# How to solve x*e^2x=4 (1 Viewer)

#### girlworld_club

##### Member
solving for x. I'm having a brain freeze with this question.

#### braintic

##### Well-Known Member
solving for x. I'm having a brain freeze with this question.
The x in front makes it impossible to get an exact solution using high school techniques.

You will have to use Newton's method to get an approximate solution.

#### girlworld_club

##### Member
this could be solved with logs right?

2^x^2+1=6

#### Squar3root

##### realest nigga
this could be solved with logs right?

2^x^2+1=6
Yes it could. The answer is sqrt[(ln6)/(ln2) -1] ~ 1.258

#### Squar3root

##### realest nigga
Back to your original question, yes hit it with newtons method. Use f(x) = 2x(lnx +1) - ln(4) = 0

#### Squar3root

##### realest nigga
Btw, haven't you already done the HSC?

#### HeroicPandas

##### Heroic!
Btw, haven't you already done the HSC?
Not everyone remembers everything

#### Squar3root

##### realest nigga
Not everyone remembers everything
Yeah lol. Half of physics slipped out of my head and all of English aswell

#### braintic

##### Well-Known Member
Yeah lol. Half of physics slipped out of my head and all of English aswell
Does ANYONE have ANY reason for remembering ANY English once the HSC is over?

#### Squar3root

##### realest nigga
Does ANYONE have ANY reason for remembering ANY English once the HSC is over?
Do you mean to say that we are never going to use techniques in an essay ever again?!?!

#### hit patel

##### New Member
Are you questioning the usefulness of HSC English?

Hahahahahaha what usefulness?

Do you mean to say that we are never going to use techniques in an essay ever again?!?!
Well you might when ur kids/ younger brother or sister or your employeee hahahahahhahah come and ask you to explain to them what an introduction is...

#### girlworld_club

##### Member
since this thread has derailed i also wanted to ask the values for which 1/((x^2)+1) is concave up and concave down? i would appreciate full working since i have spent ages and can not get the answer provided.

#### seanieg89

##### Well-Known Member
Do you mean to say that we are never going to use techniques in an essay ever again?!?!
I certainly didn't.

#### Squar3root

##### realest nigga
since this thread has derailed i also wanted to ask the values for which 1/((x^2)+1) is concave up and concave down? i would appreciate full working since i have spent ages and can not get the answer provided.
graph is always concave down. graph x^2 +1 and then recpricate the y values and you can see always concave down

I certainly didn't.
i really think bored should invest in a [sarcasm] [/sarcasm] feature to make things more clear

#### braintic

##### Well-Known Member
Are you questioning the usefulness of HSC English?

I guess HSC English is useful for anyone who believes that every novel is chock full of cryptic messages that, for some reason, the author is incapable of expressing directly.

And I guess it also teaches people the art of BSing .... for when you need to tell people what they want to hear rather than what you want to say.

I've never been good at this ... it feels to me like lying.
Wait ... it IS lying ... English teaches you how to be a great liar.
Or do I have cause and effect back to front? Do born liars have a head-start in getting a great HSC English mark?

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#### girlworld_club

##### Member
graph is always concave down. graph x^2 +1 and then recpricate the y values and you can see always concave down

That is not the answer, it needs to be solved algebraically because there are both concave down and concave up points. the answer says it is concave upwards for x<-1 and x>1 but when i differentiate and set it greater to zero i get a completely different answer. please help...

#### braintic

##### Well-Known Member
But did you differentiate TWICE?

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#### Squar3root

##### realest nigga
But did you differentiate TWICE?
umm not sure if this was directed at me? but yes i did differentate twice

#### Squar3root

##### realest nigga
hmmm, it appears that i may have been wrong. when i did it my way, i just drew a rough graph and assumed it was so *dusts hands* but now that i do it algebraically; let f(x)=1/x^2 +1

f''(x)= (2*(3*x^2-1))/(x^2+1)^3 [used wolframalpha caz i didn't want to make a mistake and i am too lazy to write it out by hand]

so it is concave up when f''(x) >0 when i solve this i get 1/root(3) and since f(x) is even hence the other side is the same

so from -infinity to -1/root(3) the graph is concave up; from -1/root(3) to 1/root(3) the graph is concave down and from 1/root(3) to infinity the graph is concave up. note that x= +- 1/root(3) are points of inflexion

Edit: this may help: http://www.wolframalpha.com/input/?i=1/(x^2++1)

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#### braintic

##### Well-Known Member
umm not sure if this was directed at me? but yes i did differentate twice
Oops, don't know how that happened. It's edited now.