Find m so the equation has exactly one solution

[gleaming perch]

i noticed some symmetry happening on here

x-1 and x-3 midpoint is 2? and 2-x feels like it has something to do with the number 2 as well

do i need logarithm for this problem? i havent learn about it in school but i can try doing it myself

[radiant plover]

#Destroyer£££

[acoustic spade]

try puttinh x=2

@gleaming perch

[gleaming perch]

m=1 in that case

[acoustic spade]

nah not 1

1.5

it comes 6/4

[gleaming perch]

oh yeah i messed up with my calc

sorry

[acoustic spade]

np 👍

[gleaming perch]

1.5 yeah

[acoustic spade]

🫡

[gleaming perch]

o7

😔 i still dont understand how i got here, besides from pure guessing/intuition. i'm pretty sure i can still get a diferent value of m if x was any other number/

[acoustic spade]

try ig

[gleaming perch]

k k

[acoustic spade]

ig u might already put x=1,3 cause they are same value

82/3 if x=1,3,

[gleaming perch]

they only meet once if m = 2

[acoustic spade]

forgot about the graph

lol good

[gleaming perch]

lord save me im in 10th grade i dont know shit about deriative

okay i'm gonna leanr

[acoustic spade]

lol

ur 10th?

[gleaming perch]

yea

[gritty shuttle]

Bro what

[gleaming perch]

studying abroad

[gritty shuttle]

I am also in 10 th but what are you doing with derivative?

[gleaming perch]

i have a friend that said id need derivative for that :[

[gritty shuttle]

This question

[gleaming perch]

yeah

[gritty shuttle]

Hmmmm I would like to try algebra in this

[gleaming perch]

o7

yessir

[acoustic spade]

dont try to use derviative if u dont know man

just solve it normally

[acoustic spade]

ur 18 years old and 10th frade wtf

[gleaming perch]

i thought so

[acoustic spade]

use log if possible dont go for dervative

[gritty shuttle]

No this year I will be 18 in 11 th grade

I was pushed back in school

Due to some ass reason

[acoustic spade]

even u cant use log in this qn ig

[acoustic spade]

which country man

[gritty shuttle]

Yea it has the plus in there

[gritty shuttle]

India

[acoustic spade]

bhai kyu fail huwe 🙁

[gleaming perch]

yeahh i thought about that too. the concept of those are still somewhat out of grasp for mr

[gritty shuttle]

FAIL NHAI HUAA

I was pushed back

I didn't fail

[acoustic spade]

teachers se

[gritty shuttle]

Bruh I know adv calc

Of jee mains and shit

[acoustic spade]

bhai tu mujese bhe bada hai our me 12th hu 😭

[gritty shuttle]

You can see me on other forum solving calculas

[acoustic spade]

ik

[gritty shuttle]

And helping other yea

[acoustic spade]

yeah ik it

[gleaming perch]

indians are so cool lol. the smartest person i personally know is indian

[acoustic spade]

yeah cause sh*t jee existss lol

[gleaming perch]

whats jee

[acoustic spade]

we have some good ass qns man

[acoustic spade]

its a exam

[gleaming perch]

oooo

yeah read indian math tests once

[acoustic spade]

i am indian -_-

[gleaming perch]

oo i know and i think thats rlly cool

[gritty shuttle]

Huh

Damn

[hallow rivet]

uhhhh

[gritty shuttle]

am a solvin it

yea no way

transdencial equation

[gleaming perch]

thought about that too

drew out too many paths

[gritty shuttle]

Approx works

[hallow rivet]

this question was about noticing the symmetricity around x = 2... it wasnt meant to be solved systematically cause welp transcendal equations......
noticing how the lhs is sum of exponent so always positive and symteric about x = 2 and similarly quadratic is also symetrical bout it 2... we just put x =2 and minmax m so the graphs of both just touch each other

[gleaming perch]

im thinking mx(4-x) = - mx^2 + 4mx + 0
axis of symmetry = -b/2a = -4m/2*m = 2 -> symmetry about x=2

[gleaming perch]

initial path

[hallow rivet]

yep u r on right track

[gleaming perch]

this is exactly what i want

i just

https://tenor.com/view/dexter-james-doakes-surprise-mf-can't-prove-it-gif-7408854170742055240

i cant prove it how the lhs' graph is quadratic?

[hallow rivet]

alright fuh it lets prove it :)

lemme do it on ms paint

[gleaming perch]

i understand the x's highest power is 2.. but it kinda is not my familiar ax^2 + bx + c :(

[gleaming perch]

cool lets do it!

i just noticed on the lhs it stays the same when i replace x with 4-x

i think

[hallow rivet]

we gotta use the x = 2 "guess" but we will prove it

if we can prove at LHS >= 6 equality holding at only x = 2 and RHS <= 6 also at x = 2... we could effectively prove LHS = RHS = 6 at only x =2... alright

Now,
lets assume 3^{(x-1)^2} + 3^{(x-3)^2} >= 6
let x-2 = t
3^{(t+1)^2} + 3^{(t-1)^2} >= 6
3^{t^2 + 1} * (3^{2t} + 3^{-2t}) >= 6
3^{t^2} (3^{2t} + 3^{-2t}) >= 2

NOW,
we know for a>0
a + 1/a >= 2
with eqality holding at a = 1
Notice
(3^{2t} + 3^{-2t}) is of same form so
(3^{2t} + 3^{-2t}) >= 2
with equality holding at 3^(2t) = 1 or at t = 0
(3^{2t} + 3^{-2t}) >= 2
multiplying both sides with 3^{t^2} -- we can do this because 3^{t^2} > 0 for all real t
3^{t^2} (3^{2t} + 3^{-2t}) >=2 * 3^{t^2}

now t^2 >= 0
implies
3^{t^2} >= 1
2* 3^{t^2} >= 2 substituting this in the above result

3^{t^2} (3^{2t} + 3^{-2t}) >= 2
this is exactly the thing we assumed

hence
3^{(x-1)^2} + 3^{(x-3)^2} >= 6

Now we know this term cant be smaller than 6 and that happens at x = 2... now to get exactly one solution with quadratic on left
mx(4-x) <= 6 at x = 2 cause welp the fuckin maxima
fuh it lets prove it ahhhhhh

f(x) = x(4-x) the maxima using f'(x) = 4-2x = 0 gives it is exists at x= 2 putting the value
max( f(x)) = 2*(4-2) or 4
now the max value of mf(x) will be 4m

we need LHS = RHS so
4m = 6
or m = 3/2

@gleaming perch

i might have fucked up somewhere lemme know

[gleaming perch]

interpreting interpreting

what is a

[hallow rivet]

thats a general result for any real number > 0

a+ 1 /a >= 2

[gleaming perch]

oo

god thank u smmmm @hallow rivet

words cant describe how grateful i am

thank u

[hallow rivet]

welcome

[gleaming perch]

[hallow rivet]

.close

@gleaming perch

[gleaming perch]

yess

.close