Asking for help with the proof of freshman If f (x) is continuous on the interval [a, b], and f (a) < A, f (b) > b, then there is f (ξ) = 0 on ξ ∈ (a, b)

This can't be proved. The closed interval is basically based on the intermediate value theorem, but it can't deduce your conclusion. The only explanation is that the problem is wrong, and the positive and negative of a and B are not explained, so we can't deduce the conclusion

It is proved that ∫ f (x + y) DXDY = ∫ [- 1,1] f (T) DT, D: | x | + | y | ≤ 1

When the integral region is D1: ∫ ∫ f (x + y) DXDY = ∫ [- 1 --- > 0] DX ∫ [- 1-x --- > 1 + x] f (x + y) dy, let x + y = t, then dy = DT, t: - 1 --- > 2x + 1 = ∫ [- 1 --- > 0] DX ∫ [- 1 --- > 2x + 1] f (...)

High numbers: F (x + y) = f (x) g (y) + F (y) g (x), f '(0) = g (0) = 1, f (0) = g' (0) = 0 prove that f (x) is differentiable on R and f '(x) = g (x)

The following limit {Δ x tends to 0} f '(x) = Lim [f (x + Δ x) - f (x)] / Δ x = Lim [f (x) g (Δ x) + F (Δ x) g (x) - f (x)] / Δ x = Lim f (x) {[g (Δ x) - 1] / Δ x} + Lim {g (x) [f (Δ x)] / Δ x} = f (x) Lim {[g (Δ x) - G (0)] / (Δ x-0)} + G (x) Lim

On the problem of discontinuous point. The discontinuous point of function f (x) = [(x ^ 2 + x) (LN | x | (sin1 / x)] / X & ˆ 2-1
The answer is three, 0,1, - 1. Why are these three breakpoints all removable,

There are three kinds of discontinuities: ① removable discontinuities = the first kind of discontinuities, left limit = limit ≠ function value (or undefined)
② Jump discontinuity = left limit of the second kind of discontinuity ≠ right limit
③ Infinite discontinuity = the third type of discontinuity limit does not exist (infinite or uncertain)
f(x)=x(x+1)ln|x|sin1/x/[x-1)(x+1)]
f(x)=xln|x|sin1/x/(x-1)
limf(1+)=1*sin1*limln|x|/(x-1)=sin1*lim(ln|x|)'/(x-1)'=sin1*1/|1|=sin1
limf(1-)=sin1*1/|1|=sin1
lim(-1+)=sin(-1)(-1)*1/|-1|=sin1
lim(-1-)=sin1
lim(0+)=limsin1/x/(x-1)*lim(xln|x|)=limsin1/x/(x-1)lim(1/|x|/(-1/x^2))=Alim(-|x|)=0
lim(0-)=0
All three discontinuities are removable

Discontinuities and types of e ^ 1 / X
What I can solve at present is:
x→0+,e^1/x→∞
x→0-,e^1/x→0
Is my answer wrong?
So, what is the type of discontinuity?
If it is e ^ [1 / (x-1)], the answer is similar, right?

Your solution is right, there is only one discontinuity, but because its right limit does not exist, it should belong to the second kind of discontinuity (infinite discontinuity). The same holds for x = 1

How to find the type of discontinuity of F (x) = 1 / [1 - (x / (x-1) power of e]?

According to the nature of the fraction, the denominator is not zero, we can know that the breakpoints are: x = 0, x = 1
When x = 0, the left limit is equal to negative infinity and the right limit is equal to positive infinity
When x = 1, the limit of F (x) is 0, so it is a removable discontinuity

Higher number! Let f (x) = (1 + 1 / x) to the power of X, find f '(1),

f(x)=(1+1/x)^x=e^(x*ln(1+1/x))=e^(x*ln(x+1)-x*lnx)f'(x)=e^(x*ln(x+1)-x*lnx) *(ln(x+1)+x/(x+1)-lnx-1)=(1+1/x)^x *(ln(x+1)-lnx-1/(x+1))f'(1)=(1+1/1)^1 *(ln(1+1)-ln1-1/(1+1))=2ln2-1

How to change the order of polar coordinates of double integral?

I suggest you take out a pen and paper and draw a picture after me before you look at the following. First y, then x, turn into a quadratic integral in this form. DX dy is to cut the area vertically from the x-axis

Transformation of polar coordinates of double integrals into rectangular coordinates

θ = 0 represents the positive direction of X axis, θ = π / 4 represents the ray, y = x (x ≥ 0), r = sec, the rectangular coordinate equation of θ is x = 1, so D is surrounded by y = 0, y = x, x = 1, d = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x}

How does the original integral curve (e.g. y = 2-x) change into R = (θ) form when double integral is calculated by polar rectangular coordinate transformation
For example, the original integral region is ∫ (lower limit 0, upper limit 1) DX ∫ (lower limit 0, upper limit 2-x) f (x) dy

Let x = RCOs θ, y = rsin θ bring in X + y = 2
RCOs θ + rsin θ = 2, r = 2 / (COS θ + sin θ)
And that's the upper limit of R
this is it.

Asking for help with the proof of freshman If f (x) is continuous on the interval [a, b], and f (a) < A, f (b) > b, then there is f (ξ) = 0 on ξ ∈ (a, b)