#Find a function $f:\bR\to\bR$ such that $\int f(f(x)+x)dx=f(f(x))+C$
1 messages · Page 1 of 1 (latest)
hmmmn
Ok so i told you i was gonna solve that equation
f'(f(x)).f'(x)=f(f(x)+x)
Wait let me send a photo
wait no latex?
What?
Find a function $f:\bR\to\bR$ such that $$\int f(f(x)+x)dx=f(f(x))+C$$
people like me dony know latex
dark matter
how do u typr latex so fast
there i thought it didnt work
oh i just copy pasted from the title lel
oh
title takes maybe 10 seconds to create
wait leme see
there are more solutions than that
And f(x) = c*x ohhhhh i see it now
i will tell in dm
I think this was the funny solution
nope
Oh
Then what was it?
Anyway let’s move on to geometry
What should we do for that?
Circle?
Triangle?
Quadrilateral?
i choose straight line
I was thinking shapes
a line is technically a circle
……..
its equidistant from infinity from all points
Ok let’s do algebra then
So what should we do here?
Ok ill choose
We’ll do real numbers
Starting off easy
$\sqrt{50} + 2\sqrt{2}$
TabMineCrafter
@steel violet @fallen gazelle?
what
7root(2)
what
who knows
Should be i derived it
assumed f(x) = a + xb
b could only be -1 or 0
do you need?
no
……
Circles?
@fallen gazelle you told me to
Give a problem
On trigonometry
We have a right triangle labeled ABC, AB is 40 and AC is 30, find the sin of the angle ABC
ok
3/4 or sqrt(7)/4
ez guess
ok
Yes
We have a normal triangle labeled ABC with AB = 12, BC = 13 and AC =15, find the area of that triangle
Ok AC is 15
its not the irght answer
yes
u solved it wrong way
wat
show work again
wa
show work bro
I fold
...
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Latent' is an ability t...
f(x) = 0 i already said
f(x) = a + bx
If you literally put everything in the identity you get
b = 0 or -1
f(x) = a or f(x) = a - x
assume a = 0 because a - x won't work at all
f(x) = 0
The initial derivation is long
Let’s do circle problems
f(x) = 0 right
well fair
Should we do another problem?@here
fr
that doesnt eliminate the constant of integration
u can write something like "the antiderivative of"
or something like "d/dx f(f(x))= stuff"
e^x
LMAO e^x WORKS
NAHAHHHHH
HAHAHAHAHAH
it's the only exponential that works
finally you found the cheese solution
i was randomly thinking about how to solve $\int e^{e^x}dx$ then stumbled on that and realized that it was cool
dark matter
Wow
fym finally, itwas my first guess
let me phrase it better: finally, someone found the cheese solution
0 is much more cheese than e^x
not really
