finding x here

>>8549023
My guess is that you're misreading the question somehow because it's not possible to explicitly solve that for x

>>8549028
Why it isn't... please I want to learn, I asked a lot of people before coming here but you are the first who replied that

>>8549028
I have the book in spanish, so I will translate:
Find the first and the second derivative(?) of the function.
Then verify them to be reasonable between their graphics.

>>8549029
Where is it in Stewart?

>>8549036
It's just asking you to differentiate the functions. Why the hell are you trying to find zeroes?

>>8549036
That's why, it's not asking you to solve for x, and as >>8549028 said it's not possible (in terms of elementary functions).

>>8549029
You can show that (restricted to the values where the function is inyective) there exists an inverse function. That doesn't mean you cabn write it as some combination of nice function.

>>8549023
You're going to need to use newtons iterative method in order to solve this. If this isn't something you have taught in your class you won't see a problem like this.

>>8549023
There is no solution.

e^x never reaches 0 and it is a increasing function. It just gets bigger.

-3x^2 highest point is 0 and then goes down, in the way quadratics do.

So they never meet.

>>8549040
Calculus, Early Trascendentals - 6th edition.
Page 181 excercise 48

>>8549043
because that way I cand find a maximum and minimums and then I can graphic

>>8549049
ur a dumb ass
f(1) = e-3 <0
f(0) =1>0
so theres a root in there

>>8549054
You can sketch a function without exact forms for the max/min points

>>8549048
Just went through Newton's iterative method and got x ~= 0.9100
There is probably at least one more root in there though.

>>8549023
it is a transcendant equation, no analytic solutions exist. Just like x=tan(x)

>>8549059
but I don't want to use a calculator
I want to do it on my own.because in the tests and exams I can't use one

>>8549023
non-elementary, might be a lambert W function case, but idk. If it's from stewart than I'm almost sure it's a numerical methods question, newton's method as someone mentioned

>>8549070
OP here, I just used Wolfram Alpha to solve it and it showed me the W thing but I don't even know that newton's method, I don't have it in my class yet

>>8549049
sure buddy

https://www.google.com/#q=y%3De%5Ex-3*x*x

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>>8549023
How do you find x? You don't.
There's no analytical solution.
The best geometric way to do that would be to use Newtons me

[math] 0 = f(x0) = f'(x0)(x1-x0) [/math]
so
[math] x1 - x0 = - \frac {f(x0)}{f'x0} [/math]
=>
[math] x1 = x0 - \frac {f(x0)}{f'x0} [/math]

>>8549097
**Newtons Method

x0 is any number you want to start with, guestimate this, the closer the guess the less approximations.

Find the tangent line to f(x) at x0
Where f'(x0) crosses the x axis, that is x1
x1 is now your new x0. Do it again. Continue until the numbers you plug back into f(x) =~0

This is all a computer does to calculate solutions.

>>8549092
https://www.google.com/#q=y%3De%5Ex-3*x*x%2Ce%5Ex%2C-3*x*x

>>8549023
[math] x = \frac{1}{\sqrt{3}} e ^ {\frac{1}{\sqrt{12}} e ^ {\frac{1}{\sqrt{48}} ...}} [/math]

>>8549104
This, but have fun evaluating that.
>inb4 this is the exact answer
His book probably wants a decimal answer to 2-3 digits, remember that calc is literally engineer tier plug and chug at this point :^(
>T. EE

>>8549097
OP here.
My teacher didn't teach me that, But it's similar to the tangent line formula

>>8549023
[math] x = \ln(3) + 2\ln(\ln(3) + 2\ln(\ln(3) + 2\ln(...))) [/math]

>>8549111
This guy manages to convey the idea in about 1 minute

https://youtu.be/1uN8cBGVpfs?t=30s

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>>8549111
In fact it's the same thing, that's what's so exciting.
[math] y = f(a) + m (x - a) [/math]
Replace x with x1 and a with x0, and m with f'(x)
y = 0 since that's literally what your equation says (and we know that happens at some point Xn,0)
[math] 0 = f(x0) = f'(x0) (x1 - x0) [/math]

It's literally using the tangent line to get very very close to the interception point, as when you get closer graphically to the point of interception, your tangent line at that point will also intercept the x axis very close to the graphical interception of your function on the x axis.

>>8549131
***
[math] 0 = f(x0) + f'(x0) (x1 - x0) [/math]

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>>8549131
Woah, I was freaking out why [math] e^x - 3x^2 =0 [/math] had three solutions on desmos, but
[math] e^x = 3x^2 [/math] only had two...

Then I zoomed out. Ecks dee.

But yeah, OP if you *didn't* learn Newton's method and this isn't a real analysis class then there'd be no way to do the problem.

>>8549097
>here's no analytical solution.
Meant algebraic
There's obviously an analytical solution.

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>>8549052
Here's how the problem looks in my book. The graphing symbol next to the problem number means you're suppose to use outside graphing software, not do it by hand. The problem is asking you to make sure to formula you compute is consistent with how the graph looks.

>>8549104
cute

had to think about it for a second

x solves the given equation iff it is a fixed point of the function [math] f(x)=e^{x/2}/\sqrt{3}[/math]

or of the function

[math] g(x)=-e^{x/2}/\sqrt{3}[/math]

the mean value theorem implies that f and g are contraction mappings on any interval [math] (-\infty, a] [/math] where f'(a)<1, i.e. [math] a<2ln(2\sqrt{3})\approx 2.5 [/math] (plotting on wolfram alpha ensures this is sufficient to include both solutions)

By the Banach fixed point theorem, the sequences f^n(1) and g^n(1) converge to the positive and negative fixed points, respectively

>>8549174
just zoomed out on the graph, and there's actually a third fixed point that you won't find with this method (it's approximately x=3.7>2ln(2\sqrt(3)), which means that f and g are not contraction mappings in a neighborhood of the point)

>>8549110
This would be from Calc I. What do you mean 'already plug and chug at this point?' What does that make the rest of the Calcs? And what would you prefer to see Calc taught as? Not arguing w/ you, just genuinely curious what you mean as I also feel the majority of college math is 'plug n chug.'

OP here
>>8549131
I actually find that fascinating
>>8549127
thank you for that video, I speak spanish but i get those terms
>>8549137
I think i just learnt it
>>8549159
I didn't know the book implied the use of a software
>>8549174
Still dont get that
>>8549196
What do you mean with 'plug n chug'
-------
Thank you everyone!! I think I'm improving here.

>>8549023
Use Newton's method. You know you can't solve any arbitrary degree 5 polynominal equation in terms of radicals, so why do you think you are going to solve e^x - 3x^2 = 0 with a single sheet of paper? What kind of solution do you think can be written on that paper? If you choose decimal approximation, then your best bet is Newton's method.

>>8549049
You're stupid. e^x = 1 when x = 0, and 3x^2 = 0 when x = 0. However, when x = 1, we have it that e^x = e < 3, but 3x^2 = 3. Using the intermediate value theorem, one can easily prove that e^x - 3x^2 = 0 for some x on (0,1).

>>8549023
Integrate at C=1 x=0 f(x)=1.

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>>8549023
The first thing you should have done is to STUDY the function exp(x)-3x^2.

You don't need to use special functions, you can first find the number of solutions. It's already a big thing. Then you can find approximations for these solutions.

You derive it once : exp(x)-6x
You derive it twice : exp(x)-6
You find variations, and you conclude thanks to the variation chart of your initial function. This is how it should be done properly.

>>8549104
>>8549123
Why not the exact solution using Lambert W function (W(z) is the solution to z=W*exp[W], see for example http://press.princeton.edu/chapters/s3-17_10592.pdf)?
x=-2W(1/(2*sqrt[3]))
And you only have to look it in a table or use one of the expansion of W.

>>8549049
close but no cigar; you want to compare e^x and 3x^2 and find the intersections.
you just provided an analysis of the equation e^x + 3x^2 = 0 for which there is no solution.

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I'm an engineer, I solve problems like this pretty much every day in my line of work.

Is that calc 1? I forget.

Either way the way the calculus classes are taught, ie differentiation (implicit, explicit, partial), integration, integration techniques and applications etc etc all throughout the calc 1-3's is extremely plug and chug. Overwhelmingly I've found that the only difference between a physics class and a math class is that sometimes the Physics class hands you a formula sheet.

Derivatives are literally plug and chug
D/dx X^n = nX^n-1..
D/dx F(g(x)) =F'(g(x))*g'(x)
"Quotient rule", "product rule".
I don't expect calc 1 studenta to derive it for themselevs but other than in math major specific calc and analysis classes I've never seen derivation or integration explained other than "rate of change and here's the formula" or "summing area / anti derivative and here's the formula".

>>8550323
>>8549196

e^x=3x^2
ln(e^x)=x
ln(3x^2) = ln(3) + ln(x^2)
Isolate, subtract, root

e^x - 3x^2 = 0

(e^x/2 - sqrt3 x) (e^x/2 sqrt3 x) = 0

(e^x/2 - sqrt3 x) = 0 or (e^x/2 sqrt3 x) = 0

Which requires the product log function

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there are no real solutions OP

must be a misprint

>>8550064
>>>/lgbt/

>>8549023
Are you familiar with the product log function?
>>8550449
wrong btw

>>8550449
engineers everyone

>>8549067
0

>>8550983
*reformatting*

[math]e^x - 3x^2 = 0[/math]
[math]e^x = 3x^2[/math]
[math]d/dx (e^x) = d/dx (3x^2)[/math]
[math]e^x = 6x[/math]
[math]d/dx (e^x) = d/dx (6x)[/math]
[math]e^x = 6[/math]
[math]ln (e^x) = ln 6[/math]
[math]x = ln 6 =1.79[/math]

Am I the only smart one in this thread? lol you guys are fucking stupid.

dont you do this pre-calc
geometric series or something.

>i never graduated highschool so i dont remember
>im only here to shitpost about global warming

>>8550064
Isn't this correct? I don't understand why this isn't right

>>8550986
if you plug in ln6 back into the equation it doesn't satisfy the initial condition you fucking traffic cone

>>8551053
0.91 works

>>8550986
can confirm this works

>>8550064
How do you just cross x and 2 out in step 2?

>>8549023
x = 0.9100075725

>>8550986
Just because two function values agree doesn't mean their derivatives have the same value...

>>8551617
That's where your wrong kiddo

if f(x) = g(x)
then f '(x) = g '(x)

>>8549049
is this bait or is /sci/ full of brainlets?

>>8549064
>probably more than 1 root

there's exactly 2 roots!!!
Pretty easy to see it desu. The parabola has a minimum at x=0 and has 2 branches pointing upwards so it will cut any function 2 times.

>>8549104
why does people use superscript notation when using exponentials (specially exponentials of matrices, functions or nested exponentials like in your case)?
It triggers my autism. Isn't it much clear to write:

[eqn]x = \frac{1}{\sqrt{3}}\exp\left(\frac{1}{\sqrt{12}}\left(\exp\left(\frac{1}{\sqrt{48}}\cdots\right)\right)\right)[/eqn]

? Or better yet

[eqn]x = \bigcirc_{i=1}^\infty\left[f_i(s)\right] \text{ with } f_i(s) = \frac{\exp(s)}{\sqrt{3\cdot4^{i-1}}}.[/eqn]

You people disgust me.

>>8551621
yes but the contrary doesn't hold true.

Fucking engineers I swear. Get off my board. GET OUT!!!

>>8549263
De que año de carrera eres? O eres underage?

En cualquier caso en primero de física hubo un chico que dijo lo mismo que este retard >>8550345
para resolver una ecuación parecida a la tuya y se quedó con el sanbenito de "el logas" para toda la carrera ya.

>>8551621
Let's say...

f(x) = 2x+1
g(x) = 4x-5

At the point where x is 3, f(x) is indeed equal to g(x). Does that mean their derivatives are the same at x=3?

>>8551653
no

[math]e^x - 3x^2 = 0[/math]
[math]e^x = 3x^2[/math]
[math]e^(x/2) = +/- 3^(1/2) * x[/math]
[math]cos(1/2) + i * sin(1/2) = +/- 3^(1/2) * x[/math]
[math]0.88 + i * 0.48 = +/- 3^(1/2) * x[/math]
[math]0.5 + i * 0.28 = x[/math]

Checking:

[math]^(0.5 + i * 0.28) = 3(0.5 + i * 0.28)^2[/math]
[math]e^(0.5) * e^(i * 0.28) = 3(0.25 + i * 0.28 - 0.08)[/math]
[math]1.65 * (cos(0.28) + i * sin(0.28)) = 0.75 + i * 0.84 - 0.24[/math]
[math]1.65 * (0.96 + i * 0.28) = 0.5 + i * 0.84[/math]
[math]1.58 + i * 0.46 = 0.5 + i * 0.84[/math]

If I didn't round the numbers would work out.

>>8551570
It just do mayne

>>8552651
>e^(x/2) = cos(1/2) + i * sin(1/2)
anon...
e^(ix) = cosx + isin(x)
let x = y/(i2)
e^(y/2) = cos(y/(i2)) + isin(y/(i2))

>>8551570
you just have to factor out the 0