Question

HELP

Answer 1

can you please explain ? step by step @hartnn

Answer 2

take f'(x) =?
f" (x) =?

Answer 3

can you explain step by step after

Answer 4

@Loser66

Answer 5

yes for both, now, relative extrema :
let f' (x) =0 , solve for x, what do you get?

Answer 6

x=0? so 0?

Answer 7

nope, it is not that, redo, please

Answer 8

idk

Answer 9

i know for the second derivative x is greater than or equal to 1/4 and x is less than 0

Answer 10

http://www.wolframalpha.com/input/?i=%28x^%28-2%2F3%29%2B8x^%281%2F3%29%29%3D0
it shows x = -1/8

Answer 11

okk

Answer 12

So, relative extrema happens when x= -1/8, now plug that value into the original function \(\huge f(x)\), to get the value of relative extrema.

Answer 13

?

Answer 14

ohh

Answer 15

so 4.011 is the relative extrema?

Answer 16

Nope, I put wrong number, x = 1/8 , not 1/4
However, x=1/4 is inflection point when we solve for second derivative, but it uses later.
Now, f'(x) =0 when x = -1/8
so, \(f(-1/8) = 3\sqrt[3]{(-1/8)}+6\sqrt[3]{(-1/8)^4}= ??\)

Answer 17

=-1.125
so, the relative extrema happens when x = -1/8 and its value is f(-1/8) = -1.125

Answer 18

Now, redo the thing for inflection by solve f"(x) =0, wolfram shows that it is =0 when x = 1/4
hence inflection point is (1/4, 4,011)
that's it

Answer 19

thank you (:

Answer 20

@mondona this is not the help I want you to get. I do all of them and you get nothing but the answer. that is not good for you.