IsTim
  • IsTim
how to graph g(x)=3x^4=3x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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IsTim
  • IsTim
I'm looking thru my old notes now, but I can't find anything that could help.
IsTim
  • IsTim
I simplified the equation so: g(x)=3x^2(x^2+1)
IsTim
  • IsTim
I was thinking apq, but I don't know if that applies to this.

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More answers

sasogeek
  • sasogeek
why do u have 2 equal signs to one function?
IsTim
  • IsTim
Oops.
IsTim
  • IsTim
g(x)=3x^4-3x^2 My bad.
IsTim
  • IsTim
\[g(x)=3x ^{4}-3x ^{2}\] If you want a cleaner version.
anonymous
  • anonymous
use a graphing calc
anonymous
  • anonymous
3x^2(x^2-1) you plug in points frankly :-/ it looks like a x^2 graph.
IsTim
  • IsTim
So I just plug in values? That's feels "brute". There's no other way?
IsTim
  • IsTim
@ Mario; I'm studying for an exam. I don't get those.
anonymous
  • anonymous
we get to use graph calcs on my exams
anonymous
  • anonymous
plug in pounts. their are zeroes at 0, 1,-1.
anonymous
  • anonymous
the rest is just brute force plugging. yes. that's how it works.
IsTim
  • IsTim
Oh well. I was looking for some equation rearranging. Would that work?
IsTim
  • IsTim
Lucky Mario.
anonymous
  • anonymous
yet i still did poorly lol
JamesJ
  • JamesJ
IsTim
  • IsTim
IF possible, please give an explaination of how to dervie the graph from teh equatioon.
JamesJ
  • JamesJ
Ok. If g(x)=3x^4-3x^2 first we'd like to know the zero; i.e., the intercepts on the x-axis. Setting g(x) = 0, we have \[ 3x^2(x^2 - 1) = 0 \] Hence \[ x = 0, \pm 1 \] Next, what's the y-intercept: y = g(0) = 0. Next, critical values ...
JamesJ
  • JamesJ
\[ g'(x) = 12x^3 - 6x = 0 \] if and only if \[ 6x(2x^2 - 1) = 0 \] i.e., \[ x = 0, \pm 1/\sqrt{2} \] The second derivative is \[ g''(x) = 36x^2 - 6 = 6(6x^2 - 1) \] ...
JamesJ
  • JamesJ
It's not hard now to show that x = 0 must be a local max and \[ x = \pm 1/\sqrt{2} \] are local mins. So now we have the behavior of the function in the interval [-1,1] What happens outside that?
JamesJ
  • JamesJ
The next thing we observe is that g(x) is an even function, g(-x) = g(x) meaning the function is symmetric about the y-axis. As \[ g(x) = 3x^2(x^2 - 1) \] it is clear that for \( x > 1 \), \( g(x) > 0 \) and as \( x \rightarrow \infty \), \( g(x) \rightarrow \infty \). We now have all the information we need to draw the graph and it's consistent with the picture I posted above. Make sense?
IsTim
  • IsTim
It's higher level information that I don't understand, but I think if I read thru it a bit more, I'll understand. Thank you very much.

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