Question

Consider the function of defined by f(x)=x^3/3-4/x

a. Find the X values for the points of inflection.
b. Determine the intervals where the function of f is concave up and concave down.

Answer 1

f(x)=x^3/3-4/x
= (1/3)x^3 - 4x^-1
f ' (x) = x^2 + 4x^-2
f '' (x) = 2x - 8x^-3
= 2x - 8/x^3

at the point of inflection, f ''(x) = 0
2x - 8/x^3 = 0
2x^4 = 8
x^4 = 4
x = ± √2

if x = √2 , then y = (2√2)/3 - 4/√2 = (-4√2) / 3

two inflections points : appr (1.4, -1.9) and (-1.4, 1.9)
check: confirmed by
https://www.wolframalpha.com/input/?i=graph+y+%3D+x%5E3+%2F+3+-+4%2Fx+from+-5+to+5

It should be straightforward to decide where the function is concave up or concave down
once you have the points of inflection.
- otherwise, the function is concave up when f ''(x) is positive, and concave down
if f ''(x) is negative. (of course if f ''(x) = 0 , you have the points of inflection)