Question

Please explain a calc question for a medal!

Answer 1

@perl

Answer 2

please, we have
\[f'(x)=x ^{-2/3}+8x ^{1/3}\]

Answer 3

and:
\[f''(x)=\frac{ 2 }{ 3 }(\frac{ -1 }{ x \sqrt[3]{x ^{2}} }+\frac{ 4 }{ \sqrt[3]{x ^{2}} })\]

Answer 4

okk i get that so far, the derivative and second derivative

Answer 5

so:
\[f''(x)\ge 0\]
when the two susequent conditions are verified:
\[x \ge \frac{ 1 }{ 4 }, x <0\]
since you have to solve two system of inequalities

Answer 6

okk following so far

Answer 7

your ituaion is that:

about sign of y''

Answer 8

so keeping in mindf'(x), you have a descending flection poin at x=0, and an ascending flection point at x=1/4
since f(x) mut be convex in (-infinity,0), and[1/4,0), and f8x) mst be concave in (0,1/4).

Answer 9

is there any relative extrema?

Answer 10

I think at x=-1/8,as you can check solvin te inequality:
\[f'(x)\ge 0\]
precisely x=-1/8 is point of minimum for f(x)

Answer 11

- and + in th above drwing, refer to the sign of the derivative of f(x), namely to the function f'(x)

Answer 12

please is it all clear?

Answer 13

im confused haha i dont get where its answering the question

Answer 14

please, you have to study the signof y''(x) or f''(x). When y''(x) is positive then y(x) or (x) is cnvex and viceversa. The points at which y''(x) is negative then f(x) is concave and vicevrsa. No, from our preceeding anaysis, we can conclude that i the subsequent interval:
\[(-\infty,0)\cup [1/4,+\infty)\]
y''(x) is positive, so our f(x) is convex, whereas in the subsequent interval:
\[(0,1/4)\]
y''(x is negaive, so our f(x)in concave.
@mondona

Answer 15

sorry ...y''(x) is negative so our function f(x) is concave