condition that f (1) = 2 ∫ XF (x) DX. Proof: there exists ∫ x ∈ (0,1), such that f (?) + F '(?) = 0

How do you think f (?) +? F '(?) = 0?

Let f (x) be continuous in [a, b] and f ′ (x) > 0, it is proved that ∫ (a, b) XF (x) DX ≥ (a + b) / 2 ∫ (a, b) f (x) DX

Constructors: F (U) = 2 ∫ [a --- > u] XF (x) DX - (a + U) ∫ [a --- > u] f (x) DX, u ∈ [a, b], obviously f (a) = 0f '(U) = 2uF (U) - ∫ [a --- > u] f (x) DX - (a + U) f (U) = UF (U) - AF (U) - ∫ [a --- > u] f (x) DX = f (U) (U-A) - ∫ [a --- > u] f (x) DX by integral mean value theorem: ∫ [a --- > u] f

Let f (x) be differentiable on [0,1] and satisfy the relation f (1) - 3 ∫ (0,1 / 3) XF (x) DX = 0. It is proved that there exists a ∫ (0,1) such that
f（§）+§f‘（§）=0

If f (x) = x * f (x), then f '(x) = f (x) + X * f' (x);
From the known f (1) = 3 * [∫ (0,1 / 3) f (x) DX],
From Lagrange's mean value theorem, we can deduce that the ratio exists T and satisfies ∫ (0,1 / 3) f (x) DX = f (T) * (1 / 3 - 0), where t belongs to (0,1 / 3)
So there exists t belonging to (0,1 / 3), f (T) = f (1)
Therefore, by roll theorem: the existence of (T, 1) is contained in (0,1), satisfying f '(?) = 0, that is, the conclusion is true

Prove that ∫ (0, a) f (x ^ 2) DX = 1 / 2 ∫ (0, a ^ 2) XF (x) DX (a > 0)

This is a mistake
Let a = 1, f (x) = 1
Then left = 1, right = 1 / 4
Obviously not

If f (x) is continuous at x = 2 and x approaches 2, the limit of F / (X-2) is 1. Is f differentiable at x = 2

Derivable,
When x approaches 2 positive, f approaches 0 positive, when x approaches 2 negative, f approaches 0 negative, there may be left limit equal to right limit equal to 0;
In addition, the function is continuous at x = 2, so f (2) = 0; so there is (f (x) - f (2)) / (X-2) = △ Y / △ x = 1;
That is, the derivative is 1
Welcome to ask

How to prove that the limit of F (x) when x approaches a is equal to the limit of F (a + H) when h approaches 0,

Let limx → a f (x) exist and be equal to L
It is necessary to deduce that limh → 0 f (a + H) also exists and is equal to L
So there must be δ (ε) such that
|f(x)-L|