More log help (1 Viewer)

[Avenger6]

Hi guys, i need some more help with logs and deriving them and stuff. Two questions I struggled with are below, I can get the first derivative for question 9 but am unsure of how I would get the second derivative and make it equal zero ect. And i'm completly lost on question 10 as far as making lnx/x =0 to find the stationary point.

Help greatly appreciated .

 

[lyounamu]

Hi guys, i need some more help with logs and deriving them and stuff. Two questions I struggled with are below, I can get the first derivative for question 9 but am unsure of how I would get the second derivative and make it equal zero ect. And i'm completly lost on question 10 as far as making lnx/x =0 to find the stationary point.

Help greatly appreciated .

9)
dy/dx = x . 1/x + ln x . 1 - 2x
= 1 + ln x - 2x
d^2y/dx^2 = 1/x - 2

Point of inflection occurs when the d^2y/dx^2 = 0.
i.e. 1/x -2 = 0
Therefore, x = 1/2

When x < 1/2, dy/dx > 0
When x > 1/2, dy/dx < 0

Therefore, there is a point of inflection at x = 1/2
Substitute x = 1/2 to find the corresponding y-value.

10)

dy/dx = (x . 1/x - ln x .1)/x^2
= 1- ln x/x^2
Stationary point occurs when dy/dx = 0.
i.e. when 1 - ln x = 0
1= ln x
Therefore, x = e

When x < e, dy/dx > 0
When x > e, dy/dx < 0
Therefore, there is a local maximum point when x = e.
Substitute x = e to find the corresponding y-value.

EDIT: I may have introduced some mistakes in my working out as I worked out at extremely quick pace. Sorry if I made any mistake but I am quite sure I got it right.

 

[Mark576]

9)
dy/dx = x . 1/x + ln x . 1 - 2x
= 1 + ln x - 2x
d^2y/dx^2 = 1/x - 2

Point of inflection occurs when the d^2y/dx^2 = 0.
i.e. 1/x -2 = 0
Therefore, x = 1/2

When x < 1/2, dy/dx > 0
When x > 1/2, dy/dx < 0

Therefore, there is a point of inflection at x = 1/2
Substitute x = 1/2 to find the corresponding y-value.

You tested the second derivative there, better just change it so the OP understands.

 

[lyounamu]

You tested the second derivative there, better just change it so the OP understands.

Thanks. Greatly appreciate your help.

EDIT: Are you asking me to remove that line? I am sorry I didn't fully comprehend that.

 

[Avenger6]

Sorry guys, im still a little confused as to how you got the first derivative. In question 9 for example i would have though the first derivative would have been y'=1.(1/x)-2x because you derive the log... Could you explain where I am going wrong?

 

[foram]

Sorry guys, im still a little confused as to how you got the first derivative. In question 9 for example i would have though the first derivative would have been y'=1.(1/x)-2x because you derive the log... Could you explain where I am going wrong?

y=xlnx -x^2

differentiating using product rule because x.lnx is a product of x and lnx

u=x
u'=1
v=lnx
v'=1/x

dy/dx = u'v+uv' - d/dx (x^2)
dy/dx = 1.(lnx) +x.x - 2x
dy/dx = 1.(lnx) + 1 -2x

 

[lyounamu]

Sorry guys, im still a little confused as to how you got the first derivative. In question 9 for example i would have though the first derivative would have been y'=1.(1/x)-2x because you derive the log... Could you explain where I am going wrong?

For question 9, you should use the product rule for the xlnx.

Which means that you will get x . 1/x + lnx . 1 - 2x = 1 + lnx -2x

 

[Avenger6]

Of course, why didn't i think of that lol. Now my problem lies in how you made y''=0 as when I use x=1/2 in the eqauation 1/x-2 i don't get 0 as an answer :S?

 

[foram]

Of course, why didn't i think of that lol. Now my problem lies in how you made y''=0 as when I use x=1/2 in the eqauation 1/x-2 i don't get 0 as an answer :S?

y'= (lnx) + 1 -2x
y''=(1/x) -2

at y''=0

0=(1/x) -2
2=1/x
2x=1
x=1/2

You find the x by letting y'' =0


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[Avenger6]

Ahk, i misread the equation thats why i didn't understand i though it read 1/(x-2)=0. Anyway this is my final question, howcome in question 10 it stated that 1 - ln x = 0 shouldn't it read (1-lnx)/x^2=0??? And with regard to determing its nature, how do you test for numbers below and above e...what is the numerical value for e?

 

[foram]

Ahk, i misread the equation thats why i didn't understand i though it read 1/(x-2)=0. Anyway this is my final question, howcome in question 10 it stated that 1 - ln x = 0 shouldn't it read (1-lnx)/x^2=0??? And with regard to determing its nature, how do you test for numbers below and above e...what is the numerical value for e?

good question. I was wondering about that myself...

e is a constant. type it into your calculator, or google it. it's irrational. so it goes on forever.

 

[tommykins]

Ahk, i misread the equation thats why i didn't understand i though it read 1/(x-2)=0. Anyway this is my final question, howcome in question 10 it stated that 1 - ln x = 0 shouldn't it read (1-lnx)/x^2=0??? And with regard to determing its nature, how do you test for numbers below and above e...what is the numerical value for e?

The reason why you test the numbers beside e is to determine what the gradient is beside the stationary point at x = e.

You do not really need to know the numerical value for e (on top of my head, it's around 2.71), you can just use e+1 and e-1.

Also, when testing numbers beside a stationary point to determine it's nature, test numbers close to it.

Another tip I can give you is say, a graph has a stationary point at x = 1 and x = 2

For testing the nature of x = 1, you should use x = 0 and x = 0.5, as if you use 2, the result will be 0, thus making it utterly useless as to what type of stationary point it is. For x = 2, you can use the result of x = 0.5 for testing the point at x = 2, as the result isn't 0 gradient.

Sorry if it seems confusing, if you have any other questions I'll try to clarify as best as I can.

 

[lyounamu]

Ahk, i misread the equation thats why i didn't understand i though it read 1/(x-2)=0. Anyway this is my final question, howcome in question 10 it stated that 1 - ln x = 0 shouldn't it read (1-lnx)/x^2=0??? And with regard to determing its nature, how do you test for numbers below and above e...what is the numerical value for e?

It's easy. Just type in e to the power of 1 in your calculator. You will see that e = 2.71828182...
It is true that (1-lnx)/x^2=0 but you disregard that x^2 because you don't want it to be 0 as it will make the whole fraction undefined. You just want to find the point that would make it 0 (i.e. 1-lnx = 0).

EDIT: Above post is quite smart. I like that. You should take his advice over mine.

 

[foram]

oh i see now. I was wondering about why 1- lnx/x^2=0 became 1-lnx=0! It's (1-lnx)/x^2=0. lol, i should have realised that sooner.

 

[lyounamu]

oh i see now. I was wondering about why 1- lnx/x^2=0 became 1-lnx=0! It's (1-lnx)/x^2=0. lol, i should have realised that sooner.

Sorry about that. I was quite vague in what I was trying to represent.

 

[Avenger6]

All makes sense now and ended up being the right answer which is always a good sign lol. Thanks for all the help.

 

[lyounamu]

All makes sense now and ended up being the right answer which is always a good sign lol. Thanks for all the help.

I wish my little help contributed to your understanding. Good luck in your further studies!