Stationary points making me stationary here :( (1 Viewer)

[Forbidden.]

I'll write down the full questions and partial answers ...

1) Find all stationary points and inflexions on the curve y=(2x+1)(x-2)⁴

y'=uv'+vu'

u=(3x+1)
v'=4(1)(x-2)³
=4(x-2)³ (My chain rule)
v=(x-2)⁴
u'=3

y'=(3x-1).4(x-2)³ + (x-2)⁴.3
=4(3x-1)(x-2)³ + 3(x-2)⁴

....

Oh god how do I extract the x-values ?

2) Find the turning point on the curve y=3x²+6x+1 and determine its nature.

y=3x²+6x+1
y'=6x+6

dy/dx=0

6x+6=0
6x=-6
x=-1

y=3(-1)²+6(1)+1
=3-6+1
=-2

.: Stationary at point (-1,-2)

x=-2
f'(x)=-6
x=-1
f'(x)=0
x=0
f'(x)=6

It is a point of inflexion


"Wrong, it's a minimum turning point." - Says the answer


But the f'(x) values increase as x increases, why not inflexion ? Use my working to tell me please, I'm quite confused between inflexion and minimum/maximum turning point.

3) Is it inflexion or inflection ? Different textbooks and study guides ...

At this rate, I wonder how well I can compete with 2-UNITers across NSW ...

 

[twilight1412]

1) Find all stationary points and inflexions on the curve y=(2x+1)(x-2)⁴

y' = u'v + v'u
u = 2x + 1
v = (x-2)⁴

then

y' = 2(x-2)⁴ + 4(2x + 1)(x-2)³
= 2(x-2)³ { (x - 2) + 2(2x + 1) }
= 2(x-2)³ { 5x }

when y' = 0 then there is a turning pt
therefore

case one
x-2 = 0
x = 2

case 2
5x = 0
x = 0

y" = 30(x-2)² + 10(x-2)³
= 10(x - 2)² { x - 1 }
y" = 0 then you have a pt of inflection
which is when
x = 1
x = 2

so x = 2 is a turning pt as well as a pt of inflection?


when x = 0 then

y" = 20 ....

which is positive therefore minimum turning pt



2) Find the turning point on the curve y=3x2+6x+1 and determine its nature. (working on your working out)

y=3x²+6x+1
y'=6x+6

dy/dx=0

6x+6=0
6x=-6
x=-1

y=3(-1)2+6(1)+1
=3-6+1
=-2

.: Stationary at point (-1,-2)

y" = 6
there fore its a minimum
note: if you get confused then think of positive as a happy face which has a minimum turning pt

3) it is pt of inflection

can someone check my working out? for part 1

 

[Forbidden.]

Yes someone check his working out to ensure consistency and accuracy ... Plus, we have not done double derivatives yet e.g f''(x) or even f'''(x) ..

......
then

y' = 2(x-2)⁴ + 4(2x + 1)(x-2)³
= 2(x-2)³ { (x - 2) + 2(2x + 1) }
= 2(x-2)³ { 5x }
....

Hold your horses, how did you turn 2(x-2)⁴ + 4(2x + 1)(x-2)³ into 2(x-2)³ { (x - 2) + 2(2x + 1) } ?
I don't simplify after first step, because I don't like solving any further, could you at least explain in-depth how that is done ?

 

[Riviet]

But the f'(x) values increase as x increases, why not inflexion ?

It can't be an inflexion point because the sign of the gradient changes from negative at x=-2 to positive at x=0, as your working shows.

aq\aa/
aa.\_/

As you go from left to right, the gradient is negative, then is zero at x=-1, then positive at x=0.

 

[Riviet]

Hold your horses, how did you turn 2(x-2)⁴ + 4(2x + 1)(x-2)³ into 2(x-2)³ { (x - 2) + 2(2x + 1) } ?
I don't simplify after first step, because I don't like solving any further, could you at least explain in-depth how that is done ?

He took a common factor of 2(x-2)³ out.

Is it inflexion or inflection ? Different textbooks and study guides ...

Both are accepted, I prefer to use inflexion.

 

[twilight1412]

hmm so i guess my part one was slightly wrong ... im still getting used to the [ sub ] >< it tends to clog it up a lil

thanks for checking rivet ^^

well second derivative is used to find which way the curve is turning so if the curve is turning upwards then it must have a minimum
so basically in most cases

y" > 0 minimum
y" = 0 inflection
y" < 0 maximum

 

[bmc]

{2) Find the turning point on the curve y=3x²+6x+1 and determine its nature.

y'=6x+6

dy/dx=0

.: Stationary at point (-1,-2)

x=-2
f'(x)=-6
x=-1
f'(x)=0
x=0
f'(x)=6

It is a point of inflexion}

BE CAREFUL nature is found by the second derivative. Make sure you differentiate the first derivative, you forgot that

 

[-pari-]

1) Find all stationary points and inflexions on the curve y=(2x+1)(x-2)4

y'=uv'+vu'

u=(3x+1)
v'=4(1)(x-2)3
=4(x-2)3 (My chain rule)
v=(x-2)4
u'=3

y'=(3x-1).4(x-2)3 + (x-2)4.3
=4(3x-1)(x-2)3 + 3(x-2)4

take out (x-2) as a common factor:

= (x-2)^3 . [4(3x-1) + 3(x-2)]
=(x-2)^3 . (12x - 4 + 3x - 6 )
=(x-2)^3 . (13x-10)

looks easier now

y'=0
.'. (x-2)^3 . (13x-10) = 0

now as you do for quadratic functions:
13x-10 = 0
x = 10/13

(x-2)^3 = 0
(x-2) = 0
x = 2.

x = 10/13 and 2

continue to find stationary point.