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Q9c (3 Viewers)

jamesfirst

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I got similar but instead, I made the end part go above the x axis..

OMGGGGGGGGGGGG I'm losing 1 mark.......
 

_pizza

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Yep exactly right, but I'd make it less pointy lol. You wouldn't get marked down for that though, its just emphasising the transition between decreasing negative grad and increasing negative grad a bit much.
 
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mitchh81

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but since x=3 its is a point of inflexion, so the concavity would change, therefore be above the x axis
 

sinsolja

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but since x=3 its is a point of inflexion, so the concavity would change, therefore be above the x axis

Nope a point of inflexion for F(x), and in turn a stationary point for Dy/dx. - and you dont know if its a inflexion point for dy/dx
Its not meant not be Vshaped:S
 

someth1ng

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x=3 was the POI on f(x) therefore it would be an STP for f'(x)
Actually, this makes sense... I think it might go above the x-axis.
What's STP --> S Turning Point...What's the S? How can it possibly go above the x-axis? The gradient is always negative.

Is it supposed to be pointed or not?
 

jamesfirst

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What's STP --> S Turning Point...What's the S? How can it possibly go above the x-axis? The gradient is always negative.
But if it's POI there should be a concavity change shouldn't it?
 

jamesfirst

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What's STP --> S Turning Point...What's the S? How can it possibly go above the x-axis? The gradient is always negative.

Is it supposed to be pointed or not?
STP = stationary points.
 

jamesfirst

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Concavity of F(x) not f, dash (x)
If the concavity changes for f(x), doesn't that mean that f '(x) will reach positive because the concavity of f(x) has changed????


Do you get what I mean ?? Argh fk it, I probably did it wrong LOL
 

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