Question

Only started the topic today and I don't quite understand it. :/ For example, how do you determine the stationary points and point of inflection for y = x^3 - 3x^2 - 9x + 7

Answer 1

Stationary points are the points the rate of change of the function is zero. It means the places the value of the function changes the least as you vary x. Try to calculate the value of \(\Delta y=f(x+\Delta x)-f(x)\). It will help you to understand it.
Also you can go to the concept of the derivative and understand it intuitively. There is a great lesson on that over at the khanacademy you should watch it:
https://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version

Answer 2

\[\Delta y=f(x+\Delta x)-f(x)\\
\Delta y=(x+\Delta x)^3-3(x+\Delta x)^2-9(x+\Delta x) +7 -(x^3-3x^2-9x +7)
\]
If you go ahead and try to solve this, you will notice that \(\Delta y\) is a function of both \(\Delta x\) and x, and there should be places where it changes very slowly.

Answer 3

Wait, so how would I have solved y = x^3-3x^2-9x+7 with that?

Answer 4

No, what I meant was to think about it. What are the values of \(x\) which \(\Delta y\) changes the least? That is a stationary point

Answer 5

Oh, okay, I kinda get it. :p

Answer 6

Can you graph that equation? Then I can show to you on the graph. Its much simpler.

Answer 7

How do I graph it on here?

Answer 8

Can you differentiate?

Answer 9

Yep! :)

Answer 10

Okay, so can you differentiate the function?

Answer 11

\[\large y = x^3 - 3x^2 - 9x+8\]

differentiate...
\[\Large \frac{dy}{dx} = ?\]

Answer 12

hello by stationary if u mean critial points
then it is asking you to find the
the points where first derivative =0
and for points of inflection where 2nd derivative = 0

Answer 13

the first derivative gives you a slope graph, so it is telling you the rate of change at every point for there to be a "stationary" point the rate of change must = 0
so the 1st derivative must = 0

Answer 14

an inflection point happens when the graphs concavity is changing or you can think of this as acceleration to deceleration and viceversa

Answer 15

or the rate of change for the rate of change is 0, because if the rate of change of the rate of change is going from positive to negative or negative to positive it must go thru the value of 0, so we calculate the 0 point of the 2nd derivative, and see how its changing to either side of the zero to determine if its a max or a min, kind of a lot to take in i know xD

Answer 16

Aha, yeah, Math's Studies is harder than I thought it would be. :p

Answer 17

just keep at it, this is where math gets fun :)

Answer 18

take one step at a time, first do you understand what the first derivative is

Answer 19

Differentiate y... have you done it yet?

Answer 20

oh i just realized the first guy posted you a khan academy link... did u watch that

Answer 21

Internet's slow so I can't really watch stuff :/ But yes I do know what the first derivative is :) and differentiating y I'm pretty sure you'll get y = 3x^2 - 6x -9? :)

Answer 22

okay, work with that... now you can equate that to zero..
\[\large 3x^2 - 6x - 9 = 0\
and solve for x

Answer 23

\[\large 3x^2 - 6x - 9 = 0\]