Originally Posted by

**calcbeg** HI

The goal is the find the local minimum and local maximum values of the following function using the second derivative test

f(x) = x^3 + 1/x

so I calculate

f'(x) = 3x^2 + -1 ... this is incorrect, should be f'(x) = 3x^2 - (1/x^2)

f"(x) = 6x ... f''(x) = 6x + (2/x^3)

so f'(x) = 0 when x = √(1/3)

f"(√1/3) = 6 √1/3

which I believe is the local minimum since f(√1/3) is >0

now I am questioning my calculations since I am suppose to also have a maximum based on the question and I don't have one

Please review my work and tell me where I have gone wrong

Thanks

calculus beginner