Question

I have a function and its derivatives....I have to determine the nature of the turning points which are (0,2) and (2,-2) where if positive it is my local minimum which is correct for (2,-2). My question is do I class 0 as negative when stating my maximum or how do I word it? I am also struggling to state the point of inflection.

Answer 1

@Kerriy what specifically are you asking?

Answer 2

I have to state the nature of the turning points and I have to find the point of inflection but I am a bit confused on the point of inflection @billj5

Answer 3

so you can take the second derivative, when the second derivative = 0, then it may be an inflection point, you also need to know if its continuous at that point

Answer 4

second derivative is 6x
therefore is my inflection point 0,0?

Answer 5

what is the function?

Answer 6

\[h(x)=x ^{3}-3x ^{2}+2\]

Answer 7

ok so the second derivative of that function is not 6x, but its close to that

Answer 8

That is the original function, the derative is \[3x ^{2}-6x\]

Answer 9

yep, and whats the derivative of 3x^2 -6x, its not 6x

Answer 10

oh poop 6x-6?

Answer 11

yep

Answer 12

so there may be an inflection point at what x value?

Answer 13

(1,0)

Answer 14

correct

Answer 15

you have to make sure its continuous there, which it is

Answer 16

thank you so very much :)

Answer 17

you're welcome

Answer 18

@billj5 when I draw the graph of the derivative why is the turning point for the quadratic at (1,-3)?

Answer 19

@Kerriy can you be more specific with what you are asking?

Answer 20

what coordinates do you think it should be at? and why?

Answer 21

Sorry I am on the same question and I had to plot the graph of the function i mentioned previously as well as its derivative and when I plot online I get that the quadratic derivative turns at (0,-3)

Answer 22

look at the graph of the original function, do you see the root at x = 1?

Answer 23

yes

Answer 24

do you see that on the original graph the slope is 0 at the point (1,0)?

Answer 25

and before that the slope is negative and after that the slope is positive?

Answer 26

yes I see that :)

Answer 27

im sorry ignore that, i was looking at the wrong graph, lol

Answer 28

ok so at the before the point, the slope of the original function is decreasing, well at the point (1,0) the slope actually begins to increase, its still negative, but it starts to become less negative or increase

Answer 29

when the slope starts to increase the derivatives output will start increasing

Answer 30

so thats why it "turns" there

Answer 31

its easier to see if you "stretch" the function a bit horizontally

Answer 32

I understand that part just can't work out why -3 if I was to plot without a calculator how would I know that?

Answer 33

well if you know the x value is 1, just plug 1 into the derivative and solve for y

Answer 34

graph doesnt pass through point (1, -3)

Answer 35

the derivative is 3x^2 - 6x, so then you have 3(1)^2 - 6(1) = 3 - 6 = -3

Answer 36

@Jack1

@Kerriy is referring to the graph of the derivative

Answer 37

apologies :(

Answer 38

dang I am so bloody stupid sometimes :( thank you :)

Answer 39

yw

Answer 40

thanks Jack :)

Answer 41

nah, think i muddied the waters, not helped

Answer 42

this series helped me out so much, check it out when you have time

https://www.youtube.com/watch?v=UcWsDwg1XwM

Answer 43

Thank you so very very much :)

Answer 44

Not at all Jack, your help is appreciated trust me :)