Stationary point Question 5 b) CSSA 2009 paper (1 Viewer)

MrBrightside · Aug 5, 2011

Hi, there

I seem to be having troubles with the factorising part after I make y' = 0.

The original question is:

Consider the curve given by y = 2x^3 - 9x^2 +12x

Find the stat points and their nature.

y' = 6x^2 - 18x +12

6x^2 - 18x +12 = 0

I divided all bases by 6

x^2 - 3x +2 = 0

when I factorise I get

(x + 3)(x - 1) = 0

x = -3 or x = 1

HOWEVER the answers do not divide their bases by 6

they just do 6(x - 2)(x - 1) = 0

x = 1 or x = 2

I'm am confused as to why I don't get the same answer, I used this method on other questions a long time ago and got them right. I don't know why its not working here.

I need an explanation as to why the 6 is left outside the brackets etc, rather than dividing everything by 6.

Thank you

SpiralFlex · Aug 5, 2011

$y = 2x^3 - 9x^2 +12x$

For stationary points, $\frac{dy}{dx}=0$

$\frac{dy}{dx}= 6x^2 - 18x +12 = 0$

$6(x^2-3x+2)=0$

$6(x-2)(x-1)=0$

$x=2, 1.$

When $x=2, y= 4$

When $x=1, y= 5$

$\frac{d^2y}{dx^2}=12x - 18$

$=6(2x-3)$

Nature:

At $x=2$

$\frac{d^2y}{dx^2}=6(2(2)-3) >0$

Hence $(2,4)$ is a minimum turning point.

At $x=1$

$\frac{d^2y}{dx^2}=6(2(1)-3) <0$

Hence $(1,5)$ is a maximum turning point.

You also had some algebra errors. They did not divide by 6 probably because you will need it to take the second derivative. It should not really matter in my opinion.

AAEldar · Aug 5, 2011

You factorised wrong.

$x^2-3x+2=0$ does NOT factorise to $(x+3)(x-1)=0$ but rather it factorises to $(x-2)(x-1)=0$ and hence that is where their answers come from.

Dividing by the common factor won't give a wrong answer.

MrBrightside · Aug 5, 2011

AAEldar said:
You factorised wrong.

$x^2-3x+2=0$ does NOT factorise to $(x+3)(x-1)=0$ but rather it factorises to $(x-2)(x-1)=0$ and hence that is where their answers come from.

Dividing by the common factor won't give a wrong answer.

but... but.. Isn't it the 3*-1 meant to = the -3x in the middle and 3 + - 1 give 2? SHIT, It's the other way 0.0 omg

I need to practice.

MrBrightside · Aug 5, 2011

SpiralFlex said:
$y = 2x^3 - 9x^2 +12x$

For stationary points, $\frac{dy}{dx}=0$

$\frac{dy}{dx}= 6x^2 - 18x +12 = 0$

$6(x^2-3x+2)=0$

$6(x-2)(x-1)=0$

$x=2, 1.$

When $x=2, y= 4$

When $x=1, y= 5$

$\frac{d^2y}{dx^2}=12x - 18$

$=6(2x-3)$

Nature:

At $x=2$

$\frac{d^2y}{dx^2}=6(2(2)-3) >0$

Hence $(2,4)$ is a minimum turning point.

At $x=1$

$\frac{d^2y}{dx^2}=6(2(1)-3) <0$

Hence $(1,5)$ is a maximum turning point.

You also had some algebra errors. They did not divide by 6 probably because you will need it to take the second derivative. It should not really matter in my opinion.

you know how you found the 2nd derivative, Do you think it is easier doing the table method using the 1st derivative to see the slopes of the gradients?

SpiralFlex · Aug 5, 2011

MrBrightside said:
you know how you found the 2nd derivative, Do you think it is easier doing the table method using the 1st derivative to see the slopes of the gradients?

Not for these types of questions. It would be convenient to find the second derivative. If you really want to test close values on the RHS and LHS, go for it.

MrBrightside · Aug 5, 2011

SpiralFlex said:
Not for these types of questions. It would be convenient to find the second derivative. If you really want to test close values on the RHS and LHS, go for it.

Nah, I see what you mean. Its quicker to do 2nd derivative for qs with easier equations. The table method is best used for log and trig qs, where decimals gets in the way of coordinates etc.

SpiralFlex · Aug 5, 2011

MrBrightside said:
Nah, I see what you mean. Its quicker to do 2nd derivative for qs with easier equations. The table method is best used for log and trig qs, where decimals gets in the way of coordinates etc.

It would be wise to also not to find the second derivative or square roots, those tricky ones. Silly mistakes can often result.

MrBrightside · Aug 5, 2011

SpiralFlex said:
It would be wise to also not to find the second derivative or square roots, those tricky ones. Silly mistakes can often result.

yeah that too.

Also when can you use the term d/dx? like to = a formula to be derived, I see it sometimes, is it when there's no y mentioned?

SpiralFlex · Aug 5, 2011

$\frac{\mathrm{d} }{\mathrm{d} x}$ is just an operator. It operates on some function of $y$ . Just take it as "simply differentiate."

MrBrightside · Aug 5, 2011

SpiralFlex said:
$\frac{\mathrm{d} }{\mathrm{d} x}$ is just an operator. It operates on some function of $y$ . Just take it as "simply differentiate."

yeah I know it means differentiate, but, i'm not sure if i'm making a mistake not using it, for example instead of saying y' could i just use d/dx, whats different between dy/dx to d/dx ? i prefer y' and y" or f'(x) and f"(x), more simpler to use, the d/dx thing throughs me off using dy/dx and d^2y/dx^2

enoilgam · Aug 5, 2011

MrBrightside said:
you know how you found the 2nd derivative, Do you think it is easier doing the table method using the 1st derivative to see the slopes of the gradients?

I did all questions in the first derivative, doing it in the second took longer for me and it increased the potential for stupid errors.

SpiralFlex · Aug 5, 2011

MrBrightside said:
yeah I know it means differentiate, but, i'm not sure if i'm making a mistake not using it, for example instead of saying y' could i just use d/dx, whats different between dy/dx to d/dx ? i prefer y' and y" or f'(x) and f"(x), more simpler to use, the d/dx thing throughs me off using dy/dx and d^2y/dx^2

When you use the operator $\frac{\mathrm{d} }{\mathrm{d} x}$ to a specific function $y$ it will result in the function of $\frac{dy}{dx}$ , the first derivative of $y$ .

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2-2x+1)$

$\frac{dy}{dx}=2x-2$

MrBrightside · Aug 6, 2011

SpiralFlex said:
When you use the operator $\frac{\mathrm{d} }{\mathrm{d} x}$ to a specific function $y$ it will result in the function of $\frac{dy}{dx}$ , the first derivative of $y$ .

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2-2x+1)$

$\frac{dy}{dx}=2x-2$

Ohhhh so d/dx is just another form of notation to say y. I get it now Thanks.

ohexploitable · Aug 6, 2011

MrBrightside said:
Ohhhh so d/dx is just another form of notation to say y. I get it now Thanks.

umm not quite

$\frac{d}{dx}$ is an operator, kinda like plus/minus/times/divide, think of it as a verb

$\frac{dy}{dx}$ is a relation, a relationship between the slope of a function and its x-coordinate, think of it as a noun

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Stationary point Question 5 b) CSSA 2009 paper (1 Viewer)

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