Question

Use any method to find the relative extrema of f(x)=2x+3x^(2/3)

Answer 1

\[2+\frac{ 2 }{ x ^{\frac{ 1 }{ 3 }} }=0,\frac{ 2 }{ x ^{\frac{ 1 }{ 3 }} }=-2,x ^{\frac{ 1 }{ 3 }}=\frac{ 2 }{ -2 }=-1\]
cubing
x=-1
\[\frac{ d^2y }{ dx^2 }=?~at~x=-1\]
?

Answer 2

Thanks!

Answer 3

So, when finding the relative extrema, you use the critical points for the first derivative and test it to the second derivative? :)

Answer 4

correct.

Answer 5

Thanks again.

Answer 6

yw