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Integration

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Original post by raheem94
We have to integrate not differentiate.


I got 32/3 (10.67) as my answer for Q)36
Original post by King-Panther
I got 32/3 (10.67) as my answer for Q)36


Correct :smile:.
Original post by just george
dydx=x2−6x+2 \frac{dy}{dx} = x^{2} - 6x + 2


should be a -3x instead of -6x i think :smile:

Yeah thats what i got, soz
Original post by raheem94
Correct :smile:.


:smile:

I should be fine for 37 as well and I got - 15/4 for that...

however, I don't know how to do 38
Original post by King-Panther
:smile:

I should be fine for 37 as well and I got - 15/4 for that...

however, I don't know how to do 38


First differentiate, y = x^3 + 2x + 1, and tell what you get.
Original post by raheem94
First differentiate, y = x^3 + 2x + 1, and tell what you get.


I got y = -1/5 x + 21/5
Original post by King-Panther
I got y = -1/5 x + 21/5


correct :smile:
Original post by raheem94
correct :smile:


41) so, i differentiate,

find when dy/dx = o

then find out if max or min....

raheem, is that an arabic name?
Original post by King-Panther
41) so, i differentiate,

find when dy/dx = o

then find out if max or min....

raheem, is that an arabic name?


Yes, for Q41, first differentiate, then set dy/dx =0 to find the values of 'x'. Then check max or min.

And yes, my name is an arabic name.
Original post by raheem94
Yes, for Q41, first differentiate, then set dy/dx =0 to find the values of 'x'. Then check max or min.

And yes, my name is an arabic name.


the two values are 3 and -1

for 3, do i sub 2 and 1 respectively into dy/dx or into dy2/dx2

Mashallah, a very nice name.. mine is also arabic, fahad, which means panther, hence the user name
Original post by King-Panther
the two values are 3 and -1

for 3, do i sub 2 and 1 respectively into dy/dx or into dy2/dx2

Mashallah, a very nice name.. mine is also arabic, fahad, which means panther, hence the user name


Yes, you have got two correct values, 3 and -1.

Sub them in d2y/dx2 to see which point is a max and which point is a min.

And remember you have to sub 3 and -1 not 2 and 1.
Original post by raheem94
Yes, you have got two correct values, 3 and -1.

Sub them in d2y/dx2 to see which point is a max and which point is a min.

And remember you have to sub 3 and -1 not 2 and 1.


yeah but for 3, you sub 2 and 1, which gave me a inflection and i got a a minimum for -1
Original post by raheem94
Yes, you have got two correct values, 3 and -1.

Sub them in d2y/dx2 to see which point is a max and which point is a min.

And remember you have to sub 3 and -1 not 2 and 1.


for 42, i got y=5x+4
Original post by King-Panther
yeah but for 3, you sub 2 and 1, which gave me a inflection and i got a a minimum for -1


d2ydx2=12x−12 \displaystyle \frac{d^2y}{dx^2} = 12x-12

Substituting x=3, gives, 12(3)-12=36-12=24>0, as we get a positive value hence the point at x=3 is a minimum.

Subbing x=-1, gives, 12(-1)-12=-12-12=-24>0, as we get a negative value hence x=-1 is a maximum.

Now find the y-coordinates of both x=-1 and x=3.

I have to do my own work now, hence i can't help you on any further questions.
Original post by King-Panther
for 42, i got y=5x+4


This answer is wrong.
Original post by raheem94
This answer is wrong.


o.k thanks....
Reply 116
Avatar for Zuzuzu
Zuzuzu
12
Original post by just george
Unparseable latex formula:

f'(x)=-64x^6^3



so turning point at x=0

Unparseable latex formula:

f''(x)=-4032x^6^2



so at x=0,f′′(x)=0x=0, f''(x)=0

so not a maximum..?

i dont understand what you're trying to say? is what i said incorrect? :L


What I was trying to say was that for a local maximum, f''(x) is not necessarily strictly less than 0. It can also take the value of 0.
Reply 117
Avatar for Astinof
Astinof
Some massive nerd debate going on here lol
Original post by Zuzuzu
What I was trying to say was that for a local maximum, f''(x) is not necessarily strictly less than 0. It can also take the value of 0.


Surely that would be a point of inflection (or something like that) if it takes the value of 0.

If you think about what f''(x) actually is, its the gradient of the gradient.
The gradient of the gradient at any maximum is clearly going to be negative :smile:
Original post by just george
Surely that would be a point of inflection (or something like that) if it takes the value of 0.

If you think about what f''(x) actually is, its the gradient of the gradient.
The gradient of the gradient at any maximum is clearly going to be negative :smile:


f′′(x)=0 \displaystyle f''(x)=0 means the point can by maximum, minimum or a point of inflection.

To confirm if it is a point of inflection or not, we find third derivative, if the 3rd derivative is not equal to zero than the point is a point of inflection.

f′(x)=0 \displaystyle f'(x)=0 and f′′(x)=0 \displaystyle f''(x)=0 , but f′′′(x)≠0 \displaystyle f'''(x) \not= 0 , then the point is the point of inflection.

Simply if you get f′′(x)=0 \displaystyle f''(x)=0 you can't say it to be the point of inflection until you check it by finding the 3rd derivative.

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