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How to find the x-coordinates of a minimum turning point P?

So I did (a) & (b) but don't know how to solve (c)
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Reply 1
Avatar for Zacken
Zacken
22
Original post by Adorable98
So I did (a) & (b) but don't know how to solve (c)


Precisely when f′(x)=0f'(x) = 0.

In this case, you have f′(x)=ex2(2x+3+2x3+6x2+2x)=0f'(x) = e^{x^2}\left(2x + 3 + 2x^3 + 6x^2+ 2x\right) = 0 so the minimum turning point occurs when:

either ex2=0e^{x^2} = 0 (invalid, reject, an exponential is never 0)

or 2x3+6x2+4x+3=02x^3 + 6x^2 + 4x + 3 = 0. Now re-arrange this equation into what they want you to show.
Reply 2
Avatar for Adorable98
Adorable98
OP
15
Original post by Zacken
Precisely when f′(x)=0f'(x) = 0.

In this case, you have f′(x)=ex2(2x+3+2x3+6x2+2x)=0f'(x) = e^{x^2}\left(2x + 3 + 2x^3 + 6x^2+ 2x\right) = 0 so the minimum turning point occurs when:

either ex2=0e^{x^2} = 0 (invalid, reject, an exponential is never 0)

or 2x3+6x2+4x+3=02x^3 + 6x^2 + 4x + 3 = 0. Now re-arrange this equation into what they want you to show.


Oh, I see!! Thank you!!
Reply 3
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Zacken
22
Original post by Adorable98
Oh, I see!! Thank you!!


No worries, can you take it from here, then? :smile:
Reply 4
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Adorable98
OP
15
Original post by Zacken
No worries, can you take it from here, then? :smile:


Well I've tried to use the factor theorem to factorise it but can't find any value that would make the equation equal to zero?
Reply 5
Avatar for Zacken
Zacken
22
Original post by Adorable98
Well I've tried to use the factor theorem to factorise it but can't find any value that would make the equation equal to zero?


There's no easy solution. You have 2x3+6x2+4x+3=02x^3 + 6x^2 + 4x + 3 = 0 and you want to show that x=−3(2x2+1)2(x2+2)x = -\frac{3(2x^2 +1)}{2(x^2 +2)}.

The obvious thing to do first, is to move the 3 to the other side to get 2x3+6x2+4x=−32x^3 + 6x^2 + 4x = -3. Then divide everything by 2 to get x3+3x2+2x=−32x^3 + 3x^2 + 2x = -\frac{3}{2}. This is start to look good!

Let's factorise an xx out of the LHS: x(x2+3x+2)=−32x(x^2 + 3x + 2) = -\frac{3}{2}.

Now, can you think about what to do to get:

x=−3(2x2+1)2(x2+1)x = -\frac{3(2x^2 + 1)}{2(x^2 +1)}?
Reply 6
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Adorable98
OP
15
Original post by Zacken
There's no easy solution. You have 2x3+6x2+4x+3=02x^3 + 6x^2 + 4x + 3 = 0 and you want to show that x=−3(2x2+1)2(x2+2)x = -\frac{3(2x^2 +1)}{2(x^2 +2)}.

The obvious thing to do first, is to move the 3 to the other side to get 2x3+6x2+4x=−32x^3 + 6x^2 + 4x = -3. Then divide everything by 2 to get x3+3x2+2x=−32x^3 + 3x^2 + 2x = -\frac{3}{2}. This is start to look good!

Let's factorise an xx out of the LHS: x(x2+3x+2)=−32x(x^2 + 3x + 2) = -\frac{3}{2}.

Now, can you think about what to do to get:

x=−3(2x2+1)2(x2+1)x = -\frac{3(2x^2 + 1)}{2(x^2 +1)}?

No, I mean if we factorise x^2+3x+2 =(x+2)(x+1) but how do you get (2x^2+1)(x^2+1)?
Reply 7
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Zacken
22
Original post by Adorable98
No, I mean if we factorise x^2+3x+2 =(x+2)(x+1) but how do you get (2x^2+1)(x^2+1)?


You're going about this the wrong way. We don't want to solve for xx, we want to get an iterative formula for xx in terms of xx.

So, from:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}2x^3 + 6x^2 + 4x +3 = 0 \Rightarrow x(2x^2 + 4) + 3(2x^2 + 1) = 0 \Rightarrow 2x(x^2 + 2) = -3(2x^2 +1) \Rightarrow \cdots\end{equation*}

Reply 8
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Adorable98
OP
15
Original post by Zacken
You're going about this the wrong way. We don't want to solve for xx, we want to get an iterative formula for xx in terms of xx.

So, from:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}2x^3 + 6x^2 + 4x +3 = 0 \Rightarrow x(2x^2 + 4) + 3(2x^2 + 1) = 0 \Rightarrow 2x(x^2 + 2) = -3(2x^2 +1) \Rightarrow \cdots\end{equation*}



Oh, I see!! Thaank you!!
Reply 9
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Zacken
22
Original post by Adorable98
Oh, I see!! Thaank you!!


No worries. This was a bit of an odd question, to be honest.

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