# Thread: Second derivative test to find local min and max

1. ## Second derivative test to find local min and max

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

f"(x) = 6x

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

2. 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
.

3. I have done some corrections - please check I am on the right track ...so

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

f"(x) = 6x + 2x^-3

f'(x) is zero when 3x^2 - x^-2 = 0

3x^2= x^-2
x^4 = 1/3
x= 4√(1/3)

so x could be + 4√(1/3) or -4√(1/3)

which means the local minimum is at 4√(1/3) since f'(4√1/3) is > 0

and local maximum is at 4√(1/3) since f"(4√(1/3) is < 0

Do I have it right now??

4. correct, $f'(x) = 0$ at $x = \pm \sqrt[4]{\frac{1}{3}}$

$f''\left(\sqrt[4]{\frac{1}{3}}\right) > 0$ ... $f(x)$ has a minimum there

$f''\left(-\sqrt[4]{\frac{1}{3}}\right) < 0$ ... $f(x)$ has a maximum there