More differentiation, lol! (1 Viewer)

[lookoutastroboy]

Hey guys,

Got stuck on another question that has 2 answers - very weird... hope someone can see the flaw in it somewhere

Anyway, the question was to find the stationary points on the curve y = x squared - x cubed.

The two points were found to be (0,) and (2/3, 4/27)

i put:

when x = 4/27 then x 0 2/3 1
y(dash) 0 0 -ve

and when x = 4/27 then x -1 2/3 -1
y(dash) -ve 0 -ve

the answer is that it is a maximum t.p but arent these two different answers?

btw i havent done horizontal points of inflection yet so is that part of the problem as well?


p.s anyone know how to differentiate this quickie: f(x) = 4x divided by 2 sqrt x



Thanks in advance,
lookoutastroboy

 

[x.Exhaust.x]

p.s anyone know how to differentiate this quickie: f(x) = 4x divided by 2 sqrt x

Thanks in advance,
lookoutastroboy

Use quotient rule:

f'(x) = vu' - uv' / v^2
f'(x) = 4(2 sqrt x) - 4x(x^-1/2) / (2 sqrt x)^2

You can do the rest.

 

[ninetypercent]

find the stationary points on the curve y = x squared - x cubed.
I think you made a careless error. Using the second derivative:





for stationary points





OR



> 0

(0,0) is a minimum turning point

< 0

(2/3,4/27) is a maximum turning point

 

[lookoutastroboy]

Use quotient rule:

f'(x) = vu' - uv' / v^2
f'(x) = 4(2 sqrt x) - 4x(x^-1/2) / (2 sqrt x)^2

You can do the rest.

the answer seems to be 2 divided by sqrt 2x

but i have got the right answer but cannot seem to simplify it to be that answer,
any tips?




thanks for the help both of you,
lookoutastroboy

 

[ninetypercent]

f(x) =

f'(x) =













the numerator



the answer seems to be 2 divided by sqrt 2x

I presume you mean 2 divided by 2rootx.

this is also equal to:

the denominator



so your answer is correct.

 

[lookoutastroboy]

no answer

 

[lookoutastroboy]

ty

 

[ninetypercent]

oh sorry. Here's the correct solution:














EDIT: oops didn't realise the OP had found the solution

 

[Dumbledore]

f(x) =

f'(x) =













the numerator



the answer seems to be 2 divided by sqrt 2x

I presume you mean 2 divided by 2rootx.

this is also equal to:

the denominator



so your answer is correct.

uhhh.... isn't 4x/2sqrt(x) just 2sqrt(x) which is easy to differentiate?

 

[ninetypercent]

uhhh.... isn't 4x/2sqrt(x) just 2sqrt(x) which is easy to differentiate?

yeah, that's right.. but the OP wanted to know how to simplify the solution which comes out of the quotient rule.

 

[Dumbledore]

yeah, that's right.. but the OP wanted to know how to simplify the solution which comes out of the quotient rule.

why would he want to use the quotient rule in the first place...?

 

[ninetypercent]

why would he want to use the quotient rule in the first place...?

I dunno.

 

[lookoutastroboy]

lol thanks for everyone's responses - i really appreciated the help and effort

just one more question - when you do maximum/minimum turning point questions,
my teacher said to try in the derivatives for x on either side of a stationary point x. I did this for the question below,

as shown above - ninetypercent - i havent used that method before - bear with me please =)

anyways i did it again using that method - this was purely experimental

anyway its y = x squared - x cubed

when x = 2/3, y = 4/27 (one of the two stationary points on the curve)

1st time:

x 0 2/3 1
y (dash) 0 0 -ve - which has no max/min t.p

and

2nd time:

x 1/3 2/3 1
y (dash) +ve 0 -ve which is indeed a max t.p and the correct answer

well both ways place lower and higher values on either side of the s.p so why does this come out with 2 different answers?? is there something wrong with zero in this case?





can someone explain this?




thanks,
lookoutastroboy

 

[ninetypercent]

umm.. you need to pick numbers that are closer (sometimes even very close depending on the graph) to the stationary point. If you pick a number that is far away (such as 1) then the result will not be correct because at x =1, the gradient will be different to what it's like at x =1/3. This is a case that occurs with functions that are not even or odd.

One of the reasons why I prefer to use the second derivative

 

[Timothy.Siu]

umm.. you need to pick numbers that are closer (sometimes even very close depending on the graph) to the stationary point. If you pick a number that is far away (such as 1) then the result will not be correct because at x =1, the gradient will be different to what it's like at x =1/3. This is a case that occurs with functions that are not even or odd.

One of the reasons why I prefer to use the second derivative

You don't HAVE to pick a number that is closer. You just can't pick a "wrong" point. e.g. if a curve has 2 stationary points, lets say one at x=2 and one at x=5, to test the stationary point at x=2, you have to pick one value on either side, BUT the value of x>2 side has to be x<5, because that is where another stationary point occurs.

The reason why you had two different answers when you tested your stationary point is because x=0 is ALSO a stationary point.