girlworld_club
Member
solving for x. I'm having a brain freeze with this question.
How to solve x*e^2x=4 (1 Viewer)
- Thread starter girlworld_club
- Start date
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
2^x^2+1=6
Squar3root
realest nigga
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
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Not everyone remembers everything
Squar3root
realest nigga
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
Do you mean to say that we are never going to use techniques in an essay ever again?!?!
hit patel
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Hahahahahaha what usefulness?
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...Do you mean to say that we are never going to use techniques in an essay ever again?!?!
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
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I certainly didn't.Do you mean to say that we are never going to use techniques in an essay ever again?!?!
Squar3root
realest nigga
graph is always concave down. graph x^2 +1 and then recpricate the y values and you can see always concave down
i really think bored should invest in a [sarcasm] [/sarcasm] feature to make things more clear
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
But did you differentiate TWICE?
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Squar3root
realest nigga
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)
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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Oops, don't know how that happened. It's edited now.