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C2 differentiation problems

Hi guys, can anyone help me with the detailed steps to solving this..I mean I kinda get the gist, I just don't properly understand what I'm doing.

so this question says

A curve has the equation y= x + 1/x

Use calculus to show that the curve has a turning point at x= 1

Show also that this point is a minimum.

Okay, so firstly I began by differentiating to get
dy/dx = 1-x^-2

and I didn't know what to do from this point.....

should I make dy/dx = 0
so I end up with 1-1/x^2 = 0
then 1/x^2 = 1
then x^2 = 1
then x = sqrt1
then x = + or - 1

?

I mean, im confused. I know that i need to find d2y/dx2 to prove that its a minimum, which i did and i got 2x^-3
then i tried to sub 1 into d2y/dx2 and i got 2 which is >0 therefore its a minimum.

I just don't understand what to properly do.

The next problem says:

A curve has gradient given by dy/dx = x^2 -6x + 9
Find d2y/dx2

Show that the curve has a stationary point of inflection when x = 3

i found d2y/dx2 as 2x-6

and then I didnt know what to do here either...

I subbed x = 3 into dy/dx and d2y/dx2 and got x = 0 for both and then said it was a point of inflection as the sign doesnt change but I just don't know how to lay it out...

Can someone please help
(edited 9 years ago)
4325963 wednesday_adyou've
Hi guys, can anyone help me with the detailed steps to solving this..I mean I kinda get the gist, I just don't properly understand what I'm doing.

so this question says

A curve has the equation y= x + 1/x

Use calculus to show that the curve has a turning point at x= 1

Show also that this point is a minimum.

Okay, so firstly I began by differentiating to get
dy/dx = 1-x^-2

and I didn't know what to do from this point.....

should I make dy/dx = 0
so I end up with 1-1/x^2 = 0
then 1/x^2 = 1
then x^2 = 1
then x = sqrt1
then x = + or - 1

?

I mean, im confused. I know that i need to find d2y/dx2 to prove that its a minimum, which i did and i got 2x^-3
then i tried to sub 1 into d2y/dx2 and i got 2 which is >0 therefore its a minimum.

I just don't understand what to properly do.

The next problem says:

A curve has gradient given by dy/dx = x^2 -6x + 9
Find d2y/dx2

Show that the curve has a stationary point of inflection when x = 3

i found d2y/dx2 as 2x-6

and then I didnt know what to do here either...

I subbed x = 3 into dy/dx and d2y/dx2 and got x = 0 for both and then said it was a point of inflection as the sign doesnt change but I just don't know how to lay it out...

Can someone please help

Hi, the first part you've done seems right. For the second part, I think you just have to equate dy/dx to 0. So 2x-6=0 and solve for x which would be 3
Original post by totoro1997
Hi, the first part you've done seems right. For the second part, I think you just have to equate dy/dx to 0. So 2x-6=0 and solve for x which would be 3



but doesnt the question already tell me its 3?
Original post by wednesday_adams
but doesnt the question already tell me its 3?

Oops sorry my bad. I think you actually equate dy/dx to 0 so x^2-6x+9=0. Then you solve this equation by factorisation which would be (x-3)(x-3). Therefore x=3. Since there are no other solutions, the point where x=3 is the only stationary point. When you substitute x into d2y/dx2, it should equal 0 suggesting that it is the point of inflection
(edited 9 years ago)
Reply 4
Original post by totoro1997
Oops sorry my bad. I think you actually equate dy/dx to 0 so x^2-6x+9=0. Then you solve this equation by factorisation which would be (x-3)(x-3). Therefore x=3. Since there are no other solutions, the point where x=3 is the only stationary point. When you substitute x into d2y/dx2, it should equal 0 suggesting that it is the point of inflection



The 2nd derivative could be 0 at a max or min, so that doesn't prove you have an inflection!

The correct idea is that dy/dx doesn't change sign - it is always +ve or 0 so the stationary point is actually a point of inflection.
Reply 5
Original post by davros
The 2nd derivative could be 0 at a max or min, so that doesn't prove you have an inflection!

The correct idea is that dy/dx doesn't change sign - it is always +ve or 0 so the stationary point is actually a point of inflection.


The correct idea is that d^2y/dx^2 is 0 and does change sign - it is a point of inflection

For the first derivative:

When dy/dx is 0 and change sign - this point is a stationary point and maximum
when chages from + to - (the d^2y/dx^2<0 here) and minmum when changes from - to +
(d^2y/dx^2>0 here)
Reply 6
Original post by ztibor
The correct idea is that d^2y/dx^2 is 0 and does change sign - it is a point of inflection

For the first derivative:

When dy/dx is 0 and change sign - this point is a stationary point and maximum
when chages from + to - (the d^2y/dx^2<0 here) and minmum when changes from - to +
(d^2y/dx^2>0 here)


You can use either condition here.

Fundamentally, I agree with you - the defining feature of an inflection is a change in the sign of the curvature, which is driven by the sign of f''(x).

However, given that the question establishes that we have a stationary point, it can only be a max, min or inflection, and since f'(x) doesn't change sign, that rules out the max and min :smile:

I'm not actually sure what they expect for the C2 syllabus - I thought inflections had been dropped because too many people found them confusing :smile:

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