Differentiation help

Hi all. Quick maths question im totally stumped on.

So the question is, use differentiation to find the stationary points of the following function and establish the nature of each point.
y=3x³-6x+5.

So what ive got written down is,
y' gives the 2 x coords, put what x equals back in to find what y is, use y'' to find min and max.

But for the life of me I just cant get it. What ive managed to do so far is differentiate to be left with y' = 9x²-6. y'' = 18x. Also I THINK x = 0.82, but this may not be correct. Does anyone have any points to help me unstick myself?

Reply 1

davros

Original post by happynova

Stationary points are where y' = 0. So solve this to get values of x. I would leave the answers as exact quantities or surds unless you are asked for decimal values!

Then put the value of x in to y'' to determine the nature of the stationary points.

y'' > 0 -> minimum
y'' < 0 -> maximum

Reply 2

happynova

Original post by davros

Ok so I work out that x = 0.82. Does this mean that when x = 0.82 y is 0?

Based on that finding then y'' would be 14.76. So it would be its minimum. But I still dont understand what im doing....

Reply 3

SamKeene

Original post by happynova

Ok so I work out that x = 0.82. Does this mean that when x = 0.82 y is 0?

Based on that finding then y'' would be 14.76. So it would be its minimum. But I still dont understand what im doing....

Note when you solved:

9x^2-6=0

0.82 (or rather,

\frac{\sqrt{6}}{2}

(keep it exact!), --- this is only one of the solutions!

But with regards to your first solution (0.82) you have shown that when x=0.82, the rate of change, that is, the gradient of the graph is 0. That is the graph is parallel to the x axis when x=0.82.

Reply 4

happynova

Original post by SamKeene

Note when you solved:

9x^2-6=0

0.82 (or rather,

\frac{\sqrt{6}}{2}

I ended up with sqrt of 6/9 = 0.82. Have I got that wrong?

Ok so, when x=0.82 thats when the gradient is 0? So the line on the graph is now no longer going up or down?

From watching a bunch of videos online I think I need to factorise

To the (x )(x ) form and use them values plugged into the equation. Am I on the right path?

Reply 5

ztibor

Original post by happynova

Ok so I work out that x = 0.82. Does this mean that when x = 0.82 y is 0?

Solving the

y'=0

equation you will get

9x^2-6=0 \rightarrow x^2=\frac{2}{3}

Taking square root of both side you will get two roots

\displaystyle |x|=\sqrt{\frac{2}{3}}

that is x1=+0.82 and x2=-0.82 are the solutions

So when x=0,82 then y'=0 and not y=0

Based on that finding then y'' would be 14.76. So it would be its minimum. But I still dont understand what im doing....

y"(0.82)=14.74 >0 implies that the original function has minimum at x=0.82
But the value of minimum is not 14.74
Value of the minimum is the value of the original function at x=0.82 that is f(0.82)

Reply 6

SamKeene

Original post by happynova

To the (x )(x ) form and use them values plugged into the equation. Am I on the right path?

remember when you have

x^2=1

, then

x=\pm 1

not just 1. Same principle applies here. You got the correct positive solution, you're missing the negative one.

Ok so, when x=0.82 thats when the gradient is 0? So the line on the graph is now no longer going up or down?

Yup, this is true.

plugged into the equation. Am I on the right path?

You've identified the x-cordinates of the stationary points. To get the y co-ordinate you sub the values of x into the original equation

y

, to get the nature (max or min) you sub them into

y''

Reply 7

happynova

Original post by SamKeene

remember when you have

x^2=1

, then

x=\pm 1

not just 1. Same principle applies here. You got the correct positive solution, you're missing the negative one.

Yup, this is true.

You've identified the x-cordinates of the stationary points. To get the y co-ordinate you sub the values of x into the original equation

y

, to get the nature (max or min) you sub them into

y''

Ok im starting to follow. I still quite unsure of how I have more than one value of x. To me x = 0.82. So subbing that into the original equation gives me 1.73.
Would I then just use -0.82 for the other solution?

Reply 8

davros

Original post by happynova

When you solve a quadratic equation like

9x^2 - 6 = 0

, you will (typically) expect to get 2 solutions. In this case you're looking for two numbers that give 6/9 (or 2/3) when squared. Can you see what the other number is going to be?

Reply 9

happynova

Original post by davros

When you solve a quadratic equation like

9x^2 - 6 = 0

, you will (typically) expect to get 2 solutions. In this case you're looking for two numbers that give 6/9 (or 2/3) when squared. Can you see what the other number is going to be?

-0.82?

Otherwise ive got no idea at all!!

Reply 10

SamKeene

Original post by happynova

-0.82?

Otherwise ive got no idea at all!!

Yes that's correct. Do you understand why?

Reply 11

happynova

Original post by SamKeene

Yes that's correct. Do you understand why?

Nope I dont, I can understand how it can be 0.82, but not -.82.

So once I have these numbers I need to plug them back into the equation and thatll give me the y coords?

So now ill have both sets of x and y coords. I then get y'', and plug both x values into that to determine the min and max, does this sound right?

Reply 12

ztibor

Original post by happynova

Nope I dont, I can understand how it can be 0.82, but not -.82.

When

x^2 =0.67

then taking the root

\sqrt{x^2}=|x|

and

\sqrt{0.67}=0.82

by definition of the square root.

So

|x|=0.82

means 2 solutions for x. When x=0.82 and x=-0.82

So once I have these numbers I need to plug them back into the equation and thatll give me the y coords?

These x values give that where are the stationary points of the function along the axis x.

Yes, plug them into the original equation, These will be the values of the stationary
points

So now ill have both sets of x and y coords. I then get y'', and plug both x values into that to determine the min and max, does this sound right?

Yes. These values (>0 or <0) give the type of the stationary points.