Mathematics Question Thread

[cxmplete]

Hi!
A few days till the maths exam and I'm still really stuck on this point:
What is the difference between a point of inflection and a point of horizontal inflection? And how would you set the working out for both?

[RuiAce]

Hi!
A few days till the maths exam and I'm still really stuck on this point:
What is the difference between a point of inflection and a point of horizontal inflection? And how would you set the working out for both?

A horizontal point of inflexion satisfies the added criteria of the first derivative equals to 0. It is a special case of the standard point of inflexion.

Because it satisfies both \( \frac{dy}{dx}=0\) AND \(\frac{d^2y}{dx^2}=0 \) it is both a stationary point AND a point where concavity changes. A typical horizontal point of inflexion is found on the curve \(y=x^3\)

[jamonwindeyer]

Thanks so much, atarnotes is incredible!

Just a PS -You are incredible! Thanks for your kind words

[Rathin]

Express y in terms of x.

[jakesilove]

Express y in terms of x.

So, we have



Assume that all 'log's are base 10

Using our log laws;



Now, recall that



So,





So,

[katnisschung]

bonjour

the derivative calculator is confusing the heck out of me so here i am!

2)find the second derivative of (first attached)

(second attached is my answer for first derivative

(third attached is my answer for the second derivative)

help can i simplify this more?!did i get it right?

thanks!

[RuiAce]

bonjour

the derivative calculator is confusing the heck out of me so here i am!

2)find the second derivative of (first attached)

(second attached is my answer for first derivative

(third attached is my answer for the second derivative)

help can i simplify this more?!did i get it right?

thanks!

Your answer is correct

[katnisschung]

could anyone show the proof that there is a minimum at (0,0)
for the graph y=x^2(e^2x)

i used the second derivative and i'm getting it to be 0.... (and its not a horizontal pt
of inflexion... )

[jakesilove]

could anyone show the proof that there is a minimum at (0,0)
for the graph y=x^2(e^2x)

i used the second derivative and i'm getting it to be 0.... (and its not a horizontal pt
of inflexion... )

Hey!



Clearly, a turning point will occur at x=0 and x=-1.

Let's test some points around x=0.




Clearly, the gradient changes signs; it goes from negative to positive! So, the point must be a minimum

[katnisschung]

hey! yeah would u be able to check that for 2nd derivative
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)

[jakesilove]

hey! yeah would u be able to check that for 2nd derivative
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)

Sure!





So, at x=0,



Since the double derivative is positive, the point is a minimum

[RuiAce]

hey! yeah would u be able to check that for 2nd derivative
i just find that its faster to find the nature than to use the first...
( i know u can use either but i'm just trying to work out where i got it wrong)

Gonna make a note on this

The second derivative is faster provided it's either easier to use, or you need to find points of inflexion later on.

If both of these conditions are not satisfied, you should just test both sides.

[cxmplete]

how would we do part iii?

[RuiAce]

how would we do part iii?

Addressed in post #351

[f_tan]

Hi, I can't seem to get the right answer for this question, can anyone please help?



Answer is $78700