g'(x)=8x^2-9x+6
what is the x value of the inflection point of g(x)

- anonymous

g'(x)=8x^2-9x+6
what is the x value of the inflection point of g(x)

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- 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

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

what is an inflection? i didn't get there yet but the concept seems simple enough

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## More answers

- apoorvk

no.. okay g'(x) doesnt have to be zero necessarily, but g''(x) will be zero.

- anonymous

9/16?

- anonymous

|dw:1332441911446:dw|

- 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

the point of inflection of a function is just it's next order derivative set equal to zero?

- anonymous

i did it correctly then... sweet :)

- 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

TheFigure, are you a teacher? lols

- 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

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

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

You're pretty smart

- anonymous

Thanks!!

- 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

Yoza! So let's start taking that first derivative!

- 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

Now for the second derivative...

- 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

and that can totally be cleaned up... :)

- 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

Now, we can set that thing equal to 0 and solve for x....

- 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

wow! this is a looong problem.

- 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

Yes, this is a long one!

- 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

\[0=3x^2-1\]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

that is incorrect

- anonymous

srsly?

- anonymous

yeah lol

- anonymous

Uh... what did we miss?

- anonymous

too much data haha

- anonymous

Hmmm... lemme check it out

- anonymous

take ur time lol

- anonymous

I found it! Ha!

- anonymous

It was a copy error! Totally my fault!!

- anonymous

ok good

- 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

thats incorrect too :(

- anonymous

ack! that doesn't look right either

- 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

its ok. it is a complex problem. thanks for the time and effort.

- anonymous

Oh wow... I've been using the chain rule incorrectly from the start.

- anonymous

Chain Rule:
F(x)=f(g(x))
F'(x)=f'(g(x)) * g'(x)

- 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

we're not on that problem

- anonymous

i posted another one in the responses above ^^^^

- myininaya

Too many responses lol

- 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

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

Now set F''=0 and solve for x

- myininaya

Hint: \[-6 \neq 0 ; e^{-3x^2} \neq 0 l but 1-6x^2=0 \text{ for two values of x}\]

- myininaya

you want that value of x that is greater than 0 according to your instructions

- myininaya

I leave solving that algebraic equation to you mario

- anonymous

+- sqrt(1/6) ???

- 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

Well is the x value of the inflection point

- myininaya

Got it?

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