Question

@Michele_Laino

Answer 1

please what is your question?

Answer 2

first you have to calculate f''(x)

Answer 3

wait, please, I'm trying...

Answer 4

yes I think there is a flection point located at x=1/4

Answer 5

can you explain the rest of the problem for me please?

Answer 6

even if there is an infinite point for f''(x) at x=0

Answer 7

your situation is:

Answer 8

that's the sign of f''(x). now in mathematical analysis there is a theorem which states:
if f''(x) is Greater than zero then f(x) is convex, and viceversa.
of course if f''(x) is less than zero, f(x) is concave, and viceversa.
So from our discussion, we see that there are two points, located at x=0, and x=1/4 respectively, at which f''(x) change its sign, then f(x), by the above theorem, changes its status of convexity or concavity

Answer 9

yes they are

Answer 10

please, before to answer , note that, when f''(x) is positive, then f'(x) is an incresing function, namely the angle between the tangent line to graph and the positive x-semi axes is incresing, viceversa if f''(x) is negative

Answer 11

oops, sorry ...the x-half axes...

Answer 12

okk gotcha

Answer 13

please substitute x=0 and x=1/4 into the formula of f(x), for example
at x=0, we have:
\[f(0)=0\]
so your first inflection point is (0,0) namely the origin,whereas at x=1/4, we get:
\[f(1/4)=3\sqrt[3]{1/4}+6\sqrt[3]{(1/4)^{4}}\]

Answer 14

that's the y-coordinate of the second inflection point located at x=1/4

Answer 15

I think so, even if better is to express it not in decimal form but in algebraic form, as I made

Answer 16

i just put it into my calculater, what did you get?

Answer 17

noo, please (0,0) and (1/4,2.83) are the inflection points, whereas in order to find the relative extrema you have to act on f'(x)

Answer 18

here is f'(x):
\[f'(x)=\frac{ 1+8x }{ x ^{2/3} }\]
now, I ask you, when f'(x) is positive?

Answer 19

noo i thought f'(x) =x^-2/3+ 8x^1/3

Answer 20

@Michele_Laino

Answer 21

it is the same, please check it

Answer 22

when f'(x)=0? solving that equatioyou illget the x-coordinates of your relative extrema point

Answer 23

noo, when 1+8x=0 so when=-1/8, now find f(-1/8) please!

Answer 24

-5?

Answer 25

right?

Answer 26

heloo ? @Michele_Laino

Answer 27

noo I got f(-1/8)=-3/4

Answer 28

how... -1/8 times 8 is -1 @Michele_Laino

Answer 29

please, note that you have to insert x=-1/8 into the formula of f(x) not f'(x)

Answer 30

WAIT does that mean 2.83 was wrong?!?! can you help me instead of leaving every two seconds..

Answer 31

please note that inflection points are point for which f''(x)=0. extrema points are points for which f'(x)=0

Answer 32

i got -15/8 lol can you explain , if not ill ask someone else

Answer 33

its been an hour and you still havnt explained @Michele_Laino

Answer 34

ok so it is -3/4

Answer 35

so my relative extrema is -3/4? end of question right?!?!??!

Answer 36

your relatie extrema is (-1/8, -9/8)
sorry I think to have made a cn error of calculus

Answer 37

where did uou get -9,8 from?!?!

Answer 38

-9/8

Answer 39

youre just pulling numbers and not explaining!!!

Answer 40

ok!, f(-1/8)=4*(-1/2)+6*(-1/8)*(-1/2)=-9/8
is it right?

Answer 41

sorry 3*(-1/2)+...

Answer 42

okk

Answer 43

reassuing, your relatie extrema point is (-1/8,-9/8), and your inflection points are:
(0,0) and (1/4, epression that I wrote)

Answer 44

thank you

Answer 45

thank you! and sorry again!