SuicideFuel Math thread problem (official)

[Intellau_Celistic]

Is that Shan, or Sam?

 

[Filthy_Reject]

you're god damn right

 

[Caesercel]

420.

 

[trying to ascend]

Solution?

 

[Caesercel]

Solution?

930

 

[trying to ascend]

930

Correct

 

[trying to ascend]

H = (x³ - x² + x) raised to the 9th power

Given that x = 2/(sqrt(5) - 1), find the least positive integer closer to H

 

[trying to ascend]

4x²/(1 - sqrt(1 + 2x))² < 2x + 9.

Find the values of x that satisfy this inequality

 

[Fallenleaves]

4x²/(1 - sqrt(1 + 2x))² < 2x + 9.

Find the values of x that satisfy this inequality

Send differentiation questions or limits

 

[trying to ascend]

Send differentiation questions or limits

Ok:

Given that

Lim f(x)/x = L
X——->0

Find (in terms of L)

Lim f(x^2 - 1)/(x - 1)

X——->0


Are you practicing it?

 

[Fallenleaves]

Are you practicing it?

Yeah we're about to get into integration now. I give up what's the answer

 

[trying to ascend]

Yeah we're about to get into integration now. I give up what's the answer

2L

 

[Fallenleaves]

2L

Would you mind telling me how you did it?

 

[Ahnfeltia]

2L

This is false. Take e.g. f(x) = sin(x). Then sin(x)/x tends to 1 as x tends to 0, yet sin(x^2-1)/(x-1) tends to sin(-1)/(-1) = sin(1) != 2 as x tends to 0.

 

[trying to ascend]

This is false. Take e.g. f(x) = sin(x). Then sin(x)/x tends to 1 as x tends to 0, yet sin(x^2-1)/(x-1) tends to sin(-1)/(-1) = sin(1) != 2 as x tends to 0.

True, it was actually x ----> 1

 

[Fallenleaves]

B

True, it was actually x ----> 1

Ruh

 

[Ahnfeltia]

True, it was actually x ----> 1

Yeah, I realized that after I had made the post. In that case 2L is correct of course.

Would you mind telling me how you did it?

Would you still like to know or can you see it now?

 

[Fallenleaves]

Yeah, I realized that after I had made the post. In that case 2L is correct of course.

Would you still like to know or can you see it now?

I don't get it

 

[Ahnfeltia]

I don't get it

Multiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.

 

[Fallenleaves]

Multiplying both numerator and denominator of f(x^2-1)/(x-1) by (x+1) yields that f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1). Now, as x tends to 1, y = x^2-1 tends to 0, so f(x^2-1)/(x^2-1) = f(y)/y tends to L, whereas x+1 tends to 2. Ergo, f(x^2-1)/(x-1) = (x+1)*f(x^2-1)/(x^2-1) tends to 2*L as x tends to 1.

Thanks

 

[Ahnfeltia]

Thanks

np

 

[trying to ascend]

Let ABCDEF be a right triangular prism, with all its edges congruent and its lateral edges AD, BE and CF. Let 0 and 0' be the barycenters of the bases ABC and DEF, respectively, and P be a point belonging to 00' such that P0' = 1/6 00' . Let π be the plane determined by P and the midpoints of AB and DF. The π plane divides the prism into two solids. Determine the ratio of the volume of the smaller solid to the volume of the larger solid, determined by the π plane

 

[Fallenleaves]

Any basic (very) integration or differentiation questions?

 

[trying to ascend]

Any basic (very) integration or differentiation questions?

Find the critical points of (x² - x)/(1 + 3x²).

Simplify the expression: I = (11/8) . e (to the power of 2x) + (3/4) . x² . e (to the power of 2x) - (3/4) . x . e (to the power of 2x) - (3/2) . integral of (x² . e to the power of 2x).

Using first principle, prove that cos'(x) = - sin(x).


Be a = limx→0 of (13x - sin(10x))/(sin(11x) + x).

and b = limx→∞ of x/sqrt(x² + x + 1).

Find a + b

 

[trying to ascend]

Given that the integral (from 0 to pi) of (f(x) + sin(2x)) = 2.

Find the integral (from 0 to pi) of (f(x) + 9(x/pi)²)

 

[trying to ascend]

Be f(x) = sqrt(sin(7x + ln(5x))).

Find the first derivative of f(x)

 

[Fallenleaves]

Find the critical points of (x² - x)/(1 + 3x²).

y prime= (3x-1)(x+1)/[(1+3x^2)^2]
f(1/3)= -1/6
f(-1)=1/2

Be a = limx→0 of (13x - sin(10x))/(sin(11x) + x).

and b = limx→∞ x/sqrt(x² + x + 1).

Find a + b

Gave up but a=1/4 by l'hospital's rule

 

[Fallenleaves]

I'm sleeping

 

[Fallenleaves]

Find a + b

5/4?

 

[trying to ascend]

y prime= (3x-1)(x+1)/[(1+3x^2)^2]
f(1/3)= -1/6
f(-1)=1/2

Gave up but a=1/4 by l'hospital's rule

Correct!

 

[Khanivore]

2+2 in spanish when it is tuesday

 

[trying to ascend]

5/4?

Correct

 

[Fallenleaves]

Be f(x) = sqrt(sin(7x + ln(5x))).

Find the first derivative of f(x)

Just used the chain rule only and got the wrong answer

 

[trying to ascend]

S = (xy + xz + yz + 2(x + y + z) + 3)/((x + 1)(y + 1)(z + 1)).

Given that x = log ab (in the base c).

y = log ac (in the base b)

z = log bc (in the base a).

Find the value of S

 

[trying to ascend]

What's the biggest value of n, such that 21 (to the power of n), divides 2020! ?

 

[Fallenleaves]

Evaluate the limit

lim [x^2(3+sinx)]/[(x+sinx)^2]

x--->0

 

[trying to ascend]

Evaluate the limit

lim [x^2(3+sinx)]/[(x+sinx)^2]

x--->0

3/4?

I just expanded the square in the denominator and then divided everyone by x²

 

[Fallenleaves]

3/4?

I just expanded the square in the denominator and then divided everyone by x²

Nice but the question was introduced to me when I was learning how to use sinx/x=1 as x tends to 0 in questions

 

[trying to ascend]

Find all functions, R ---> R, such that:

f(x)f(y) = f(x + y) + xy.

 

[Caesercel]

What's the biggest value of n, such that 21 (to the power of n), divides 2020! ?

334

 

[Caesercel]

Given that the integral (from 0 to pi) of (f(x) + sin(2x)) = 2.

Find the integral (from 0 to pi) of (f(x) + 9(x/pi)²)

3pi + 2

 

[Caesercel]

Be f(x) = sqrt(sin(7x + ln(5x))).

Find the first derivative of f(x)

Cos(7x + ln(5x)). (7 + 1/x) /2.sqrt(sin(7x + ln(5x)))

 

[Caesercel]

Using first principle, prove that cos'(x) = - sin(x).

Cos'(x) = lim h ->0 (Cos(x+h) - Cosx) /h

= (CosxCosh - SinxSinh - Cosx) /h

= lim h->0 -SinxSinh/h

= - Sinx

 

[Ahnfeltia]

Find all functions, R ---> R, such that:

f(x)f(y) = f(x + y) + xy.

First, plugging in y = 0 yields that f(x)*f(0) = f(x) which means that either f(x) = 0 for all x or f(0) = 1. One can easily check that f = 0 is not a valid solution however.

In case f(0) = 1, plug in x = 1 and y = -1 to find that f(1)*f(-1) = f(0) - 1 = 0. Ergo, either f(1) = 0 or f(-1) = 0. If f(1) = 0, then plugging in y = 1 yields that 0 = f(x+1) + x. Letting z = x+1 and rearranging then yields that f(z) = 1-z. Similarly the case f(-1) = 0 leads to f(z) = 1+z.

All in all, the solutions are f(z) = 1±z.

 

[Caesercel]

Simplify the expression: I = (11/8) . e (to the power of 2x) + (3/4) . x² . e (to the power of 2x) - (3/4) . x . e (to the power of 2x) - (3/2) . integral of (x² . e to the power of 2x).

e^2x

 

[trying to ascend]

3pi + 2

First, plugging in y = 0 yields that f(x)*f(0) = f(x) which means that either f(x) = 0 for all x or f(0) = 1. One can easily check that f = 0 is not a valid solution however.

In case f(0) = 1, plug in x = 1 and y = -1 to find that f(1)*f(-1) = f(0) - 1 = 0. Ergo, either f(1) = 0 or f(-1) = 0. If f(1) = 0, then plugging in y = 1 yields that 0 = f(x+1) + x. Letting z = x+1 and rearranging then yields that f(z) = 1-z. Similarly the case f(-1) = 0 leads to f(z) = 1+z.

All in all, the solutions are f(z) = 1±z.

Cos'(x) = lim h ->0 (Cos(x+h) - Cosx) /h

= (CosxCosh - SinxSinh - Cosx) /h

= lim h->0 -SinxSinh/h

= - Sinx

334

e^2x

Cos(7x + ln(5x)). (7 + 1/x) /2.sqrt(sin(7x + ln(5x)))

Correct .

What about the logarithm problem?

 

[Ahnfeltia]

S = (xy + xz + yz + 2(x + y + z) + 3)/((x + 1)(y + 1)(z + 1)).

Given that x = log ab (in the base c).

y = log ac (in the base b)

z = log bc (in the base a).

Find the value of S

S = 1. This has to be the most convoluted way of expressing 1 I've ever seen.

 

[trying to ascend]

S = 1. This has to be the most convoluted way of expressing 1 I've ever seen.

Correct

 

[Jitowtel]

Now you slay weights at the gym

It is better

then slay stacys and niggers irl(in vidya game) irl(in vidya game)

 

[trying to ascend]

i think this is it
View attachment 682105View attachment 682106

quite a long way to write 1

Correct