anonymous
  • anonymous
g'(x)=8x^2-9x+6 what is the x value of the inflection point of g(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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apoorvk
  • apoorvk
|dw:1332441574646:dw| so at point of inflection of g(x), both g'(x) and g''(x) will be zero (it is basically the point where the curve's slope is changing its tendency or sign)
anonymous
  • anonymous
Take a second derivative and set it equal to zero. Recall, the second derivative set equal to zero describes inflections points in the original function.
anonymous
  • anonymous
what is an inflection? i didn't get there yet but the concept seems simple enough

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apoorvk
  • apoorvk
no.. okay g'(x) doesnt have to be zero necessarily, but g''(x) will be zero.
anonymous
  • anonymous
9/16?
anonymous
  • anonymous
|dw:1332441911446:dw|
anonymous
  • anonymous
\[g'(x)=8x^2-9x+6 \]Then: \[g''(x)=16x-9\]\[0=16x-9\]add 9 to both sides \[9=16x\]divide 16 on both sides \[9/16=x\] So the point of inflection occurs when x=9/16
anonymous
  • anonymous
the point of inflection of a function is just it's next order derivative set equal to zero?
anonymous
  • anonymous
i did it correctly then... sweet :)
anonymous
  • anonymous
An inflection point is a location on a curve where the concavity shifts from concave up to concave down or vice versa.|dw:1332442065876:dw| Notice, on the left - tangent lines are drawn above the curve and on the right tangent lines are drawn below the curve. On the left of the inflection point the graph is concave up and on the right of the inflection point the graph is concave down. Remember - "concave down makes a frown" :)
anonymous
  • anonymous
TheFigure, are you a teacher? lols
anonymous
  • anonymous
"the point of inflection of a function is just it's next order derivative set equal to zero?" Yes. :) Because it is the rate of change of the rate of change. If we're talking speed in a vehicle, the rate of change of position is velocity (first derivative), the rate of change velocity is acceleration (second derivative).
anonymous
  • anonymous
I am a teacher, although my certification is only good through Algebra 1. :) I think I'm right with the answer that I gave above.
anonymous
  • anonymous
Crap - and I said that backwards! What I wrote above. On my picture the left is concave down and the right is concave up! Oops! :)
anonymous
  • anonymous
You're pretty smart
anonymous
  • anonymous
Thanks!!
anonymous
  • anonymous
Here's another question I have: The function F(x)=e^(-3x^2) has 2 inflection points. Find the x-coordinate of the inflection point when x>0, correct to 2 decimal places.
anonymous
  • anonymous
Yoza! So let's start taking that first derivative!
anonymous
  • anonymous
\[F(x)=e^{-3x^2}\]\[F'(x)=e^{-3x^2}\times(-3x^2)\times(2)\]and we can clean that up a bit\[F'(x)=-6x^2e^{-3x^2}\]
anonymous
  • anonymous
Now for the second derivative...
anonymous
  • anonymous
\[F'(x)=-6x^2e^{-3x^2}\]Time for a product rule!\[F''(x)=(-6x^2)\times (e^{-3x^2})\]\[F''(x)=(-6x^2)\times d/dx(e^{-3x^2})+(e^{-3x^2})\times d/dx(-6x^2)\]
anonymous
  • anonymous
and that can totally be cleaned up... :)
anonymous
  • anonymous
\[F''(x)=-6x^2e^{-3x^2}\times(-3x^2)\times(2)+(-12xe^{-3x^2})\] \[F''(x)=36x^4e^{-3x^2}-12xe^{-3x^2})\] Whew!
anonymous
  • anonymous
Now, we can set that thing equal to 0 and solve for x....
anonymous
  • anonymous
\[0=36x^4e^{-3x^2}-12xe^{-3x^2}\] Well, let's pull a GCF first... \[0=12xe^{-3x^2}(3x^3-1)\]Now we can use the Zero Product Property and set each factor to 0.
anonymous
  • anonymous
wow! this is a looong problem.
anonymous
  • anonymous
\[0=12xe^{-3x^2}\]\[0=12\times x \times e^{-3x^2} \]We cause use the Zero Product Property some more here \[0=12\]\[0=x\]\[0=e^{-3x^2} \]\[0=3x^3-1\] So, let's examine what we've got... 0=12 we throw out because it is false x=0 would be great, but the question asked x>0 so we have to throw it out too.
anonymous
  • anonymous
Yes, this is a long one!
anonymous
  • anonymous
As for the last two answers: \[0=e^{-3x^2}\]When we try to solve for x:\[ln(0)=ln(e^{-3x^2})\]We end up undefined, so we have to throw this one out too...
anonymous
  • anonymous
\[0=3x^2-1\]So the answer we're looking for must be hidden here. Solve for x. \[0=3x^2-1\]Add 1 to both sides \[1=3x^2\]Divide 3 on both sides \[1/3=x^2\]Square root both sides \[\pm \sqrt{1/3}=\sqrt{x^2}\]And, because they asked x>0, (a million years ago when we first started this question) we throw out the negative answer. So finally \[x=\sqrt{1/3}\] \[x \approx .5773502692\]Although they just wanted it correct to 2 decimal places, so: \[x \approx .58\]
anonymous
  • anonymous
that is incorrect
anonymous
  • anonymous
srsly?
anonymous
  • anonymous
yeah lol
anonymous
  • anonymous
Uh... what did we miss?
anonymous
  • anonymous
too much data haha
anonymous
  • anonymous
Hmmm... lemme check it out
anonymous
  • anonymous
take ur time lol
anonymous
  • anonymous
I found it! Ha!
anonymous
  • anonymous
It was a copy error! Totally my fault!!
anonymous
  • anonymous
ok good
anonymous
  • anonymous
It should be: \[0=3x ^{3}-1\]and I read it as:\[0=3x ^{2}-1\] So try: \[x=\sqrt[3]{1/3}\]\[x \approx .6933612744\]or just \[x \approx .69\]
anonymous
  • anonymous
thats incorrect too :(
anonymous
  • anonymous
ack! that doesn't look right either
anonymous
  • anonymous
I've got to head out now, but I'll come back to take a look at it later. I'd say - ask it again in the thread and see what others say. It's a good question, I'm sorry I couldn't figure it out in a timely manner... lol :) Good luck, friend!!
anonymous
  • anonymous
its ok. it is a complex problem. thanks for the time and effort.
anonymous
  • anonymous
Oh wow... I've been using the chain rule incorrectly from the start.
anonymous
  • anonymous
Chain Rule: F(x)=f(g(x)) F'(x)=f'(g(x)) * g'(x)
myininaya
  • myininaya
So g' is given to us g'(x)=8x^2-9x+6 right? And we want to find the inflection points of g, correct?
anonymous
  • anonymous
we're not on that problem
anonymous
  • anonymous
i posted another one in the responses above ^^^^
myininaya
  • myininaya
Too many responses lol
anonymous
  • anonymous
The function F(x)=e^(-3x^2) has 2 inflection points. Find the x-coordinate of the inflection point when x>0, correct to 2 decimal places.
myininaya
  • myininaya
Ok thanks... \[F(x)=e^{-3x^2}\] \[F'(x)=(-3x^2)'e^{-3x^2} =-6xe^{-3x^2}\] \[F''(x)=-6(xe^{-3x^2})'=-6(1 \cdot e^{-3x^2}+x \cdot (-3x^2)'e^{-3x^2})\] \[F''(x)=-6(e^{-3x^2}+x(-6x)e^{-3x^2})=-6e^{-3x^2}(1+x(-6x))\] \[F''(x)=-6e^{-3x^2}(1-6x^2)\]
myininaya
  • myininaya
Now set F''=0 and solve for x
myininaya
  • myininaya
Hint: \[-6 \neq 0 ; e^{-3x^2} \neq 0 l but 1-6x^2=0 \text{ for two values of x}\]
myininaya
  • myininaya
you want that value of x that is greater than 0 according to your instructions
myininaya
  • myininaya
I leave solving that algebraic equation to you mario
anonymous
  • anonymous
+- sqrt(1/6) ???
myininaya
  • myininaya
\[1-6x^2=0\] \[1=6x^2 \] \[\frac{1}{6} =x^2\] \[x=\pm \sqrt{\frac{1}{6}}=\pm \frac{\sqrt{1}}{\sqrt{6}}=\pm \frac{1}{\sqrt{6}}=\pm \frac{\sqrt{6}}{6}\] f''(0) f''(1) ---|-----|----- -.41 .41 - + SO since the concavity switches at x approx. .41 then \[x=\frac{\sqrt{6}}{6} \text{ is an inflection point } \]
myininaya
  • myininaya
Well is the x value of the inflection point
myininaya
  • myininaya
Got it?

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