Please help on this concavity problem!

Question

anonymous · Answer

For the function, do the following.
$$f(x) = x ^{4/3}-6x^{1/3}$$
(a) Determine the intervals on which the function is concave up and on which it is concave down. (Enter your answer using interval notation. If an answer does not exist, enter DNE. Round your answers to three decimal places.) 

concave up:     ______________
concave down: ______________

(b) Find any points of inflection. (If an answer does not exist, enter DNE. Round your answers to three decimal places.) 

(x,y) = ( _____________ ) smaller x-value
(x,y) = ( _____________ ) larger x-value

amistre64 · Answer

gonna need to derive it 2 times

amistre64 · Answer

since its just power rules .. that should be pretty basic if your covering concavity

anonymous · Answer

I've derived it twice..now do I just find the zeros and points at which the second derivative is undefined, then make a number line..?

amistre64 · Answer

those are the basic steps yes

the first derivative at 0 or undefined are our critical ... they are points of interest and tell us where the function has slowed down to zero, this usually indicates a bend in the graph.

the second derivative at 0 or undefined are critical points as well, this usually indicates that the graph has changed it concavity, an inflection point.

amistre64 · Answer

$$f = x ^{4/3}-6x^{1/3}$$
$$f' = \frac43 x ^{1/3}-\frac63x^{-2/3}$$
$$f'' = \frac{4}{9} x ^{-2/3}+\frac{12}{9}x^{-5/3}$$