Need help with graph HSC question (1 Viewer)

BlueGas · Sep 2, 2015

I have no clue on how to do questions like these with no equation.

AztecWarrior · Sep 2, 2015

When f'(x)=0, stationary points occour,
I think POI occour when there is a turning point on f'(x), meaning POI at x=1

psyc1011 · Sep 2, 2015

AztecWarrior said:
When f'(x)=0, stationary points occour,
I think POI occour when there is a turning point on f'(x), meaning POI at x=1

AztecWarrior strikes again! It should be 'occur'. It's not like 'humour', 'favour' or 'valour'.

AztecWarrior · Sep 2, 2015

psyc1011 said:
AztecWarrior strikes again! It should be 'occur'. It's not like 'humour', 'favour' or 'valour'.

Spelling is never my strong point, maths is

psyc1011 · Sep 2, 2015

AztecWarrior said:
Spelling is never my strong point, maths is

Please demonstrate that! Provide an adequate answer that would guide OP into the light!

Your answer has an odour of uncertainty, thereby illuminating your incompetence in this topic

AztecWarrior · Sep 2, 2015

psyc1011 said:
Please demonstrate that! Provide an adequate answer that would guide OP into the light!

Your answer has an odour of uncertainty, thereby illuminating your incompetence in this topic

Whose a bit salty tonight???

AztecWarrior · Sep 2, 2015

AztecWarrior said:
When f'(x)=0, stationary points occour,
I think POI occour when there is a turning point on f'(x), meaning POI at x=1

And with f'(x) increasing, so does f(x)

InteGrand · Sep 2, 2015

BlueGas said:
I have no clue on how to do questions like these with no equation.

$The given graph is a graph of the derivative of the function $f$. In other words, the given graph tells us the slope of the function at any point. Also, it might help you to notice that the graph given looks like that of a cubic, so our graph of $y=f(x)$ should look like a quartic (you should be familiar with graphs of polynomials). Also keep in mind that the slope of the $f^\prime $ graph is the second derivative of $f$.$

$We have $f(0)=0$, so the graph of $y=f(x)$ passes through the origin. From the given graph of $y=f^\prime (x)$, we can see that the zeros of the function $f^\prime$ are $1$ and $3$. Therefore, $y=f(x)$ has stationary points at $x=1$ and $x=3$. We can see that $f^\prime (x)>0$ when $x$ is just less than $3$, and $f^\prime (x)<0$ when $x$ is just greater than 3, so $x=3$ must be a local maximum of $y=f(x)$.$

$We can see that $f''(0)=0$ (i.e. the slope of the derivative is 0 at $x=0$), and $f'$ changes slope (so $f$ changes concavity (sign of its second derivative)) about this point, so the stationary point of $f$ at $x=0$ is actually a stationary point of inflexion (similar in shape to that of $y=x^3$ at $x=0$).$

$The other potential inflexion point for $y=f(x)$ is at $x=2$, because we can see that here, $f''(0)=0$ (since the slope of the derivative graph is 0 here). Since the slope of $f^\prime$ clearly changes sign about this point, $f$ changes concavity about this point, so we have a point of inflexion for $f$ at $x=2$.$

$Note that as $x\to -\infty$, $f^\prime (x) \to +\infty$, so the slope of $f$ gets arbitrarily large and positive the more you go along the negative $x$ axis. This means $f(x)$ will be very negative in value as $x\to -\infty$, because $f(0)=0$ and $f$ is increasing for $x<0$. Also, as $x\to \infty$, $f^\prime (x) \to -\infty$, so the slope of the graph of $y=f(x)$ becomes arbitrarily large and negative as $x\to \infty$. Also, make sure you draw $f$ as increasing for all $x\leq 3$, and decreasing for all $x>3$, due to the sign of $f^\prime (x)$ in these intervals. Further, the slope of the $y=f(x)$ graph is 2 at $x=0$. Your final graph should have this sort of shape (I have sketched the derivative in blue, and the function $y=f(x)$'s graph in red):$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

psyc1011 · Sep 2, 2015

AztecWarrior said:
Whose a bit salty tonight???

You know I speak the truth

#verum #tautology

The additions to your initial answer implies defeat!!

GGWP

AztecWarrior · Sep 2, 2015

psyc1011 said:
You know I speak the truth #verum #tautology

The additions to your initial answer implies defeat!!

GGWP

It shows that I learn from my mistakes unlike you.

thanks @InteGrand for a great solution

Ello1234 · Sep 2, 2015

psyc, my friend, please be nice. Remember 'sparkling smiles'. hahaha

AztecWarrior · Sep 2, 2015

Ello1234 said:
psyc, my friend, please be nice. Remember 'sparkling smiles'. hahaha

No chance there, he's saltier than hot chips tonight

psyc1011 · Sep 2, 2015

AztecWarrior said:
It shows that I learn from my mistakes unlike you.

thanks @InteGrand for a great solution

Please enlighten me on what mistakes I done?

Ello1234 said:
psyc, my friend, please be nice. Remember 'sparkling smiles'. hahaha

Yes, that (and everything y'day) got out of hand. Apologies for my immaturity and childish behaviours. I take all direct and indirect insults back.

I'll strive to stop reviling harmless internet friends

AztecWarrior · Sep 2, 2015

psyc1011 said:
Please enlighten me on what mistakes I done?

being salty and your inconsistencies, you want to correct grammar, but not punctuation

Drsoccerball · Sep 2, 2015

psyc1011 said:
AztecWarrior strikes again! It should be 'occur'. It's not like 'humour', 'favour' or 'valour'.

You're fun at parties...

psyc1011 · Sep 2, 2015

Drsoccerball said:
You're fun at parties...

Of course, everyone appreciates my presence.

DatAtarLyfe · Sep 2, 2015

psyc1011 said:
Of course, everyone appreciates my presence.

You remind me of Sheldon XD

Need help with graph HSC question (1 Viewer)

BlueGas

Well-Known Member

AztecWarrior

Member

psyc1011

#truth

AztecWarrior

Member

psyc1011

#truth

AztecWarrior

Member

AztecWarrior

Member

InteGrand

Well-Known Member

psyc1011

#truth

AztecWarrior

Member

Ello1234

Active Member

AztecWarrior

Member

psyc1011

#truth

AztecWarrior

Member

Drsoccerball

Well-Known Member

psyc1011

#truth

DatAtarLyfe

Booty Connoisseur

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