limits differentiability

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[quick lichen]

d

[halcyon hare]

have you figured out what f(x) is?

[quick lichen]

5-|x|

[quick lichen]

is that right?

[halcyon hare]

does f(x)''

mean second derivative

we usually write f''(x)

[quick lichen]

that's what I assumed

but it was eliminated so didn't matter much

[halcyon hare]

hmm
i tried x=y
ended with f(x)=f''(x)

but your f(x) doesnt satisfy this

[quick lichen]

I used the definition of derivative

f'(x) = f(x+h)-f(x)/h

[halcyon hare]

lim h->0

[quick lichen]

yeah

[halcyon hare]

anyways how does that help

[quick lichen]

then put x and h as x/2 and y/2

also in the initial rule, i put y=0 and got f(x/2)

then on substituting it f"(x) was eliminated

and I was left with f'(x/2) = f(y)-f(0)/y

[quick lichen]

mb this is 5-x

[halcyon hare]

hmm

nope

[quick lichen]

why not?

[halcyon hare]

if we consider this...
f(x) can be ke^x or ksinhx or equiv

[halcyon hare]

it doesnt satisfy f(x)=f''(x)

[quick lichen]

yeah

but how would you get f(x) then?

[halcyon hare]

think about this...

equivalent to sinh(x) can be $\frac{a^x-a^{-x}}{2}$

[round trail]

pratham

[quick lichen]

isn't it e?

[halcyon hare]

yes

wait others dont satisfy

[quick lichen]

yeah

the other 2 are satisfied by 5-x

[quick lichen]

any ideas?

[halcyon hare]

thinking....

how do you solve a second degree DE

[quick lichen]

no clue, I've mainly dinner first degree till now

*done

[halcyon hare]

yeah even me

the reason i am just guessing functions

[hoary crest]

maybe function is a second degree polynomial

[halcyon hare]

but 2nd degree polynomial doesnt satisfy this

[hoary crest]

like ax^2 +5 lol

idk haha

maybe-1/2* x^2 +5

no I am dumb nvm

[quick lichen]

f"(x) will be -1

[hoary crest]

lol yeah

[unkempt skiff]

iirc the solutions of f'' = f are exp(x) and exp(-x)

And linear combinations of them

So sinh and cosh are another basis of the solution space

Wdym do I sleep, it's 9:30AM

[halcyon hare]

and sinhx, coshx too

[quick lichen]

but this fns won't satisfy f(0)=5 and f'(5)=-1

[unkempt skiff]

You can fit initial conditions

You know that the solutions of f'' = f are in the form:
f(x) = A exp(x) + B exp(-x)

[quick lichen]

wdym?

[halcyon hare]

yes like ksinh(x+t)

[unkempt skiff]

So using those two conditions you can solve for A and B

[quick lichen]

oh ok

[halcyon hare]

ahh this is perfect

[quick lichen]

so a+b=5

[unkempt skiff]

Yep, and with the derivative you can do something too

Gives you a 2x2 system to solve

[quick lichen]

ohh

ill try to solve it

[hoary crest]

man what exercise is this

[halcyon hare]

jee aspirant?

[quick lichen]

idk Q's on lcd

[quick lichen]

yeah

[halcyon hare]

noice

[quick lichen]

is this right?

[halcyon hare]

haha now thats complex indeed

[hoary crest]

lol can you test it

[halcyon hare]

but we arent supposed to find f(x)

[quick lichen]

I have no idea

[quick lichen]

oh yeah 😂

that's ae^x + be^-x

[halcyon hare]

why i asked for f(x) because we can next do functional analysis

[hoary crest]

lol

nic3

[quick lichen]

it looks like it satisfies it

[hoary crest]

nice get the limit 🤣

[halcyon hare]

a,b>1

[quick lichen]

yeah

[quick lichen]

I can't 💀

@unkempt skiff what do I do now?

[halcyon hare]

hey @quick lichen if both a,b are +ve we can assume graph resembles coshx

or not maybe

we need to find how function looks like
dont use desmos

lets analyse

[quick lichen]

okay

[halcyon hare]

the domain doesnt contain negative no.s (except integers)

makes sense?

[quick lichen]

yeah

wait what?

isn't the domain R

[hoary crest]

yeah like x^2 🤣

[halcyon hare]

oh no

well serious mistake my bad

[hoary crest]

all values positive?

[quick lichen]

yeah

[halcyon hare]

https://www.desmos.com/calculator/279tz469gv
i just want to theoretically prove that graph looks like this

[hoary crest]

nice

[quick lichen]

you calculated a and b?

also

[halcyon hare]

no

they are random values

[quick lichen]

I don't think f'(5)=-1 is being satisfied

[quick lichen]

oh ok

[quick lichen]

i got 1

[hoary crest]

lol

so...

can we do lim with derivative lol

[halcyon hare]

i am sorry i am kinda stuck

@unkempt skiff what do you say
we need to analyse the function now
how does the graph look

||sorry for ping||

[hoary crest]

stuuuck xd

[quick lichen]

on a side note

I've got another question

[halcyon hare]

oh ok make another forum

[quick lichen]

okay

[halcyon hare]

lets leave this for this particular question only

[unkempt skiff]

No worries

[halcyon hare]

i think we have to see how function is using derivatives and the check how modulus moves the function

like |f(x)| reflects function below x axis to top

[unkempt skiff]

The graph of A exp(x) + B exp(-x) is very dependent on the sign of A and the sign of B

At infinity, it behaves like A exp(x), whereas at minus infinity, it behaves like B exp(-x)

But this is not very relevant to the question being asked

f(x)^( 1 /(x-4)) = exp(ln(f(x))/(x-4))

[halcyon hare]

he wants explaination for option d) so..

[quick lichen]

i thought the answer is d

but I was wrong about f(x) itself then

so idk about all the options rn

[halcyon hare]

multiple correct question?

[unkempt skiff]

Nondifferentiabilities are located where |x|= 0 or f(|x|) = 0

[quick lichen]

not sure

[unkempt skiff]

So you count the number of different solutions

Which are x = 0 and whatever x that verifies f(|x|) = 0

If A > 0 and B > 0, there are not too many choices

[quick lichen]

will there be any value that satisfies it?

[unkempt skiff]

For the first condition, well x = 0

For the second condition, seeing that A > 0 and B > 0, no

[quick lichen]

so... what next?

[unkempt skiff]

Well it means that answer d is unlikely to be correct

Since we only found 1

[quick lichen]

which one's correct?

[unkempt skiff]

Who knows, you can try confirming the other answers and see where it leads you

[quick lichen]

im still confused

[unkempt skiff]

in my honest opinion the only answer that has a chance to be correct is c